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montebianco
2006-Jan-11, 01:13 AM
I'm going to argue this, even though I agree that 0.9~ = 1.

0+1, 1!, etc. are expressions, not numbers. When you evaluate those expressions, you find they all equal 1, but each expression itself is a different creature, consisting of both operators and numbers.

The symbol '1' itself is an expression. If the argument is that 0.999~ and 1 are different symbols, but represent the same number, then I agree with that. That's using the standard definition of the real number system. I would like for some of the not-equal people to explain how they're defining their numbers. I think a great deal of the problem is people interpreting <different symbols> as implying <different numbers>...

montebianco
2006-Jan-11, 01:15 AM
snarkophilus - actually, it's a myth. Computer will always get 1.0 when dividing two identical real numbers. Furthermore, it will always get (2.0+7.0)/9.0 to be exactly 1.0, because you can represent these numbers in binary without any approximation. The situation changes, when the expression needed to exactly represent the number in binary is longer than allowed by format used. Then you get approximation and therefore the result might not be correct. Very much like if you were restricted to eight decimal digits and tried to calculate 1 / 3 * 3 = ? First you would get 0.3333333 (and stop there, because you have no more space), then you'd multiply for the final result 0.9999999, which is not 1, as it has a finite number of nines.

I even get away with testing real numbers for equality on computers, even though you're not supposed to do that :D One must be very careful when playing such games, though...

Vaelroth
2006-Jan-11, 01:27 AM
2/4 and 1/2 are not the same numerals either but I have no problem saying that they are equal, in the sense that we are talking about.

Would you say that they are not the same either?

I would assert that one is 2/4 and the other is 1/2. One is a representation in fourths and the other is a representation in halves. Mathematically they are the same, yes. Their representation is different.

01101001
2006-Jan-11, 01:42 AM
I would assert that one is 2/4 and the other is 1/2. One is a representation in fourths and the other is a representation in halves. Mathematically they are the same, yes. Their representation is different.
They are different names, representations, for exactly the same concept.

I'm going to argue this, even though I agree that 0.9~ = 1.

0+1, 1!, etc. are expressions, not numbers. When you evaluate those expressions, you find they all equal 1, but each expression itself is a different creature, consisting of both operators and numbers.
You're thinking at the wrong level.

Sure context is important. "One" doesn't represent a particular number, is just a bunch roman letters on your screen, or a bunch of squiggly lines at another level, or a pattern of dots at another level, until you're thinking in English.

When your are thinking in "expressions" then those different combinations of squiggly lines all represent a particular number.

The journeys are different, but the destinations are the same.

Vaelroth
2006-Jan-11, 01:48 AM
They are different names, representations, for exactly the same concept.

Thats what I've been saying since my first post.

hhEb09'1
2006-Jan-11, 08:31 AM
Thats what I've been saying since my first post.Does that mean, in the sense you meant in this post (http://www.bautforum.com/showthread.php?p=650318#post650318), that you would say that in reality 2/4 and 1/2 are both equal and inequal?

Fram
2006-Jan-11, 09:02 AM
U2: "We're one, but we're not the same"

Carnifex
2006-Jan-11, 10:11 AM
I even get away with testing real numbers for equality on computers, even though you're not supposed to do that :D One must be very careful when playing such games, though...

Actually, you can get away with it just as long as your number may be expressed like a / 2^b. For example, 0.3 has an infinite number of digits in binary form, starting 0.0100110... I think it goes periodic after some time, I just didn't have enough time to convert more digits.

hhEb09'1
2006-Jan-11, 10:39 AM
Actually, you can get away with it just as long as your number may be expressed like a / 2^b. For example, 0.3 has an infinite number of digits in binary form, starting 0.0100110... I think it goes periodic after some time, I just didn't have enough time to convert more digits.I betcha montebianco has even got away with using 0.3 :)

Every ratinal number is periodic (in the spirit of this thread :) ), which is more or less a result of having a finite number of remainders in the long division (they have to be smaller than the denominator). Once you get one you've encountered before, the digits will start repeating.

In the case of 0.3, you're guaranteed to have already started repeating. In fact, 0.3 is 0.0(1001)...

Carnifex
2006-Jan-11, 10:54 AM
Again, talking about computers, the main problem is limited space. However, serious approximation problems in calculations may be prevented by leaving numbers in "a / b" form for as long as possible.

agingjb
2006-Jan-11, 11:20 AM
I wonder if calculators have reached the point where (internal) representation by continued fractions is a possibility (not that I'd know where to start).

Carnifex
2006-Jan-11, 11:28 AM
Actually, most of the calculators dubbed "scientific" (which have trigonometric, logarythmic, exponential etc functions) and Windows Calculator (use Start - Run - "calc" to get it) keep the numbers in "a / b" form instead of changing into "a1...an,b1...bn..." form. Of course, irrational numbers, such as pi or e, get approximated, but that would happen anyway, so it's OK.

Vaelroth
2006-Jan-12, 12:30 AM
Does that mean, in the sense you meant in this post (http://www.bautforum.com/showthread.php?p=650318#post650318), that you would say that in reality 2/4 and 1/2 are both equal and inequal?

Yes, that would be correct. Equal in mathematical concept, but inequal in representation.

HenrikOlsen
2006-Jan-12, 12:53 AM
That should clinch it for you then, "=" means equal in mathematical concept, so for you the answer to the original question is yes.

snarkophilus
2006-Jan-12, 07:45 AM
Perhaps the problem is that most real numbers only have one decimal representation, so it seems strange that a few should have two.

So here's a question: are there more real numbers with one decimal representation than there are real numbers with two? Because there are an infinite number of both.... How do those infinities stack up?

I'd say more have one, because my gut feeling is that all transcendentals have only one... but I could be wrong. Anyone got any ideas?

snarkophilus
2006-Jan-12, 08:43 AM
snarkophilus - actually, it's a myth. Computer will always get 1.0 when dividing two identical real numbers. Furthermore, it will always get (2.0+7.0)/9.0 to be exactly 1.0, because you can represent these numbers in binary without any approximation.

I have a couple dozen processors I'd like you to tell that to. :) The truth is that sometimes you'll write something like
a = 3.0
maybe some other stuff, like a = a*2. , a = a / 2. , a = a + 15. , a = a - 15.
print a

and the output will be
2.99999999999999

That's partly because IEEE 754 keeps extra bits, called guard and round bits (and I think it has a sticky bit, too, but I can't remember for sure), along with each number. If your program does a multiplication by 1, for instance, one of those bits might be flipped (it shouldn't be if you have strict IEEE math flags on in your compiler, but if you try to use the fast math routines it almost invariably happens), and a subsequent addition or test might trigger a rounding error.

Actually, I think the 7.0 + 2.0 example might cause a small rounding error if it was actually performed (and not simplified at compile-time), because there is a bit shift required to perform the addition, and the guard and round bits would be required to perform the calculation. You might not see the error when you printed the output of 7. + 2., but those bits would be used when dividing by 9, and the error might show up there, giving you 1.0000001 or 0.999999. It does happen sometimes, but it depends on which numbers you use.

In addition, the division algorithm usually implemented in modern processors is not exact. The only requirement is that the last digit be within a half unit of the correct answer. So more or less anything goes with the guard and round bits, and you get more FLOPS if you don't worry about building your processor to accomodate them, and that error is carried over to future calculations. I bet if you try 9.0/9.0 on your home machine, forcing the compiler to not optimize it out, and you look at the error bits, you'll find that your machine hasn't quite given you the correct answer. (Unless you have a Cray machine on your desk, in which case I will have to eat my words :D ) It's even worse if you try to do 0.3/0.3, for instance.

Montebianco: testing reals for equality == bad! *slaps Montebianco on the wrist*

HenrikOlsen
2006-Jan-12, 10:40 AM
The natural numbers (up to a maximum size) has a precise representation in IEEE 754 and addition, subtraction and multiplication of them will also yield the correct precise representation, as will division in most cases where the result is correctly representable.
Your series of operations should not be expected to result in other than 3.0000~, as all values are representable.

Incidentally, in the absence of a specified rounding strategy, IEEE7544 specifies rounding to even which mitigates errors.

For more info have a look at the files here (http://www.cs.berkeley.edu/~wkahan/ieee754status/) or follow the links here (http://http.cs.berkeley.edu/~wkahan/) they are from the site of William Kahan who was the arhitect behind IEEE 754.

Carnifex
2006-Jan-12, 11:48 AM
Huh :eh: All three default floating point number formats are as follows (at least my Assembler for Pentium III book says so)

Sign-bit - (1/1/1)
Characteristic - (7/10/15)
Normalised mantiss (I'm not sure about the spelling) without initial 01 - (24/53/64)

No other bits are there :eh: And these formats hold since the very first i8087 FPU :eh: And all my statements hold true in these formats :eh:

Of course, it's possible, that all my teachers were lying and all my books too, then I'll go and purge them all with a fire so no more lies are spread in this world.

HenrikOlsen
2006-Jan-12, 12:05 PM
guard bits are used for handling roundoff correctly, they are never seen in the actual result, so you are both correct.

Moose
2006-Jan-12, 12:35 PM
I wonder if calculators have reached the point where (internal) representation by continued fractions is a possibility (not that I'd know where to start).

I flirted in college with writing a matrix math library where each cell held a fraction object, not a float. As far as I went, it was possible, effective, but not quick because of the need to factor down the fraction periodically to keep the digits from blowing out the long int variables.

I never got it to the point where I'd be wanting to use it on a very large matrix.

Carnifex
2006-Jan-12, 01:00 PM
Yes, you have to factor the numbers down once in a while, and that is not a very fast operation, that's why floating point numbers are still around.

montebianco
2006-Jan-12, 01:23 PM
So here's a question: are there more real numbers with one decimal representation than there are real numbers with two? Because there are an infinite number of both.... How do those infinities stack up?

I'd say more have one, because my gut feeling is that all transcendentals have only one... but I could be wrong. Anyone got any ideas?

As you say, both sets are infinite. But, it looks to me like the only numbers that have two representations are rational numbers p/q where q is of the form (2^m)*(5^n) (or where p/q have common factors, and everything except the 2s and 5s can be eliminated from the denominator). This is a subset of the rationals, and on an interval, the rationals have Lebesgue measure of zero, so by that definition of "more", there are more real numbers with one representation...

Carnifex
2006-Jan-12, 03:42 PM
I can put it even more simply:

There is a way to assign rational numbers a natural number. That makes rational numbers countable albeit infinite.
There is no way to assign irrational numbers a natural number. That makes irrational numbers a continuum set.
(Or in another way: real numbers are a continuum set, rational numbers are a countable set, irrational numbers are real numbers except those that are rational, continuum set minus countable set is a continuum set)
From this we can make a conclusion that there will always be more numbers that are represented in a single way than numbers with two available representations, despite what positional system (binary, ternary, decimal, hexidecimas, octal, you-name-it) we use.

Relmuis
2006-Jan-12, 04:45 PM
I would like to point out that the real numbers are not independent of the rational numbers (or indeed of the natural numbers). Natural numbers are being used to define rational numbers, and rational numbers are being used to define real numbers.

Therefore, asserting that a real number like 0.999999999.... is (or is not) equal to the natural number 1 is not meaningless. The assertion will be definitely true or definitely false.

In this case, it is possible to prove that the assertion is true.

Soltras
2006-Jan-15, 07:52 AM
0.999999999999........ is, by definition of it being a repeated decimal expansion, is just an infinite sum.

Specifically, it is 9/10 + 9/100 + 9/1000 + 9/10000 + ....

Now, if we were dealing with finite terms, then of course it would not equal 1, but it would also not be expressed as a never-ending decimal expansion.
For example, 0.9999 = 9/10 + 9/100 + 9/1000 + 9/10000. Nobody debates this.

In other words, a decimal expansion with n repeating 9's can be expressed (imagine sigma notation): sum ( i=1..n , 9/10^i ).

So 0.99999999... (repeating) is simply the limit of the above taken as n approaches infinity. It is 1.

montebianco
2006-Jan-15, 09:05 AM
Hi Soltras, welcome to the board. It needs more people who understand convergent infinite series.

jfribrg
2006-Jan-15, 04:37 PM
Hi Soltras, welcome to the board. It needs more people who understand convergent infinite series.

There's plenty here who understand convergent infinite series. Most of us though have bad memories of this thread from last year and are ignoring it in the hope that it will die real soon.

Soltras
2006-Jan-15, 05:26 PM
Thanks montebianco.

And sorry for bringing back painful memories jfribrg. I have a few of my own from last year that I'd like to forget.

worzel
2006-Jan-15, 11:44 PM
There's plenty here who understand convergent infinite series. Most of us though have bad memories of this thread from last year and are ignoring it in the hope that it will die real soon.
Yes, because it's so hard and time consuming to see a thread title in "New Posts" and not click on it.

snarkophilus
2006-Jan-16, 06:02 AM
As you say, both sets are infinite. But, it looks to me like the only numbers that have two representations are rational numbers p/q where q is of the form (2^m)*(5^n) (or where p/q have common factors, and everything except the 2s and 5s can be eliminated from the denominator). This is a subset of the rationals, and on an interval, the rationals have Lebesgue measure of zero, so by that definition of "more", there are more real numbers with one representation...

That's a nice way of putting it, actually. Now, I wonder if there is a proof.... It should be easy to show the only rationals with two representations have that form, but maybe not so easy to show that no irrational can have two representations. It looks obvious, but what's rigorous enough to actually *show* it?

hhEb09'1
2006-Jan-16, 03:01 PM
Yes, that would be correct. Equal in mathematical concept, but inequal in representation.? But, I don't think anyone on this board is going to make the mistake of thinking "1" looks just like ".999..."! :)
Yes, because it's so hard and time consuming to see a thread title in "New Posts" and not click on it.D*ng you worzel, you made me read another one of your posts
That's a nice way of putting it, actually. Now, I wonder if there is a proof.... It should be easy to show the only rationals with two representations have that form, but maybe not so easy to show that no irrational can have two representations. It looks obvious, but what's rigorous enough to actually *show* it?All terminating decimals have a fractional representation where the only factors in the denominator are two and five, obviously. If an irrational had two representations, then those two representations must differ at some point, at some digit. If two decimals differ at some digit, then the only way they can be equal is if the rest of the lower decimal approaches the next increment--and the only way that can happen is if it would approach the next increment, and be terminating.

jlhredshift
2006-Jan-16, 04:38 PM

The Supreme Fascist is laughing....

montebianco
2006-Jan-17, 03:22 AM
There's plenty here who understand convergent infinite series.

I agree with that. But the board still needs more.

Most of us though have bad memories of this thread from last year and are ignoring it in the hope that it will die real soon.

Was it really so bad?

Dragon Star
2006-Oct-25, 11:59 PM
Sorry everyone, but I just had to post this...

http://en.wikipedia.org/wiki/0.999...

In mathematics, 0.999… (also denoted http://upload.wikimedia.org/math/b/0/f/b0f7e84cc3b14140ce7c25bceb919e1a.png or http://upload.wikimedia.org/math/d/e/0/de0175f40b34536008a8752b6b2dc5c7.png ) is a recurring decimalexactly equal to 1. In other words, the symbols '0.999…' and '1' represent the same real number. Mathematicians have formulated numerous proofs of this identity, which vary as to their level of rigor, preferred development of the real numbers, background assumptions, historical context, and target audience.

:D

pghnative
2006-Oct-26, 01:05 AM
If nobody posts, maybe this will die out quickly.

......oops

Chuck
2006-Oct-26, 03:44 AM
Maybe some newcomers haven't debated this yet and might want to join in. If not, maybe I'll change sides.

mickal555
2006-Oct-26, 06:44 AM
Ye gods...

LayMan
2006-Oct-26, 08:51 AM
Maybe some newcomers haven't debated this yet and might want to join in. If not, maybe I'll change sides.

:D You rang, m'lord? :D

As some of you already know, I'm not high on maths, but to ignore a direct invitation would be rude, so...

I'm inclined to say they are not exactly equal, but I'm not sure, and remember, if you're going to point out that I know nothing about maths, then you're wasting you're breath: I already know that! :D

I'm just curious what people would say about 1.000~1, basically adding 1 to the number you get when dividing 1 infinitely by 10? Would you say that 1 infinitely divided by 10 doesn't yield a 'valid' mathematical number? And if it does, then the number of 0's between the comma and the one would become infinite also, no? Would you say 0.00~1 equals 0?
And that 0.00~1 plus 1 equals 1? And what would the difference be between 1 divided infinitely by 10 and 2 divided infinitely by 10?
0.00~2 - 0.00~1 = ? (Sorry if this has already been mentioned, but there's a lo-h-ot of pages in this thread, so I haven't read all posts - me and maths, you know... :p )

Looking at it from that perspective, you'd end up with 3 notations (definitions?) of 1:

1) 1
2) 0.99~9
3) 1,00~1

To be honest, the second notation differs slightly from the original one, 0.99~. I added the 'last' 9 to 'comply' with the notation of the third: 0.00~1, where I couldn't simply end with the '~', since then I would 'loose' the 1...:think: :D
If I hadn't done that, the 3 notations would have been like this:

1) 1
2) 0.99~
3) 1,00~1

And then I'm pretty sure some of the 'are-equal'-supporters would point out that there's something fishy going on. And I'm just taking a stab in the dark here, but I'm guessing it'll have something to do with the difference between infinite and finite: the third notation (in the second representation) does look like it's implying finity (because it ends with 1, not the '~' sign)...

But I would object to the logic of:

Also:

1/3 = 0.3333....
0.3333.... * 3 = 0.9999....
Therefore, 0.9999.... = 1

If 1 divided by 3 equals 0.33~, then (by definition) 0.33~ times 3 does not equal 0.99~, but 1. It depends on the 'validity' of the second line: if you (a priori) accept that 0.99~ is exactly equal to 1, then the second premis of the logic is valid, and so is the conclusion. However, if you (again, a priori) don't accept that 0.99~ is exactly equal to 1, then you won't accept the validity of the second premis. And therefor not the conclusion...

Anywayzzzzz~, I feel a splitting headache coming on, so if people want to put this puppy to sleep, I won't object to it.

Disinfo Agent
2006-Oct-26, 11:05 AM
Hi, LayMan. It's nice to see more people interested in math, but the problem is that some of the objections you've raised have been dealt with already in this thread (and others). I suggest you start by reading it from the start.

If 1 divided by 3 equals 0.33~, then (by definition) 0.33~ times 3 does not equal 0.99~, but 1.6 divided by 4 equals 1.5. Does that mean it can't equal 3/2?

Nicolas
2006-Oct-26, 11:10 AM
0.00~1

This is equal to zero. In fact anything specific behind an infinite repetition is of no use, and therefore not done.

LayMan
2006-Oct-26, 12:39 PM
Hi, LayMan. It's nice to see more people interested in math, but the problem is that some of the objections you've raised have been dealt with already in this thread (and others). I suggest you start by reading it from the start.

Hello Disinfo Agent, you're right, reading the entire thread before posting is the smart thing to do, but as stated in my post, I'm not very good at math, so I didn't feel like plowing through all these posts...

6 divided by 4 equals 1.5. Does that mean it can't equal 3/2?

Of course it can and it does, but now you're asking a different question ("Does 1.5 equal 1.5?"). The analogical question to the one in the OP here would be: "Would you also say 3/2 = 1.499~ , indicating that 1.5 = 1.499~?"

LayMan
2006-Oct-26, 12:41 PM
Ouch, just became responsible for adding page 44 to this thread, sorry 'bout that... :eek:

:D

hhEb09'1
2006-Oct-26, 12:50 PM
Of course it can and it does, but now you're asking a different question ("Does 1.5 equal 1.5?"). The analogical question to the one in the OP here would be: "Would you also say 3/2 = 1.499~ , indicating that 1.5 = 1.499~?"It's not a different question, because in your previous post, you claimed that 3 time 0.333... (1/3) would not equal 0.999... because it equaled 1. However, the argument is, that any time you multiply a digit by 3, you do not get a carry, so 3 times 0.333... would seem to equal 0.999...

Disinfo Agent was just pointing out the weakness in your arguement

Got rid of extraneous formatting marks

SeanF
2006-Oct-26, 01:59 PM
0.00~1
This is equal to zero. In fact anything specific behind an infinite repetition is of no use, and therefore not done.
I disagree that this is equal to zero. I would say that this is meaningless.

There's no way to put a one after the infinite number of zeroes, because the zeroes never end, so there's no place to put the one.

Frog march
2006-Oct-26, 02:10 PM
that's the second time I voted in that poll.:D

LayMan
2006-Oct-26, 02:15 PM
It's not a different question, because in your previous post, you claimed that 3 time 0.333... (1/3) would not equal 0.999... because it equaled 1. However, the argument is, that any time you multiply a digit by 3, you do not get a carry, so 3 times 0.333... would seem to equal 0.999...

Disinfo Agent was just pointing out the weakness in your arguement

Got rid of extraneous formatting marks

Sorry, I disagree: I did not claim that 3 times 0.33~ would not equal 0.99~ because it equaled 1, I claimed (correctly I believe) that...

IF 1 divided by 3 equals 0.33~
THEN 0.33~ times 3 equals 1
(AND THUS 1 = 1).

(Or more formally: if x/y=z, then z*y=x - and I doubt that anyone will argue that 1 ≠ 1).

In this form, there's not even mentioning of any 0.99~... That is what I was getting at, the equality of 1 and 0.99~ was introduced by adding a second premis within the formal model (thus creating a logical model instead of a mathematical axioma):

IF 1 divided by 3 equals 0.33~
AND 0.33~ times 3 equals 0.99~
THEN 1 = 0.99~.

I was just questioning whether you can do that, I mean mathematically speaking... After all, like you say yourself, if you don't get a carry when multiplying 3 with 0.33~, then 3*0.33~=0.99~, but it would not equal 1, since this would eventually require a carry?? What does that logically say about premis 1? Does 1/3=0.33~? :D

Just kidding, but according to me, the math only allows for the formal "if x/y=z, then z*y=x". Which was the point I was trying to make: logic doesn't care about the validity of the premises themself, but about the validity of their relationship. Take this logical example:

Premis 1: IF all flunks are flenks
Premis 2: AND all flenks are flonks
Conclusion: THEN all flunks are flonks

Is premis 1 'true' or valid? Is premis 2? I don't know, since I have no idea what exactly flunks, flenks and flonks are... :D

So I wouldn't say that there's a weakness in my 'argument' as you state, since I did not use it as one, instead I was asking whether or not it could be used/would be allowable to be used as an argument...

In other words:

IF 0.33~ times 3 equals 0.99~, AND 1/3=0.33~,THEN by definition of the formal "if x/y=z, then z*y=x", 0.99~ and 1 would indeed be exactly the same.
But that would also mean that 0.99~/3=0.33~, and not 1/3=0.33~.
So Disinfo Agent was asking a different question, since he asked:
"6 divided by 4 equals 1.5. Does that mean it can't equal 3/2?"

In other words:

IF 6 divided by 4 equals 1.5
THEN 1.5 can't equal 3/2?

And like I said, it can:

IF 6 divided by 4 equals 1.5
AND 3/2 equals 1.5
THEN 6 divided by 4 equals 3/2 (and yes, 1.5 equals 1.5)

Not sure this comes out right, let me try it from a different perspective:

Does 3.14 = π (Pi)? Does 3,1415? Basically, I was trying to point out that the question in the OP can be related to the question "What is the exact value of π?"...

But forget it, I'll settle for "1 equals 0.99~ (but not exactly)"... :razz:

worzel
2006-Oct-26, 02:23 PM
Try actually dividing 1 by 3 the long way by hand and see what you get :)

LayMan
2006-Oct-26, 02:35 PM
Try actually dividing 1 by 3 the long way by hand and see what you get :)

You mean the ~ way? :D

Nicolas
2006-Oct-26, 02:49 PM
I disagree that this is equal to zero. I would say that this is meaningless.

There's no way to put a one after the infinite number of zeroes, because the zeroes never end, so there's no place to put the one.

And that's why I said it's equal to zero :). I know there's no point in placing something after an infinite series -which I said-, but rather than interpreting it as rubbish you can also look at it as if the part after the infinite repitition wasn't there. After all, there's no place to put it as you said. :)

Let it be clear that I fully understand why it is rubbish. I've had my share of infinities in mathematics, and not only in the grades ;) :D

jlhredshift
2006-Oct-26, 03:04 PM
Penrose said that if a number has an infinite exponent that we need not to "worry too much about it", but if the infinte exponent has an infinite exponent then "that we need to worry about" REF:Princeton lectures.

peter eldergill
2006-Oct-26, 03:50 PM
I just wanted to make this trhead have 1300 posts..it's currently at 1299

Pete

worzel
2006-Oct-26, 04:02 PM
I just wanted to make this trhead have 1300 posts..it's currently at 1299

Pete
It took you 46 minutes to type that? :)

Frog march
2006-Oct-26, 04:06 PM
It took you 46 minutes to type that? :)

Ive been fooled by that before. He probably looked at the 'reply' number when looking at the thread titles. The 'post number' when you are actually reading the thread is 1 higher as it includes the OP.

hhEb09'1
2006-Oct-26, 04:35 PM
Sorry, I disagree: I did not claim that 3 times 0.33~ would not equal 0.99~ because it equaled 1, I claimed (correctly I believe) that...Weird, I'm not sure how else to interpret this comment:
If 1 divided by 3 equals 0.33~, then (by definition) 0.33~ times 3 does not equal 0.99~, but 1.

But forget it, I'll settle for "1 equals 0.99~ (but not exactly)"... No, the point is that it is "exactly". That is the issue at hand.

Moose
2006-Oct-26, 04:50 PM
LayMan, this has already been debated ad nauseum. Is it really necessary to make us write this all up yet again because you don't feel like doing a little reading? It's pretty much all been said by page seven anyway.

Please just read part of the thread. If you still have questions after you've done so, we'll be considerably more pleased to help you figure it out.

Nicolas
2006-Oct-26, 05:32 PM
It's pretty much all been said by page seven anyway.

So 7 does equal 44 in the end...

peter eldergill
2006-Oct-26, 06:40 PM
Ive been fooled by that before. He probably looked at the 'reply' number when looking at the thread titles. The 'post number' when you are actually reading the thread is 1 higher as it includes the OP.

Indeed, I was hoping that no one noticed that...oh well, I wasn't the 1300th poster, the 1300th response...maybe I'l be online of 2000?

Pete

peter eldergill
2006-Oct-26, 06:41 PM
It took you 46 minutes to type that?

You've never seen my type!! That's an improvement

Pete

Disinfo Agent
2006-Oct-26, 09:47 PM
IF 1 divided by 3 equals 0.33~
AND 0.33~ times 3 equals 0.99~
THEN 1 = 0.99~.The red bit is where you're failing to understand. We don't just assume that 0.33~ times 3 equals 0.99~. We can prove it.

And since we can also prove mathematically that 0.33~ x 3 = 1 (you've done it yourself), we reach the inescapable conclusion that we must have 0.99~ = 1. QED.

0.00~1I disagree that this is equal to zero. I would say that this is meaningless.

There's no way to put a one after the infinite number of zeroes, because the zeroes never end, so there's no place to put the one.I would qualify that: it's meaningless in the context of the decimal representation of real numbers. In short: it does not stand for a number.

worzel
2006-Oct-26, 10:24 PM
You've never seen my type!! That's an improvement

Pete
Well I can type at 50 words a minute. Unfortunately I can only think at a rate of 1 word per minute :o

Disinfo Agent
2006-Oct-26, 10:41 PM
Can you type .99999... digits in a second? :p ;)

worzel
2006-Oct-26, 10:42 PM
Can you type .99999... digits in a second? :p ;)

ToSeek
2006-Oct-26, 10:49 PM
I would qualify that: it's meaningless in the context of the decimal representation of real numbers. In short: it does not stand for a number.

I would say that the only defensible interpretation of 0.000....001 is that it is the limit of the value as you keep adding zeroes - which is obviously 0.

Frog march
2006-Oct-27, 12:52 AM
I've just been to the end of all the nines (at infinity)and there was a rather obese drunk transvestite mouse smoking a cigar who told me to go away(although not in those words), if that is any help?

ToSeek
2006-Oct-27, 01:34 AM
Was the mouse left-handed or right-handed? This is crucial.

PhantomWolf
2006-Oct-27, 01:55 AM
It's not exactly equal to 1, but it is within experimental error. ;)

Frog march
2006-Oct-27, 02:13 AM
Was the mouse left-handed or right-handed? This is crucial.

from the gesture he gave me, I think he was lefthanded.

edit-I had to think for a moment.

PhantomWolf
2006-Oct-27, 02:33 AM
And since we can also prove mathematically that 0.33~ x 3 = 1

This isn't actually true. 0.333333~ x 3 is 0.999999~

1/3 ~ 0.333333333~ not 1/3 = 0.33333333~

You can see this by doing the division, you'll always end up with a remainder. because of that we can can say that 1/3 is only approximately 0.33333~ rather than that 1/3 is actually equal to 0.333333~.

hhEb09'1
2006-Oct-27, 02:41 AM
because of that we can can say that 1/3 is only approximately 0.33333~ rather than that 1/3 is actually equal to 0.333333~.Maybe you can't say it, but I can. :)

1/3 is actually equal to 0.333333~

That's the effect of the meaning of the tilda at the end

PhantomWolf
2006-Oct-27, 03:03 AM
That's the effect of the meaning of the tilda at the end

No, that means that it continues on forever, the reason being is that you can never divide 10 by 3 exactly and get a whole number. Because of that you have to keep going and will always have a remainder. Mathematically even taken to infinity it still won't exactly equal 1/3. There is no exact decimal answer to 1/3, only one that is approximate to within the bounds of reason. Hence why my original comment was that it is not exactly 1, but is within experimental error.

mike alexander
2006-Oct-27, 04:46 AM
I was just reading a book about Kepler, the Counter Reformation and the Thirty Years War.

Now, I actually UNDERSTAND it.

Next:

Pluto IS NOT!

IS SO!

yuzuha
2006-Oct-27, 05:57 AM
I can write the one, don't have time to write the other ^-^

Nicolas
2006-Oct-27, 07:04 AM
I would say that the only defensible interpretation of 0.000....001 is that it is the limit of the value as you keep adding zeroes - which is obviously 0.

Agreed. I know that it makes no real sense to put the one there, but I would classify it just above "meaningless" as the number does represent 0 in a way.

However, I doubt the people using that notation realize that :).

Nicolas
2006-Oct-27, 07:06 AM
Mathematically even taken [0.3333...] to infinity it still won't exactly equal 1/3.

Not to be blunt, but this shows that you don't get the mathematical meaning of infinity.

1/3 = 0.333... would in fact be a good definition of the concept of infinity in mathematics.

You are thinking at practical bounaries close to infinity, but mathematics takes the concept of infinity as truly endless and beyond all limitations of precision. The only thing that keeps 0.3333333333333 frmo being equal to 1/3 is precision. Same goes for 0.333333333333333333333333333333. The mathematical concept of infinity states that it improves the precision to absolute equality, per definition. "very very close to" is not existing in the mathematical concept of infinity. Infinity is all about making an approximation exactly equal (as one of the applications of infinity). It is the theoretical limit, beyond practical limitations of precision. Just like an asymptote is the value a function will theoretically reach, but for all practical amounts of time will not. Even if you wait really reallly long, it's still not exactly at that value. But if you'd wait truly infinitely long, it would be.

within experimental error.

The whole point of mathematical infinity is that it takes a progression beyoond all practical limitations, towards the theoretical limit value. For 0.999... this is 1, and I really can't help it ;).

Think of it the other way around: mathematically speaking, truly infinitely small becomes zero. The word "becomes" points out that the value will only be reached in infinity. Another nice definition of infinity is the point where 1/x is exactly equal to zero. Of course you can say&#239;t will get really reallly reallly small but not exactly equal to zero" and I see your reasoning in that, but then again you forget about "infinity" being a concept, a concept that defines the pooint where a limit value is actually reached.

LayMan
2006-Oct-27, 07:38 AM
LayMan, this has already been debated ad nauseum. Is it really necessary to make us write this all up yet again because you don't feel like doing a little reading? It's pretty much all been said by page seven anyway.

Please just read part of the thread. If you still have questions after you've done so, we'll be considerably more pleased to help you figure it out.

I started out doing that, but there was just sooo much of it... Sorry 'bout that (now where's that "Shame-on-me-oh-boy-I-wish-I-could-unpost-things-smiley" when you need it??:D ;)

Maybe we can start a new thread about Cantor's diagonal proof? We can use the incompleteness of the list of smiley's and work our way up so it covers the wide range of human emotions? :p

Weird, I'm not sure how else to interpret this comment:

The emphasis was on the word because... I was trying to distinguish between the mathematical causality

IF x/y=z
THEN z*y=x",

and the logical causality

IF 1 divided by 3 equals 0.33~
AND 0.33~ times 3 equals 0.99~
THEN 1 = 0.99~.

But as Moose pointed out, I better stop posting here before reading the entire thread... But not today: it's friday, the weekend's knocking on the door and I ain't gonna let it get spoiled by a little thing called infinity... (well, unless anyone can proof that since time is undividable, it'll last forever...). Maybe next week... ;)

worzel
2006-Oct-27, 07:46 AM
I would take both mathematical and logical causality to be the same thing (although I'd rather use "necessity" in place of "causality"). So if 1/3*3=1 mathematically and 1/3*3=0.999~ logically then I would say that 0.999~=1, both mathematically and logically.

Disinfo Agent
2006-Oct-27, 09:56 AM
I would say that the only defensible interpretation of 0.000....001 is that it is the limit of the value as you keep adding zeroes - which is obviously 0.In the context of decimal expansions of real numbers, yes.

hhEb09'1
2006-Oct-27, 10:27 AM
That's the effect of the meaning of the tilda at the end

No, that means that it continues on forever, the reason being is that you can never divide 10 by 3 exactly and get a whole number. Because of that you have to keep going and will always have a remainder. Mathematically even taken to infinity it still won't exactly equal 1/3. There is no exact decimal answer to 1/3, only one that is approximate to within the bounds of reason. Hence why my original comment was that it is not exactly 1, but is within experimental error.Mathematically, taken to infinity, it does exactly equal 1/3.

That's the mainstream interpretation of it, today. :)

c * (1 - an+1)/(1 - a)) = c + c * a + c * a2 + ... + c * an

The ellipsis in the above expression does not stand for infinity. But if we let n grow without bound, we can let it stand for infinity. That means the right hand side goes out forever with no last term, and the an+1 term disappears from the left hand side. You can verify that the equality is true by just doing the division 1/(1 - a) by longhand division.

There are problems when you try to use numbers like c=1 and a=2, for then it says
1/(1-2) = 1 + 2 + 4 + 8 + 16 + ...
and the problem is obviously that the an term grows without bound, and we really couldn't ignore the an+1 term on the left, even though the long division still works out. But if a is less than 1 (and greater than negative one), then the terms do not grow without bound--they actually get smaller, with a bound of zero. So, with c=3/10, and a=1/10, we get
1/3 = .333...

The "experimental error" is that term an+1. That's how much the left hand side deviates from true equality. But pick any value for your "allowable experimental error"--we can take it out to n places, and find that our experimental error is less than whatever allowable value you choose. It's smaller than any value, except zero. And that's pretty much the definition of zero. Our experimental error is zero.

jlhredshift
2006-Oct-27, 11:46 AM
R. P. Feynaman said; "for any delta you can always find an epsilon that is smaller"

yuzuha
2006-Oct-28, 09:49 AM
You know the old bit about rotating the graph of y = 1/x where x>1 around the x axis resulting in a horn with a finite volume but infinite surface area? Always wondered what would happen if you filled it with paint and then turned it inside out... think about it!

hhEb09'1
2006-Oct-28, 10:38 AM
You know the old bit about rotating the graph of y = 1/x where x>1 around the x axis resulting in a horn with a finite volume but infinite surface area? Always wondered what would happen if you filled it with paint and then turned it inside out... think about it!I already have :)

The paint on the last little (infinite) bit would not be of industry standard thickness. :)

mugaliens
2006-Oct-28, 11:55 AM
Also:

1/3 = 0.3333....
0.3333.... * 3 = 0.9999....
Therefore, 0.9999.... = 1

Actually, 0.33... (infinate number of 3's) * 3 = 1.0

hhEb09'1
2006-Oct-28, 11:58 AM
Actually, 0.33... (infinate number of 3's) * 3 = 1.0Why "actually"? That's part of Lycus's point.

cjl
2006-Oct-29, 03:17 AM
Actually, 0.33... (infinate number of 3's) * 3 = 1.0
If you work it out mathematically, it equals .999999...

However, as is clear from the way that number was obtained, it also equals 1.0

The only way this is possible is if .999... = 1.0.
It's quite simple really (and there are about 6 different ways of proving it).

afterburner
2006-Oct-29, 06:55 PM
I think this is more a question of whether infinity can be reached or not...

If it can...then sure 0.999... = 1, if not...:naughty:

But don't tell me that the Universe is not infinite, if infinity can't be reached. :D

hhEb09'1
2006-Oct-29, 07:04 PM
I think this is more a question of whether infinity can be reached or not...It's easy to do just using the old Zeno technique: go .9 meter in .9 seconds, then .09 meters in .09 seconds, then .009 meters etc.. After, one, second, you will have covered an infinite number of such intervals.

Frog march
2006-Oct-29, 07:23 PM
as you bend you arm at your elbow, it will have gone through an infinite number of angles

agingjb
2006-Oct-29, 08:29 PM
I think this is more a question of whether infinity can be reached or not...

If it can...then sure 0.999... = 1, if not...:naughty:

But don't tell me that the Universe is not infinite, if infinity can't be reached. :D

Limits. Either people accept the concept and rigorous definition of a limit or they don't. I assume, here, that those who do not accept the concept understand it, but still reject it. And really there is little more to say.

Nicolas
2006-Oct-30, 08:54 AM
I think this is more a question of whether infinity can be reached or not...

It apparently is very difficult to put physical limitations aside and think of a mathematical construct only. Mathematical infinity states infinity is reached. Point. Forget about physical limitations. Mathematics also can define a perfect circle. Forget about physical limitations. Don't make your head explode ;).

LayMan
2006-Oct-31, 10:04 AM
It apparently is very difficult to put physical limitations aside and think of a mathematical construct only. Mathematical infinity states infinity is reached. Point. Forget about physical limitations. Mathematics also can define a perfect circle. Forget about physical limitations. Don't make your head explode ;).

Yes, but 'set theory' is also a mathematical construct. Would you say that the sets [0,1], ]0,1], [0,1[ and ]0,1[ are all exactly the same? If not, what about the elements they contain? It's clear that 0.5 and 0.78 for instance are both elements of all 4 sets, but what about 0? And 1? And 0.999~? (Sorry Moose, haven't yet read all posts, I got to about page 15 this time... :D ).

Nicolas
2006-Oct-31, 10:12 AM
Following the strict rules of mathematics, there is a clear answer to all of your questions.

Would you say that the sets [0,1], ]0,1], [0,1[ and ]0,1[ are all exactly the same?

No, they differ in their bounds (inclusive vs exclusive) and hence in the elements they contain (see below).

0 is only contained in [0,1] and [0,1[
1 is only contained in [0,1] and ]0,1]

0.999~ is equal to 1, and hence the same answer as above applies.

As long as you stick to clear mathematical definitions, there is no prblem in answering these questions. 0.999~ implies that the boundary value 1 has been reached, hence simply the boundary rules apply, nothing like "very very close to it"

LayMan
2006-Oct-31, 10:18 AM
Following the strict rules of mathematics, there is a clear answer to all of your questions.

No, they differ in their bounds (inclusive vs exclusive) and hence in the elements they contain (see below).

0 is only contained in [0,1] and [0,1[
1 is only contained in [0,1] and ]0,1]

0.999~ is equal to 1, and hence the same answer as above applies.

As long as you stick to clear mathematical definitions, there is no prblem in answering these questions. 0.999~ implies that the boundary value 1 has been reached, hence simply the boundary rules apply, nothing like "very very close to it"

Well, maybe it works for you, but not for me: I simply can't squeeze 0.999~ into the set [1,2]... Then again, I never liked math anyway... :p :D

Nicolas
2006-Oct-31, 10:40 AM
It's a nice one, is 0.999~an element of [1,2]

It is, but I can see the pain one has to go through to accept that :). It only is becasuse of the "purity" of mathematical infinity. The problem with visualising the problem is that you visualize the transient, but not the end result, the only thing to which mathematical infinity is applicable. And when visualizing the transient, indeed 0.999~ does not seem to be a part of [1,2]

LayMan
2006-Oct-31, 10:50 AM
It's a nice one, is 0.999~an element of [1,2]

It is, but I can see the pain one has to go through to accept that :). It only is becasuse of the "purity" of mathematical infinity. The problem with visualising the problem is that you visualize the transient, but not the end result, the only thing to which mathematical infinity is applicable. And when visualizing the transient, indeed 0.999~ does not seem to be a part of [1,2]

:evil:

So 0.999~ would not be an element of [0,1[, but it actually belongs to [1,2]?
Makes you wonder what the 'last' element of [0,1[ would be...: 0.99~8?

;)

hhEb09'1
2006-Oct-31, 11:09 AM
So 0.999~ would not be an element of [0,1[, but it actually belongs to [1,2]?Because 0.999~ is equal to 1

By definition--by how we define what we mean by 0.999~

LayMan
2006-Oct-31, 12:15 PM
Because 0.999~ is equal to 1

By definition--by how we define what we mean by 0.999~

I was under the impression that it needed to be proved it was, not defined... :think:

Why? Where is it defined to be the case? Do you agree that the infinite series
1 + 1/2 + 1/4 + 1/8 + 1/16 + ... = 2 ?

Actually, this is getting interesting: according to the same 'logic' (or mathematical definition, if you will), the 'last' summation that you need to 'get to' 2 would be 1/∞. Which would be 0 by the same definition...

So looking from the point of view of the series, you 'miss' the last summation. But looking at it from the point of view of 2, the difference is 2 - 1/∞, or 2 - 0 = 2... Still leaves the question wether or not 1/∞ = 0...

Actually, I failed my last math exam. I wish I could see my math teachers face if he knew what I was doing here... Would drive him nuts! :D :D

Or in the words of John Locke:

"Whatever positive ideas we have in our minds of any space, duration, or number, let them be never so great, they are still finite; but when we suppose an inexhaustible remainder, from which we remove all bounds, and wherein we allow the mind an endless progression of thought, without ever completing the idea, there we have our idea of infinity... yet when we would frame in our minds the idea of an infinite space or duration, that idea is very obscure and confused, because it is made up of two parts very different, if not inconsistent. For let a man frame in his mind an idea of any space or number, as great as he will, it is plain the mind rests and terminates in that idea; which is contrary to the idea of infinity, which consists in a supposed endless progression."

Moose
2006-Oct-31, 01:09 PM
I was under the impression that it needed to be proved it was, not defined... :think:

... by how we define the tilde character (to mean an infinite progression).

Actually, this is getting interesting: according to the same 'logic' (or mathematical definition, if you will), the 'last' summation that you need to 'get to' 2 would be 1/∞. Which would be 0 by the same definition...

But there are an infinite number of increasingly small terms before you get to that point. That's where most people get the problem wrong. There seems to be an instinctive assumption built into our conciousness that infinity has to end somewhere. It doesn't. That's what infinity means. There's no end.

And because infinity doesn't end, 0.999~ must equal 1, by the various proofs and demonstrations, documented in this thread, that show this.

LayMan
2006-Oct-31, 01:27 PM
... by how we define the tilde character (to mean an infinite progression).

That's the definition for 0.99~ as such, not the fact whether it equals 1 or not.

And because infinity doesn't end, 0.999~ must equal 1, by the various proofs and demonstrations, documented in this thread, that show this.

That's basically the whole point of the discussion: those who voted "yes" in the poll, would say the 'neverending story' proofs that they are equal, those that voted "no" will say that, since the progression never ends, you can never 'get to the point' to say that they are equal (since that would imply that you actually 'stopped' adding 9's and 'moved on' to say "See, I've reached 1..."). Which has been basically said ad infinitum... :D And how you're gonna fit a number starting with 0.xxx~ into the set [1,2] is still beyond me...

Then again, if you insist that you are right by definition, well... ;)

hhEb09'1
2006-Oct-31, 01:51 PM
That's the definition for 0.99~ as such, not the fact whether it equals 1 or not.That's what I meant, when I said by definition. Once you accept that definition, the result (1 = 0.999~) is straightforward.
Then again, if you insist that you are right by definition, well... ;)By mathematics :)

Moose
2006-Oct-31, 02:09 PM
those that voted "no" will say that, since the progression never ends, you can never 'get to the point' to say that they are equal (since that would imply that you actually 'stopped' adding 9's and 'moved on' to say "See, I've reached 1...").

And that's what the rest of the people commonly get wrong: confusing the various "proofs by induction" with the reality of infinity. You don't have to go through an iterative process to be infinite. The whole point of limit theory is to use iterative/recursive processes to prove a result that would otherwise require infinity to compute.

Also, displaying 0.999~ isn't the same thing as saying 0.999[~]999. The former is an infinite progression. The latter is necessarily finite. They simply cannot be the same thing.

Remember, infinity cannot be iterated. It's infinity. It never ends. Iteration must end. So while you can't use iteration to display or compute 0.9 + 0.09 + 0.009 + ..., you can use it as part of one of several proofs by induction to know that 0.999~ = 1. This is what limit theory does.

Tell you what: simply show us a number x that lies between 0.999~ and 1. Any number at all (and there would necessarily be an infinite number of them if you're right.) If 0.999~ < x < 1, then 0.999~ must be something other than 1. That's all you have to do to prove us wrong. Should be simple if x exists.

LayMan
2006-Oct-31, 02:10 PM
By mathematics :)

The days of enforcing mathematics upon me are over!!!

I'm a big boy now!

Am too, am too!!!!!

:p :D

Anywayzz, I might be tempted to accept the fact about the average of 0.999~ and 1, since I wouldn't know how to answer that one...

But not the mentioning of Zeno's paradox. That sounds a bit too 'practical' to me... After all, if you keep dividing the distance between you and the wall, there's always going to be some distance between you and the wall. You can't divide by 2 and not end up with 2 halfs, right?...

jlhredshift
2006-Oct-31, 02:14 PM
The days of enforcing mathematics upon me are over!!!

I'm a big boy now!

Am too, am too!!!!!

:p :D

Anywayzz, I might be tempted to accept the fact about the average of 0.999~ and 1, since I wouldn't know how to answer that one...

But not the mentioning of Zeno's paradox. That sounds a bit too 'practical' to me... After all, if you keep dividing the distance between you and the wall, there's always going to be some distance between you and the wall. You can't divide by 2 and not end up with 2 halfs, right?...

How did the song go..."take it to the limit"

Nicolas
2006-Oct-31, 02:23 PM
A good question raised was

What is the last element of

[0,1[

I can't give a mathematical expression for that number straight away.

Moose
2006-Oct-31, 02:23 PM
But not the mentioning of Zeno's paradox. That sounds a bit too 'practical' to me... After all, if you keep dividing the distance between you and the wall, there's always going to be some distance between you and the wall. You can't divide by 2 and not end up with 2 halfs, right?...

Right, in an iterative process.

Keep in mind, though, that as you're dividing the distance, you're also dividing the time it takes to travel that distance. As the running time approaches zero, so does the distance. At the limit of this iterative process, the time reaches zero, and so does the distance.

The thing to remember is that Zeno necessarily must reach the wall. (We know this because we do similar tasks every day. We walk somewhere, we get there.)

Limit theory shows why Zeno's "paradox" isn't really a paradox at all.

LayMan
2006-Oct-31, 02:32 PM
Tell you what: simply show us a number x that lies between 0.999~ and 1. Any number at all (and there would necessarily be an infinite number of them if you're right.) If 0.999~ < x < 1, then 0.999~ must be something other than 1. That's all you have to do to prove us wrong. Should be simple if x exists.

Deal, I'll fry my brains on that until they drag me off to the institute, kicking 'n screaming... But only if you then tell me what the last number of π is. Should be equally simple if π exists... :D:D

Chuck
2006-Oct-31, 02:36 PM
A good question raised was

What is the last element of

[0,1[

I can't give a mathematical expression for that number straight away.

There is no last element. For any number in the set you can always find a greater number that's still less then 1 so there can't be a highest member.

Nicolas
2006-Oct-31, 02:41 PM
This implies an infinite series converging to one but not reaching one. How to express that in maths?

Because if we'd take the iteration for i = 1..inf of x(i+1) = x(i) + 0.5*(1-x(i)), we would take a number closer to one but smaller than one on every iteration, though in the limit we would arive at x(inf) = 1, hence outside [0,1[.

And x(inf-1) = x(inf), so that doesn't help us.

LayMan
2006-Oct-31, 02:44 PM
There is no last element. For any number in the set you can always find a greater number that's still less then 1 so there can't be a highest member.

Interesting. But the same goes the other way around: there's always going to be a smaller number that still is bigger then 0. So by the same argumentation, there can't be a lowest member to the set [0,1[. Which seems to imply that 0 isn't a member... ;)

Moose
2006-Oct-31, 02:44 PM
Deal, I'll fry my brains on that until they drag me off to the institute, kicking 'n screaming... But only if you then tell me what the last number of π is. Should be equally simple if π exists... :D:D

Deal. You go first. ;)

LayMan
2006-Oct-31, 02:50 PM
Deal. You go first. ;)

Rats, should have seen that coming... :wall:

SeanF
2006-Oct-31, 05:21 PM
Interesting. But the same goes the other way around: there's always going to be a smaller number that still is bigger then 0. So by the same argumentation, there can't be a lowest member to the set [0,1[. Which seems to imply that 0 isn't a member... ;)
No, it means there can't be a lowest member to the set ]0,1], and 0 is not a member of that set.

The lowest member of [0,1] is 0, and the last member is 1.

]0,1[ has neither a lowest nor highest member.

:)

Nicolas
2006-Oct-31, 06:41 PM
Can anyone answer if one can express a series converging to 1 but never reaching 1 mathematically? (that is, the right side of the [0,1[ interval)?

(yesyes I already know 0.999~ = 1, but I wonder how to express the kind of (non)convergence described above)

hhEb09'1
2006-Oct-31, 06:51 PM
Can anyone answer if one can express a series converging to 1 but never reaching 1 mathematically? (that is, the right side of the [0,1[ interval)?

(yesyes I already know 0.999~ = 1, but I wonder how to express the kind of (non)convergence described above)Not sure what you are asking for. If it converges, it is convergent, of course.

Why doesn't 0.999~ satisfy your request? What other property are you looking for?

Nicolas
2006-Oct-31, 06:57 PM
Well, the problem is that 0.999~ = 1, hence is no member of [0,1[

I wonder how to mathematically express the rightmost side of the interval [0,1[

(that's what I meant with converging to 1 but never reaching it, indeed in fact this is not converging :) )

hhEb09'1
2006-Oct-31, 07:00 PM
I wonder how to mathematically express the rightmost side of the interval [0,1[Mathematically, there is no rightmost side of that interval.

A mathematician would say that the set is bounded, but has no largest member.

Nicolas
2006-Oct-31, 08:27 PM
So the correct answer to my question would be "1[" as there is no shorter correct mathematical expression for "everything upto 1, excluding 1" :)

BUt isn't this contrary to limit theory? I mean, you can always find a number closer to 1 in the interval [0,1[, but even when doing infinitely so you still have not arrived at 1 in the interval [0,1[

Disinfo Agent
2006-Oct-31, 08:44 PM
BUt isn't this contrary to limit theory? I mean, you can always find a number closer to 1 in the interval [0,1[, but even when doing infinitely so you still have not arrived at 1 in the interval [0,1[I think that's what many people get wrong about limit theory, unfortunately. The limit of a sequence is not a number that you always get closer to, but never reach, even "when doing infinitely". It's the number you do reach when you "do it infinitely".

BTW, one possible sequence is sn = 1 - 0.1n. There are infinitely many of them. :)

hhEb09'1
2006-Oct-31, 09:06 PM
So the correct answer to my question would be "1[" as there is no shorter correct mathematical expression for "everything upto 1, excluding 1" :)

BUt isn't this contrary to limit theory? I mean, you can always find a number closer to 1 in the interval [0,1[, but even when doing infinitely so you still have not arrived at 1 in the interval [0,1[I can't see anything really wrong in your post. "1[" is not a standard expression, but I see what you mean.

Except I'm not sure why you think that it is contrary to limit theory? Why would you say that?

Nicolas
2006-Oct-31, 09:43 PM
Because any series I can think of that always finds a number smaller than 1 but closer to 1, will reach 1 in the limit, so will fall out of the interval [0,1[ in the limit.

I don't mean it is contrary to limit theory as in "it shows limit theory wrong", I'm just trying to find a series that shows the behaviour of going closer to 1 but never reaching it, not even in the limit, hence never going outside the [0,1[ interval.

Nicolas
2006-Oct-31, 09:45 PM
I think that's what many people get wrong about limit theory, unfortunately. The limit of a sequence is not a number that you always get closer to, but never reach, even "when doing infinitely". It's the number you do reach when you "do it infinitely".

BTW, one possible sequence is sn = 1 - 0.1n. There are infinitely many of them. :)

I think you missed the point of my question :)

Your series does reach 1 at infinity, hence does not fall in the interval [0,1[ at infinity.

I'm looking for a series that results in a value as high as possible, but never going outside of [0,1[

hhEb09'1
2006-Oct-31, 10:20 PM
Because any series I can think of that always finds a number smaller than 1 but closer to 1, will reach 1 in the limit, so will fall out of the interval [0,1[ in the limit.

I don't mean it is contrary to limit theory as in "it shows limit theory wrong", I'm just trying to find a series that shows the behaviour of going closer to 1 but never reaching it, not even in the limit, hence never going outside the [0,1[ interval.Oh! that's easy to do. These numbers: .3, .33, .333, .3333, .33333, etc., are all smaller than one, and they get closer to one, but do not reach one in the limit. And their limit is still in the interval.

Moose
2006-Oct-31, 10:36 PM
Any approximation of (Pi/10), or (e/10) would meet those requirements as well.

Nicolas
2006-Oct-31, 10:39 PM
OK, and now an expression for a number as close as possible to 1 but never reaching it :)

I thought that was clear...

------------

I know that there is no single number that is "the closest possible to 1 but not equal to one". Any series that comes as close as possible to one, will per definition arrive at 1 in infinity. That in itself shows that [0,1[ has no defined upper value inside its interval (even though all values are smaller than 1).

That's another one for some to blow their head over: while you can pick any number smaller than one, you can never say exactly how large the largest possible number then is.

spacemanspiff
2006-Nov-01, 02:29 AM
I am not mathematician and I know absolutely nothing about calculus. Speaking from a laymans point of view I would say no they are not.

Mathematicaly speaking they may be equal, but from a purley practical point of view they can't be can they. I mean 0.999999999~ is continualy smaller fractions of 1 right? So even if it is infinite it would never equal 1.

Fortis
2006-Nov-01, 02:46 AM
I am not mathematician and I know absolutely nothing about calculus. Speaking from a laymans point of view I would say no they are not.

Mathematicaly speaking they may be equal, but from a purley practical point of view they can't be can they. I mean 0.999999999~ is continualy smaller fractions of 1 right? So even if it is infinite it would never equal 1.
A simple way to look at this is to ask yourself, "What number lies between 0.999~ and 1?" If they are different, then there must be an infinite number of reals that lie between them.

cjl
2006-Nov-01, 03:40 AM
Here's a proof that shows it must equal 1. It has probably been shown before in this thread, possibly even by me, but I don't want to search through, and it is a good way of explaining it.

Lets define x as 0.999~
10x is therefore equal to 9.999~ just by shifting the decimal place.
If we subtract x (0.999~) from 10x (9.999~), we get 9. This eliminates the endless decimal. If 9x = 9, then x must necessarily equal 1. Therefore, 0.999~ can be mathematically proven to equal 1.

This probably won't end the discussion, but at least it's another way to look at it...

spacemanspiff
2006-Nov-01, 04:20 AM
OK I'm probably going to make a fool out of myself but here 'goes. If I'm not mistaken 9x with x=0.999~ is the same as 9 times 0.999~. I did this on a calculator(I told you I'm not good at math) and I did not get 9, I got 8.991. I may be missing somthing more complicated here if I am sorry. I hate to sound like a doofus but I just don't understand complicated math so I may simply be unable grasp the concepts involved in this equation.

A simple way to look at this is to ask yourself, "What number lies between 0.999~ and 1?" If they are different, then there must be an infinite number of reals that lie between them.

Wouldn't there be an infinite number of nines between them?

Corgon
2006-Nov-01, 04:49 AM
OK I'm probably going to make a fool out of myself but here 'goes. If I'm not mistaken 9x with x=0.999~ is the same as 9 times 0.999~. I did this on a calculator(I told you I'm not good at math) and I did not get 9, I got 8.991. I may be missing somthing more complicated here if I am sorry. I hate to sound like a doofus but I just don't understand complicated math so I may simply be unable grasp the concepts involved in this equation.
A calculator can't hold an infinite number of nines any better than a piece of paper can. It multiplied 9 by 0.999 (not 0.999~) and got the proper answer for that equation.

Wouldn't there be an infinite number of nines between them?
0.999~ is already "an infinite number of nines", though. How would you add more nines to 0.999~ ?

hhEb09'1
2006-Nov-01, 04:51 AM
Mathematicaly speaking they may be equal, but from a purley practical point of view they can't be can they. I mean 0.999999999~ is continualy smaller fractions of 1 right? So even if it is infinite it would never equal 1.
OK I'm probably going to make a fool out of myself but here 'goes. If I'm not mistaken 9x with x=0.999~ is the same as 9 times 0.999~. I did this on a calculator(I told you I'm not good at math) and I did not get 9, I got 8.991.Now try it with 0.999999999 :)

"What number lies between 0.999~ and 1?" Wouldn't there be an infinite number of nines between them?OK, make it "What number lies between 1 and 0.999~?" :)

cjl
2006-Nov-01, 05:30 AM
OK I'm probably going to make a fool out of myself but here 'goes. If I'm not mistaken 9x with x=0.999~ is the same as 9 times 0.999~. I did this on a calculator(I told you I'm not good at math) and I did not get 9, I got 8.991. I may be missing somthing more complicated here if I am sorry. I hate to sound like a doofus but I just don't understand complicated math so I may simply be unable grasp the concepts involved in this equation.

Wouldn't there be an infinite number of nines between them?

As already stated, the calc cannot hold enough nines. Think of it this way. Multiplying any number by ten (in base ten) is simply moving the decimal point one time to the right. For example, 8.57 times 10 is 85.7

So, if you have 0.999~ with the ~ representing an infinite sequence of nines, multiplying it by ten gives 9.999~, which also has an infinite number of nines after it. By then subtracting the original number, you get 9 (simply because 9.anything minus 0.anything [provided the anythings are the same] is 9). Therefore, 9 times the original number is 9. Is that any clearer?

agingjb
2006-Nov-01, 12:47 PM
It is clear that there are many posters here who reject the traditional axiomatic development of the reals, and the concept of limits. But shouldn't this be in "Against the Mainstream"?

jlhredshift
2006-Nov-01, 01:03 PM
It is clear that there are many posters here who reject the traditional axiomatic development of the reals, and the concept of limits. But shouldn't this be in "Against the Mainstream"?

I do not think so. When explaining rudimentry math concepts such as a point, or a line that is infinetly small, infinetly thin and long is tuff enough. Whereas the concept of limit or a Dedakind cut is even tougher for most students when first presented with these concepts. After all, this debate has been ensuing for thousands of years and I think it is language that fails us more than anything else.

Nicolas
2006-Nov-01, 03:21 PM
It is clear that there are many posters here who reject the traditional axiomatic development of the reals, and the concept of limits. But shouldn't this be in "Against the Mainstream"?

For the record, I accept the concept of limits and 0.999~ = 1.

That said, I do think it's interesting that [0,1[ has no quanitfyable upper value, not even expressed as a limit, even though we know we can choose from all values smaller than and not including 1 to find that upper value. (and not below zero of course).

Donnie B.
2006-Nov-01, 07:36 PM
I don't know why the idea that [0, 1[ has no "highest value" should be so unnerving. The same is true for the set of reals, or positive integers.

I've brought this up before in this thread's first go-round: I think what trips up many people on the 0.999~ question is that they misunderstand the ~ (or ...) symbol as indicating an operation. '~' in this context does not mean "keep adding nines forever". If it did, of course you'd never get to exactly 1.0 no matter how fast you added nines or how many you added. We poor mortals don't have the luxury of waiting forever!

Rather, the '~' means that an infinite number of nines is already there. We don't have to execute any operation. We don't have to do anything. As soon as we tack on the '...' those nines already exist all the way to infinity.

Maybe this will help those who can't accept the other arguments and proofs. But somehow I doubt it! :shifty:

SMEaton
2006-Nov-01, 07:41 PM
May I say that this is one of the rare multi-page threads that hasn't regressed into a Firefly discussion? ;)
I did just see a comedian do an asymptote joke. I laughed and then had a calculus flashback.

Nicolas
2006-Nov-01, 07:44 PM
I don't know why the idea that [0, 1[ has no "highest value" should be so unnerving. The same is true for the set of reals, or positive integers.

The big difference is that you know that any value contained in [0,1[ is smaller than 1, while for reals or positive integers you can't state any value that is higher than any real or integer.

That's what makes the unexistence of the highest value of [0,1[ far more strange than that of the integers. The integers have no upper bound whatsoever, [0,1[ has one. Only due to the excluding character of the bound itself, the highest value between the bounds becomes undefined.

I've brought this up before in this thread's first go-round: I think what trips up many people on the 0.999~ question is that they misunderstand the ~ (or ...) symbol as indicating an operation. '~' in this context does not mean "keep adding nines forever". If it did, of course you'd never get to exactly 1.0 no matter how fast you added nines or how many you added. We poor mortals don't have the luxury of waiting forever!

Rather, the '~' means that an infinite number of nines is already there. We don't have to execute any operation. We don't have to do anything. As soon as we tack on the '...' those nines already exist all the way to infinity.

I think that indeed this is the biggest hurdle in accepting that 0.999~ = 1. One must see that ~ is a mathematical principle, not an operation with some transient behaviour. Even though it appears to be stated as a process (x goes to n), also a limit is not refering to the process but to the result. ONe should better read it as "x has arrived at n". (n can also be infinity)

spacemanspiff
2006-Nov-02, 05:37 AM
A calculator can't hold an infinite number of nines any better than a piece of paper can. It multiplied 9 by 0.999 (not 0.999~) and got the proper answer for that equation.

Yes I know a calculator can't hold an infinite number.(I pretty much thought that would be a given), I did exactly that. I multiplied 9 by 0.999 and I still get 8.991.

So, if you have 0.999~ with the ~ representing an infinite sequence of nines, multiplying it by ten gives 9.999~, which also has an infinite number of nines after it. By then subtracting the original number, you get 9 (simply because 9.anything minus 0.anything [provided the anythings are the same] is 9). Therefore, 9 times the original number is 9. Is that any clearer?

I'm sorry but it isn't. I still don't understand how 9 times 0.9999999~ comes out to 9.

It is clear that there are many posters here who reject the traditional axiomatic development of the reals, and the concept of limits.

OK I am obviously WAY out of my depth here. Because I have no idea what reals are in math and I don't understand why not having anything between 0.9999999~ and 1 makes them equal. It probably would do no good for you guys to try and explain it to me because I simply don't understand the basic concepts. Your talkin to a guy who flunked high school algebra here.
I'm more of a language and art person.
Just pretend I didn't say anything.

cjl
2006-Nov-02, 05:47 AM
I'm sorry but it isn't. I still don't understand how 9 times 0.9999999~ comes out to 9.

OK, lets try one more time then.

First of all, do you agree that 10 time 0.9999~ is 9.9999~?

If not, what is the problem? If so, great.

Second, can you see that 9.999~ - 0.999~ is 9? If not, try looking at it this way:

Line them up vertically in the standard way for doing longer subtraction problems. Both will have infinite nines stretching off, all of which are identical and will therefore cancel. The only digit that does not cancel is the 9 before the decimal.

Esseintially,
0.999~ -
9.999~
_______
9.000~

Is that better?

Now, assume x is equal to 0.999~ Clearly, 10x is 9.999, and if you subtract 1x from 10x, you have by definition created 9x. So, 9x is equal to 9. Is this better?

rahuldandekar
2006-Nov-02, 06:14 AM
As I understand it,

0.99999999999999~ is not a number.
It is a sequence, define by xn = 9 x 10-1 + 9 x 10-2 +... + 9 x 10-n

and the "number" 0.99999~ is the limit of the sequence.

Now, it can be proved that the limit is 1.

Given e > 0, we choose we need to choose no > 0 such that
|xn - 1| < e for all n>no .

It is easy to choose no. Given e , just choose n such that 10-no < e .

So, we have lim xn = 1.

Thus, 0.9999~, which is the limit of the sequence, is 1.

I have not read all the 47 pages here, has this been stated before?

spacemanspiff
2006-Nov-02, 06:15 AM
"First of all, do you agree that 10 time 0.9999~ is 9.9999~?"

Yep, got that.

"Second, can you see that 9.999~ - 0.999~ is 9?"

Yep got that too.

"Now, assume x is equal to 0.999~ Clearly, 10x is 9.999"

Yep.

"if you subtract 1x from 10x, you have by definition created 9x"

This is where I get lost. Because when I subtract 1x from 10x I don't get 9x, I get 9. Now if I am not mistaken, 9x(with x being 0.9999~) is the same as 9 times 0.9999~.
If you actually do that equation(9 times 0.9999~) you don't get 9, you get 8.9999999~. Except with a calculator which of course can't go to infinity you get a 1 on the end.

I'm sorry rahuldandekar, but I didn't understand any of the equations you posted. As I said I no nothing about calculus. So it may be a waste of every ones time to try and explain this to me. I am trying to understand, I guess my mind just doesn't work that way.

cjl
2006-Nov-02, 06:21 AM
I want you to ignore that for right now however, as the calculator cannot do that. Also, it can be mathematically proven that 8.9999~ is 9 for the same reason that 0.999~ is 1.

If you subtract 1x from 10x, you get 9x, regardless of the value of x. That is simple algebra.

Example: 10x - 1x when x = 5 is 50 - 5, which is 45. This is 9 times 5, or 9x. If you attempt to get 9x through direct multiplication with infinite decimals, you can run into that error that you had above. So, you have to get it through the slightly more roundabout method.

Hence, 9.999~ - 0.999~ is 9 and 10x - 1x = 9x, and therefore, there is no other possible conclusion other than that 9x = 9.

rahuldandekar: Yes, I understand that, and all is fine and good if you understand limits. The problem arises when attempting to explain it to those who do not fully understand the concept of limits (and yeah, that's been stated about 16 million times, and unfortunately, like all of the other proofs and explanations, has not worked).

spacemanspiff
2006-Nov-02, 06:34 AM
"Example: 10x - 1x when x = 5 is 50 - 5, which is 45. This is 9 times 5, or 9x."

OK well that makes more sense. I Think I can see what your saying now.

"If you attempt to get 9x through direct multiplication with infinite decimals, you can run into that error that you had above. So, you have to get it through the slightly more roundabout method."

I guess it's the roundabout method I don't quite get. It sounds to me like one of those convoluted mathematical problems which doesn't sound right but works anyway.

Thank you cjl for your patience with this poor mathematically challenged soul. Your explanation really did help.

rahuldandekar
2006-Nov-02, 06:51 AM
Well, ok, I'll try and explain it without limits... er, but the idea is the same.

See, it's hard to understand 0.9999~ as a number. think of it as the infiniteth term of a sequence, 0.9, 0.99, 0.999, 0.9999 , and so on. (The concept of the infiniteth term may be hard to understand, but if you can't get around the concept of a number with infinite digits, this is an easier way).

What actually IS an "infiniteth term"? Well, here's how it's defined:

We cannot actually write infinite terms. So, we write a finite number of 9's ... like, 0.999999....999. Now, we want to see if 0.999~ is equal to 1. You can say it's equal to one, if you can make the difference between 0.999...99 and 1 less than any number after a certain number of 9's (the number of 9's depends upon the number which you are aiming for).

For example, suppose I want to make the difference between 1 and 0.999...999 less than 0.0000000000000047 (14 zeros before 47). So, I write 15 9's ... 0.999999999999999 . the difference between this number and 1 is 0.000000000000001 , which is 14 zeros before the 1. So it is less than the number we've taken.

Now, it is obvious that the difference between 0.9999...999 and 1 can be made smaller than any number, provided we take a sufficient number of 9's.

Thus, if we write an infinite number of 9's, we can understand that if we take any number, say 0.000...0001, with any number of zeros before it, the difference between 0.999~ and 1 is smaller than it (since there are infinite 9's).

So, since there is essentially no difference between 0.9999~ and 1, so 1 - 0.999~ = 0 , with an infinite number of 9's. So, 0.99999~ = 1.

In other words...

What we're looking at is the difference 1 - 0.999~ . It can be positive, it can be zero (it cant be negative). If it is positive, it's greater than some postive number.

Now, consider the numbers 1- 0.9, 1-0.99, 1-0.999, 1-0.9999, etc. Obviously the numbers are getting smaller. Now, we saw above that if we take any positive number, after a sufficient number of 9's, the difference can be made smaller than it. So, the difference 1- 0.999~ is smaller than every positive number. What else can be be except zero?

It's anti-intuitive... how can you replace an infinite number of 9's by a single one? ;) Well, you can. If you think this is silly, you ain't seen nothing yet.

Hope you understood it. :-)

rahuldandekar
2006-Nov-02, 07:36 AM
I edited my above post... added a section "In other words...".

hhEb09'1
2006-Nov-02, 09:53 AM
As I understand it,

0.99999999999999~ is not a number.

::snip::

Thus, 0.9999~, which is the limit of the sequence, is 1. You've contradicted yourself there. If as you say 0.9999~ is 1, then it is a number.

hhEb09'1
2006-Nov-02, 09:57 AM
This is where I get lost. Because when I subtract 1x from 10x I don't get 9x, I get 9.

::snip::

I'm sorry rahuldandekar, but I didn't understand any of the equations you posted. As I said I no nothing about calculus. So it may be a waste of every ones time to try and explain this to me. I am trying to understand, I guess my mind just doesn't work that way.First you have to learn algebra :)

The "x" stands for any number, and "10x" stands for ten times that number. What if x were 3? Then 10x would be 30, and 10x minus 1x would be 27. Which is 9x, right? Not 9.

rahuldandekar
2006-Nov-02, 10:03 AM
You've contradicted yourself there. If as you say 0.9999~ is 1, then it is a number.

Whoops. :silenced: I meant to say it is the limit of a sequence... so it is a number. The "not a number" line is wrong. I stand corrected.

Thanks for noticing :clap:.

LayMan
2006-Nov-02, 10:22 AM
"Example: 10x - 1x when x = 5 is 50 - 5, which is 45. This is 9 times 5, or 9x."

OK well that makes more sense. I Think I can see what your saying now.

"If you attempt to get 9x through direct multiplication with infinite decimals, you can run into that error that you had above. So, you have to get it through the slightly more roundabout method."

I guess it's the roundabout method I don't quite get. It sounds to me like one of those convoluted mathematical problems which doesn't sound right but works anyway.

Thank you cjl for your patience with this poor mathematically challenged soul. Your explanation really did help.

Oh no, we're losing one to the Dark Side of Mathemagics... :D :D

rahuldandekar: Yes, I understand that, and all is fine and good if you understand limits. The problem arises when attempting to explain it to those who do not fully understand the concept of limits (and yeah, that's been stated about 16 million times, and unfortunately, like all of the other proofs and explanations, has not worked).

Tell me, do you think 0.99~ is an element of the set of natural numbers? I know 1 is... And if there both 'exactly' the same, they should belong to the same set...

I also have problems with the 'proof' involving "1/3"...

Does 1/3 = 0.33~? I say "no". That's the definition of 'not being divisable' by a number, which is precisely why we have the set of real numbers in the first place. You can't divide an angle into 3 equally large parts, 1 of them will always be infinitesimally larger than the other 2... :). That's the problem with dividing an angle into exactly 3 equally large parts: in order to be able to do that, you would need to know the exact value of Pi, right down to the last decimal. You can only exactly divide numbers by 3 (or any other number) if those numbers are 'divisable' by 3. So 27 divided by 3 will equal exactly 9, hence 9 times 3 = 27.

But 1 divided by 3 does not exactly equal 0.33~. If it did, then 2 divided by 3 would need to equal exactly 0.66...67 (with an infinite number of 6's between the comma and 7).

0.67 + 0.33 = 1
0.66667 + 0.33333 = 1
...

Or:

0.66 + 0.34 = 1
0.66666 + 0.33334 = 1
...

Again, that is the whole point of 'not being divisable' by 3.

Take the following problem: a farmer has 17 cows and 3 sons. He wants his oldest son to get half of them, his middle son a third and his youngest a ninth... How does he do that? Easy, he borrows 1 cow from his neighbour, giving a total of 18 cows. Now the oldest son gets 9 cows (18/2), the middle one 6 (18/3) and the youngest 2 (18/9). 9 + 6 + 2 = 17, so he gives back the borrowed (18th) cow to his neighbour.

Question, did he actually divide 17, or would it be more correct to say he was dividing 18? It seems to me that mathematicians have borrowed an infinitely small cow, but then refused to give it back and claim it never existed in the first place... :D

After all, mathematically speaking, the whole idea behind calculus is that the d in dx becomes infinitesimally small, but (and that's the important bit) never exactly equal to 0... If it ever actually did become exactly 0, you end up adding an infinite amount of 0's (or multiplying a finite given number by 0) and any surface would be 0.

I think that's what many people get wrong about limit theory, unfortunately. The limit of a sequence is not a number that you always get closer to, but never reach, even "when doing infinitely". It's the number you do reach when you "do it infinitely".

If you do it infinitely, you never reach the number.

BTW, one possible sequence is sn = 1 - 0.1n. There are infinitely many of them. :)

No, according to the same logic that would make 1/3 = 0.33~ , 0.1n would actually reach 0 as n 'goes/converges/disappears' to infinity. So there would only be 1 element, namely 1 - 0 = 1.

And yes, I'm aware of Cantor and his exploits, I just choose to disagree with him...:razz:

That's the whole problem with mathematicians, as soon as they realize someone is having some innocent fun with infinities, they start kicking 'n screaming...:D ;)

I think it's all part of a big conspiracy that...

[KNOCK, KNOCK]...

OH NO, THEY'VE FOUND ME...

[Hides behind a smiley]

LayMan
2006-Nov-02, 10:23 AM
;-)

rahuldandekar
2006-Nov-02, 10:38 AM
Arghhh... infinities....

Tell me, do you think 0.99~ is an element of the set of natural numbers? I know 1 is... And if there both 'exactly' the same, they should belong to the same set...

Well, why not? 0.5 + 0.5 is a natural number. Just because it's expresssed in decimals, that doesn't mean it's not a natural number.

I also have problems with the 'proof' involving "1/3"...

[snip]
But 1 divided by 3 does not exactly equal 0.33~. If it did, then 2 divided by 3 would need to equal exactly 0.66...67 (with an infinite number of 6's between the comma and 7).

Where will you put the 7? The thing with infinity is, you cant put the 7 anywhere! It's just 0.6666... *keeps on writing* ... 6666... *more writing*...

0.67 + 0.33 = 1
0.66667 + 0.33333 = 1
...

Or:

0.66 + 0.34 = 1
0.66666 + 0.33334 = 1
...

What about 0.666666~ + 0.33333~ = 0.9999~ = 1 ?

After all, mathematically speaking, the whole idea behind calculus is that the d in dx becomes infinitesimally small, but (and that's the important bit) never exactly equal to 0... If it ever actually did become exactly 0, you end up adding an infinite amount of 0's (or multiplying a finite given number by 0) and any surface would be 0.

Yes, it never exactly becomes zero. Just as you can never do infinite sums. dx doesn't reach zero, and the terms don't ever reach infinity. The limit form of calculus, and the definition of limit, clarify this, don't they?

If you do it infinitely, you never reach the number.

The thing is, you can never do it infinitely. So, why argue about what happens when you do? You just cannot.

That's the whole problem with mathematicians, as soon as they realize someone is having some innocent fun with infinities, they start kicking 'n screaming...:D ;)

Infinity is hard to grasp. You can never reach it, but you always use it. That's why we have rigid definitions of limits.

hhEb09'1
2006-Nov-02, 10:42 AM
But 1 divided by 3 does not exactly equal 0.33~. If it did, then 2 divided by 3 would need to equal exactly 0.66...67 (with an infinite number of 6's between the comma and 7).O no, 2 divided by 3 only needs to equal 0.666~

That's the beauty of it, they add up to 0.999~, which is one, as you would expect. :)
Take the following problem: a farmer has 17 cows and 3 sons. He wants his oldest son to get half of them, his middle son a third and his youngest a ninth... How does he do that? Easy, he borrows 1 cow from his neighbour, giving a total of 18 cows. Now the oldest son gets 9 cows (18/2), the middle one 6 (18/3) and the youngest 2 (18/9). 9 + 6 + 2 = 17, so he gives back the borrowed (18th) cow to his neighbour.

Question, did he actually divide 17, or would it be more correct to say he was dividing 18?If you add up the numbers one half, one third, and one ninth, you get 17/18, so it seems he originally intended to not only divide by 18, but also not give the entire herd to his three sons. Looks like he gave away the farm.
It seems to me that mathematicians have borrowed an infinitely small cow, but then refused to give it back and claim it never existed in the first place...Cute. You got your cow back now. It was in yesterday's mail.

LayMan
2006-Nov-02, 11:30 AM
Cute. You got your cow back now. It was in yesterday's mail.

Thanks, I've been looking everywhere for it! Funny how it's always in the last place you look... :) Her milk's gone off, though...

I'm having waaaay to much fun with this, can't be good for my mental health... :D

Nicolas
2006-Nov-02, 12:03 PM
Just a question of terminology:

For the half-open interval [0,1[ the lower bound is 0. What do we say about the upper bound? Is it still called 1 for this open interval or is it called undefined? The highest value is undefined (we can only say that it is smaller than 1, but we can't specify it), but I wonder what terminology to use for the bound itself.

rahuldandekar
2006-Nov-02, 12:21 PM
The upper bound is defined mathematically... that's what counts, we know it exists. Why do we want to give a decimal form to it? It isn't necessary... we can't give an exact decimal form to pi too, but we know it exists, and is irrational, transcendental, etc.

And it is NOT 0.9999~ . That's because, well, it doesn't belong to the set. I know I'll have to prove what I say... I'm looking into it. ;-) The most clear answer is that it is 1, and 1 is not in the set.

Also, almost always, the most important thing is the supremum, and here we know it doesn't belong to the set, and that it is 1.

LayMan
2006-Nov-02, 12:23 PM
Just a question of terminology:

For the half-open interval [0,1[ the lower bound is 0. What do we say about the upper bound? Is it still called 1 for this open interval or is it called undefined? The highest value is undefined (we can only say that it is smaller than 1, but we can't specify it), but I wonder what terminology to use for the bound itself.

Maybe you should ask the person who makes the last post in this thread... :evil: :D

Naturally (bad pun), since I've voted "no", I would say it equals 0.99~, which would make 0.99~ and 1 adjacent numbers, not equal... There are no elements between the 2 sets [0,1[ and [1,2], otherwise we'd end up with double values in the set [0,2]... Just like there is no element between the two natural values 1 and 2. Doesn't mean 1 = 2...

Which answers this question:

Tell you what: simply show us a number x that lies between 0.999~ and 1. Any number at all (and there would necessarily be an infinite number of them if you're right.) If 0.999~ < x < 1, then 0.999~ must be something other than 1. That's all you have to do to prove us wrong. Should be simple if x exists.

There's no number between 0.99~ and 1 in R (R being the set of real numbers, the actual symbol is a bit different - R with a double vertical line, but I can't get in in here...). Neither is there a number between 1 and 2 in N (N being the set of natural numbers). Doesn't mean there equal, though.

You're turn Moose, what's the last decimal of Pi? We made a deal, remember?

Deal. You go first. ;)

:D ;)

LayMan
2006-Nov-02, 12:39 PM
The upper bound is defined mathematically... that's what counts, we know it exists. Why do we want to give a decimal form to it? It isn't necessary... we can't give an exact decimal form to pi too, but we know it exists, and is irrational, transcendental, etc.

And it is NOT 0.9999~ . That's because, well, it doesn't belong to the set. I know I'll have to prove what I say... I'm looking into it. ;-) The most clear answer is that it is 1, and 1 is not in the set.

Also, almost always, the most important thing is the supremum, and here we know it doesn't belong to the set, and that it is 1.

Quoted from Wikipedia (my bold):

"Informally, a logical statement is vacuously true (or trivially true) if it is true but doesn't provide any useful information; examples are statements of the form "anything that has property A must also have property B", where no entities with property A exist.

It is tempting to dismiss this concept as silly. It does, however, have useful applications. One example is the empty product—the fact that the result of multiplying no numbers at all is 1—which is useful in a variety of mathematical fields including probability theory, combinatorics, and power series. The mathematician Gian-Carlo Rota, speaking before an audience of mathematicians in Baltimore in January 1998, stated that physicists in particular like to dismiss this idea as ivory-towerish (Rota's rhetorical style often used exaggeration and hyperbole), and mentioned as an example that the elementary symmetric polynomial in no variables at all is 1. He went on to say that this led to the remarkable discovery that the Euler characteristic is one of the finitely additive "measures" treated in Hadwiger's theorem, so that "pure" mathematicians who attach importance to this kind of "vacuity" have the last laugh.

Outside of mathematics, vacuously true statements are often used to mislead others without technically lying. For example, a child might tell his parents "I ate every vegetable on my plate," when there were no vegetables on the child's plate. The statement is technically true, because there were zero vegetables, and the child ate zero vegetables, which was all of them."

No fair... :p

rahuldandekar
2006-Nov-02, 12:52 PM
Naturally (bad pun), since I've voted "no", I would say it equals 0.99~, which would make 0.99~ and 1 adjacent numbers, not equal... There are no elements between the 2 sets [0,1[ and [1,2], otherwise we'd end up with double values in the set [0,2]... Just like there is no element between the two natural values 1 and 2. Doesn't mean 1 = 2...

There's no number between 0.99~ and 1 in R (R being the set of real numbers, the actual symbol is a bit different - R with a double vertical line, but I can't get in in here...). Neither is there a number between 1 and 2 in N (N being the set of natural numbers). Doesn't mean there equal, though.

:D ;)

I'm just getting interested... :boohoo:

Well, you may have known that the real numbers are dense... that is, between any two distinct real numbers, there exists another. That is why, if there is no number between 0.999~ and 1, they are equal. That is what one uses in the limit definition too... that their difference is smaller than any real number, which means there are no reals between them (otherwise there would be a finite difference between 0.999~ and 1).

LayMan
2006-Nov-02, 01:16 PM
I'm just getting interested... :boohoo:

Well, you may have known that the real numbers are dense... that is, between any two distinct real numbers, there exists another. That is why, if there is no number between 0.999~ and 1, they are equal. That is what one uses in the limit definition too... that their difference is smaller than any real number, which means there are no reals between them (otherwise there would be a finite difference between 0.999~ and 1).

I know I'm gonna get loads of criticism on this, but the way I read that, that means all real numbers are equal, since the equality goes for all pairs of real numbers... :D

And yet another contradiction: IF you say "that between any two distinct real number, there exists another" AND you say "That is why, if there is no number between 0.999~ and 1, they are equal" then either 0.99~ and 1 aren't distinct real numbers, or they are not equal... ;)

LayMan
2006-Nov-02, 01:20 PM
And yet another contradiction: IF you say "that between any two distinct real number, there exists another" AND you say "That is why, if there is no number between 0.999~ and 1, they are equal" then either 0.99~ and 1 aren't distinct real numbers, or they are not equal... ;)

OOUUCCHH!!! Just got hit by the meaning of the word distinct... :D :D :D :D

OK, I conceide... :silenced: :D

Fun while it lasted, though...

rahuldandekar
2006-Nov-02, 01:22 PM
I know I'm gonna get loads of criticism on this, but the way I read that, that means all real numbers are equal, since the equality goes for all pairs of real numbers... :D

And yet another contradiction: IF you say "that between any two distinct real number, there exists another" AND you say "That is why, if there is no number between 0.999~ and 1, they are equal" then either 0.99~ and 1 aren't distinct real numbers, or they are not equal... ;)

:P Between any two distinct real numbers, there exists another. This is such a clear statement. What is your problem? :p Are you just playing dumb? :wall:

If you have two distinct real numbers, then you will get another in between.

A logical statement. It is true. The contrapositive istrue too:

If you do not get any number between two real numbers, they are not distinct.

:wall:

rahuldandekar
2006-Nov-02, 01:24 PM
OOUUCCHH!!! Just got hit by the meaning of the word distinct... :D :D :D :D

OK, I conceide... :silenced: :D

Fun while it lasted, though...

Er... more :wall: .

:lol:

Moose
2006-Nov-02, 01:26 PM
LayMan, it would be a good time to now reiterate my challenge: please show us a number, any number, that lies between 0.999~ and 1. If you can do that, you'll have been right all along, and we'll concede.

I'll then get right to work getting you the last digit of Pi.

LayMan
2006-Nov-02, 01:35 PM
LayMan, it would be a good time to now reiterate my challenge: please show us a number, any number, that lies between 0.999~ and 1. If you can do that, you'll have been right all along, and we'll concede.

I'll then get right to work getting you the last digit of Pi.

Never mind, deal's off... I'm officially an idiot, therefor any deals made by me are void of any binding disposition.
:p

Nicolas
2006-Nov-02, 02:47 PM
I'll then get right to work getting you the last digit of Pi.
Mathematically expressed, that number would be the limit value of the BBP algorithm.

There, I've solved your problem. ;)

I can mathematically express the last digit of PI, but not the highest number in [0,1[

It's just not fair :D

Moose
2006-Nov-02, 02:49 PM
Mathematically expressed, that number would be the limit value of the BBP algorithm.

There, I've solved your problem. ;)

I was simply going to say it was 6 and leave it as an exercise to the reader to verify it. ;)

rahuldandekar
2006-Nov-02, 02:51 PM
Can you really calculate the last digit of a number with unending digits? I mean, isn't pi random? How can you get a limit value of a random sequence?

Moose
2006-Nov-02, 03:02 PM
No, Pi is quite deterministic. At best you can say the "final digit" is pseudo-random and assign probabilities. And that assumes Pi is rational. If Pi truly is irrational (and I don't know if that's been formally proven or not), then you can't even do that much. An irrational number has no final digit.

Nicolas
2006-Nov-02, 03:09 PM
I was simply going to say it was 6 and leave it as an exercise to the reader to verify it.

Now you sound like a maths book. :)

LayMan
2006-Nov-02, 03:16 PM
I was simply going to say it was 6 and leave it as an exercise to the reader to verify it. ;)

That's 9 to the people living in the southern hemisphere... :D

I mean, isn't pi random?

Depends on where you live and the time of year. With the holidays rapidly coming, my guess would be pumpkin pi... :)

No, Pi is quite deterministic. At best you can say the "final digit" is pseudo-random and assign probabilities. And that assumes Pi is rational. If Pi truly is irrational (and I don't know if that's been formally proven or not), then you can't even do that much. An irrational number has no final digit.

Would you say irrational numbers are dense?:D

Ooops, sorry, I forgot I already conceded...:silenced: ;)

hhEb09'1
2006-Nov-02, 04:27 PM
Just a question of terminology:

For the half-open interval [0,1[ the lower bound is 0. What do we say about the upper bound? Is it still called 1 for this open interval or is it called undefined?1 is the upper bound, even though it is not a member of the set. Closed intervals contain their boundary.

Moose
2006-Nov-02, 05:00 PM
That's 9 to the people living in the southern hemisphere... :D

It's P to Grapes, here. ^^^ ;)

Frog march
2006-Nov-02, 05:24 PM
If you state the value of pi in the number base pi then the last digit of pi is 1.

Moose
2006-Nov-02, 05:33 PM
That's 9 to the people living in the southern hemisphere... :D

It just occurred to me to ask: wouldn't that make the rest of the thread about whether or not 1 = ~6666.0 ?

hhEb09'1
2006-Nov-02, 05:37 PM
It just occurred to me to ask: wouldn't that make the rest of the thread about whether or not 1 = ~6666.0 ?I'm thinking 1 = ~6666'0

Moose
2006-Nov-02, 05:49 PM
Yeah, good point. :)

Nicolas
2006-Nov-02, 06:45 PM
1 is the upper bound, even though it is not a member of the set. Closed intervals contain their boundary.

Thanks. I knew the part about being (not) contained in the interval, but I wasn't sure on calling the uncontained boundary value really the bound. Apparently that is good practice. I will be able to sleep tonight.

agingjb
2006-Nov-02, 08:19 PM
Well I'm obviously in a minority of one (as usual), but I still regard the "No it is not equal" assertions as Against the Mainstream. OK, it's mathematics, not astronomy or physics, but I'd say it's a close enough discipline.

Obviously the "No it is not equal" group are not (as some might uncharitably claim) motivated by ignorance or mischief. Even to propose that possibility is an ad hominem which would merit swift banning.

No, these folks (2 out of every 5 of a good sample of Baut contributors, some of the more enlightened people in a dark age) assert with considerable force, and after considerable reflection, that a substantial part of accepted mathematics is false or meaningless. That's interesting.

Yes, I know that I (and others) actually find the conclusions of the tradition even more interesting, and it is precisely because of that interest that I would welcome the sort of point by point attack and defense (with the usual sanctions) that people encounter if they propose some "Against the Mainstream" theory in Astronomy.

I said "minority of one"; perhaps I ought to take the simple step that would make it "minority of zero".

Nicolas
2006-Nov-02, 08:20 PM
Well I'm obviously in a minority of one

I say you're in a minority of 0.999~

agingjb
2006-Nov-02, 08:33 PM
I say you're in a minority of 0.999~

Well at least it's an integer.

Nicolas
2006-Nov-02, 08:33 PM
I don't know how decimal you feel :)

Moose
2006-Nov-02, 08:39 PM
Keep it real, folks, keep it real. No floaters.

jlhredshift
2006-Nov-02, 09:35 PM
Keep it real, folks, keep it real. No floaters.

Well, not exactly, I mean e^(pi)i = 1, and that's complex!

Moose
2006-Nov-02, 09:48 PM
Man, all of these puns are making me numb(er).

I hope I don't get obtuse and derive myself down the stairs a bunch of times, like a log, rythmically, acutely bisecting a sin-bone so that I'm in angly, and forcing them to add me to the list for a lengthy term in (sub)traction. Cos that'd be an unfortunate tangent to integrate in my life.

(Did I get all of them?)

hhEb09'1
2006-Nov-02, 09:57 PM
Well, not exactly, I mean e^(pi)i = 1, and that's complex!Not to be negative, but that should be e^(pi)i = -1

jlhredshift
2006-Nov-02, 10:15 PM
Not to be negative, but that should be e^(pi)i = -1

That's correct, but if I become trancendental, e^(pi)i +1 = 0, it means nothing! I am trying to not be irrational.

pghnative
2006-Nov-02, 10:18 PM
Punny posts are a sine that you are irrational.

jlhredshift
2006-Nov-02, 10:21 PM
Punny posts are a sine that you are irrational.

But I dropped a sign, you see!

jlhredshift
2006-Nov-02, 10:25 PM
Besides, Feynman would drop signs all the time and have the grads go back and fix it up. At least there is one thing that I can do like Feynman.

rahuldandekar
2006-Nov-03, 01:20 AM
Well I'm obviously in a minority of one (as usual), but I still regard the "No it is not equal" assertions as Against the Mainstream. OK, it's mathematics, not astronomy or physics, but I'd say it's a close enough discipline.

Obviously the "No it is not equal" group are not (as some might uncharitably claim) motivated by ignorance or mischief. Even to propose that possibility is an ad hominem which would merit swift banning.

No, these folks (2 out of every 5 of a good sample of Baut contributors, some of the more enlightened people in a dark age) assert with considerable force, and after considerable reflection, that a substantial part of accepted mathematics is false or meaningless. That's interesting.

Yes, I know that I (and others) actually find the conclusions of the tradition even more interesting, and it is precisely because of that interest that I would welcome the sort of point by point attack and defense (with the usual sanctions) that people encounter if they propose some "Against the Mainstream" theory in Astronomy.

I said "minority of one"; perhaps I ought to take the simple step that would make it "minority of zero".

People reject limit theory?

Nicolas
2006-Nov-03, 08:57 AM
In the end...

LayMan
2006-Nov-03, 09:00 AM
Well I'm obviously in a minority of one (as usual), but I still regard the "No it is not equal" assertions as Against the Mainstream. OK, it's mathematics, not astronomy or physics, but I'd say it's a close enough discipline.

There's something to say for that, of course, but I see it more as a discussion about the bounderies of mathematics versus meta-mathematics (or more accurately, the difference between science itself versus 'the science of science'): it's mostly about definition and semantics: if you define 1 divided by infinity as equal to 0, then by definition you (have to) accept that 1/3 = 0.33~, and that 1 = 0.99~. If you don't accept that a-priori definition, you would say that 1/3 ≈ 0.33~ and that 1 ≈ 0.99~. :D

I guess it's only 'fair' to let the mathematicians handle the definitions of 'their' maths, since that is what makes it work... It's just that, as you get to the very edge of your discipline, you start knocking on other peoples doors, and you just never know who will answer it...

Might just be a cow-thief threatening your herd... :lol:

Obviously the "No it is not equal" group are not (as some might uncharitably claim) motivated by ignorance or mischief. Even to propose that possibility is an ad hominem which would merit swift banning.

Thanks, very much appreciated!

No, these folks (2 out of every 5 of a good sample of Baut contributors, some of the more enlightened people in a dark age) assert with considerable force, and after considerable reflection, that a substantial part of accepted mathematics is false or meaningless. That's interesting.

Again, it's really about the definition of 1/∞, I wouldn't call that a substantial part of anything... And a difference in opinion on 1 (very, very, very, ~ small) aspect of a well tested system doesn't necessarily imply falseness or meaninglessness(nesslessness or something) on the whole (or most parts).

Yes, I know that I (and others) actually find the conclusions of the tradition even more interesting, and it is precisely because of that interest that I would welcome the sort of point by point attack and defense (with the usual sanctions) that people encounter if they propose some "Against the Mainstream" theory in Astronomy.

Astronomy relies more on observable facts, maths more on reasonable facts (after all, no one has seen or can construct a perfect circle - except Plato perhaps, but he was drowning in numbers when Atlantis went down)... :D

I said "minority of one"; perhaps I ought to take the simple step that would make it "minority of zero".

The majority of zero would object to that...

Anyway, we had a nice chuckle, a cheap bundle of laughs and a lot of smileys. No harm done... :razz:

worzel
2006-Nov-03, 09:49 AM
Again, it's really about the definition of 1/∞, I wouldn't call that a substantial part of anything... And a difference in opinion on 1 (very, very, very, ~ small) aspect of a well tested system doesn't necessarily imply falseness or meaninglessness(nesslessness or something) on the whole (or most parts).
All of calculus relies on 1/x exactly equalling 0 in the limit as x->infinity.

rahuldandekar
2006-Nov-03, 11:11 AM
All of calculus relies on 1/x exactly equalling 0 in the limit as x->infinity.

Yep. One thing I'd like to point out, though, is that there is no "number" called infinity. I've said that before, and let me say it now... "infinity" may be a concept, but it isn't defined in calculus. (In set theory, the cardinality of the set of natural numbers is Aleph nought, so that's an infinity, but it isn't"used anywhere".)

Talking about calculus, x --> infinity is just notation for "as x becomes very large". "As x --> infinity, 1/x --> 0" means that as you keep on increasing the value of x, 1/x approaches zero. That much is obvious, and to reject it would be, I think, a rejection of the archimedean principle (or whatever principle that is which states that there always is a natural number greater than any real you take.)

So how can anyone say 1/infinity is not 0 ?? Except to argue that the statement doesn't imply 1/x --> 0 as x --> infinity.

Nicolas
2006-Nov-03, 12:03 PM
Yep. One thing I'd like to point out, though, is that there is no "number" called infinity. I've said that before, and let me say it now... "infinity" may be a concept, but it isn't defined in calculus.

illustrated by infinity - 1 = infinity - 5465798465873514984165574567485798 = infinity.

Disinfo Agent
2006-Nov-03, 12:41 PM
If you do it infinitely, you never reach the number.Yes, you do.

No, according to the same logic that would make 1/3 = 0.33~ , 0.1n would actually reach 0 as n 'goes/converges/disappears' to infinity.And it does.

And yes, I'm aware of Cantor and his exploits, I just choose to disagree with him...:razz:O.K., so you've admitted that you have an against-the-mainstream stance on this. It's been nice talking to you.

Disinfo Agent
2006-Nov-03, 12:44 PM
Just a question of terminology:

For the half-open interval [0,1[ the lower bound is 0. What do we say about the upper bound? Is it still called 1 for this open interval or is it called undefined? The highest value is undefined (we can only say that it is smaller than 1, but we can't specify it), but I wonder what terminology to use for the bound itself.Supremum. (Or "least upper bound".)
It's easily shown that the limit of an increasing sequence of real numbers is its supremum.

Nicolas
2006-Nov-03, 12:46 PM
What's the term for an inclusive upper bound then?

Disinfo Agent
2006-Nov-03, 12:49 PM
"Maximum." But not all sets of real numbers have a maximum.

Disinfo Agent
2006-Nov-03, 12:49 PM
No, Pi is quite deterministic. At best you can say the "final digit" is pseudo-random and assign probabilities. And that assumes Pi is rational. If Pi truly is irrational (and I don't know if that's been formally proven or not), then you can't even do that much. An irrational number has no final digit.Absolutely, some 200 years ago, if memory serves me.

Disinfo Agent
2006-Nov-03, 12:54 PM
Yep. One thing I'd like to point out, though, is that there is no "number" called infinity. I've said that before, and let me say it now... "infinity" may be a concept, but it isn't defined in calculus.Ya wanna see me define it? :p

Moose
2006-Nov-03, 01:14 PM
Absolutely, some 200 years ago, if memory serves me.

There we go, then. No final digit.

I'd be curious to see how that proof goes. Does the proof have a special name I might google?

LayMan
2006-Nov-03, 01:16 PM
All of calculus relies on 1/x exactly equalling 0 in the limit as x->infinity.

I know I already conceded in this thread, so I hope I'm not violating any rules...

Worzel, this is exactly the thing that always bothered me in calculus as it was explained to me in high school: the whole principle of calculus is using rectangles to determine the surface of a curved plane-segment (well, at least that's how I understand it)... So being the curious, smartie-pants that I was (and not being much of a fan of maths, guilty as charged there), I deviously (and quite cleverly, I thought) asked my math teacher, "What's that funny dx doing in the equation?". Remember, this was high shool, I was about 15 years and studied the 'old school' (Latin-Greec, where the main portion of classes was languages, we only had about 2 hours maths a week), and from what I had learned from math, I got used to the idea that whenever someone presented me a problem such as '4x=y', I was supposed to enter a value for x in order to determine the value of y... So when confronted with this 'bizar' looking "...times 'd times x' (the .dx portion)", I thought I was supposed to fill in the d value too... My math teacher at the time (trust me, I don't envy him trying to teach math to a couple of youngsters who -in their minds - had better and more fun things to do then math :D) told me that the 'd' wasn't a specific value, just a very, very, very,... small number that tends to 0 as x goes to infinity. So I asked him - again, quite cleverly from where I was sitting - "But as d becomes 0, and it is part of a multiplication, doesn't the entire equation become 0? After all, multiply any number or value by 0, and you get 0, no?"...

Unfortunately, his only reply was that, given only 2 class-hours a week, he didn't have the time to go into it.

Ever since then, I've always been, well, a bit suspicious about calculus, since I was convinced he didn't know it himself :D .

But anyway, that is the point I was trying to make: in calculus, you want d to become infinitely small, since that means the (incalculated) error for the value of the surface would also become infinitely small (with 'incalculated', I mean that math 'knows' it's making an error in determining a curved surface by using smaller, non-curved surfaces, but that's the point of calculus: use enough (near-infinite) small surfaces, and the error becomes equally very, very small).

However, I still feel that calculus can't allow the d value to 'really' become exactly 0, since that would mean that dx becomes 0.

When reversing the logic behind that (yes, I know, I'll regret putting it like that :p ), the difference between 1 and 0.99~ gets smaller and smaller, without becoming 0.

That's the whole problem, I think. And the reason this question will keep popping up on discussion boards all over the place. And why people still vote differently in polls regarding the subject.

EDIT [Because] it will always sound like math is 'cheating':

- in calculus, math says the value (for d, stated as 'as small as possible', which I always interpret as being 1/∞) will go to 0, without ever reaching it (otherwise they have a 0 in their multiplications)...

- wrt the difference between 1 and 0.99~, it states that 1/∞ is in fact exactly 0.

Disinfo Agent, just noticed you're reply wrt my ATM stance as I was previewing this rather lenghty post: you may have noticed the smiley there. This isn't about me telling math what to do or how to do it, it's about me not getting something... And there seem to be others besides me, strungling with this too.

Moose
2006-Nov-03, 01:31 PM
Okay, you've got some pretty serious misconceptions there, I think. I'm not sure how to explain it to you either, because my calculus is pretty shaky after having not done any for about fifteen years.

The basic idea of differential calculus is to be able to deal with problems where the functions themselves are variable. What's the slope of a curved line, for example. There are some applications of this in modelling stock markets, for example. As well as in engineering. The d you're talking about refers to a derivitive function. Not a limit, and certainly not something you can attempt to multiply numerically.

Now integral calculus, on the other hand, is where you take infinitely thin rectangles to compute, say, the surface area of an irregular object. It has some applications in statistics as well. Integration involves computing derivatives (because the width of the irregular object is itself a variable function.)

LayMan
2006-Nov-03, 01:46 PM
I didn't know there were 2 different forms of calculus, that could explain a few things. I believed that the d value was in fact a limit. (don't worry, I'll blame my teacher for the misconceptions - shame on me, I know) :o ...

Disinfo Agent
2006-Nov-03, 02:29 PM
There we go, then. No final digit.

I'd be curious to see how that proof goes. Does the proof have a special name I might google?Googling for "irrationality of pi" turns up several proofs.

Moose
2006-Nov-03, 02:43 PM
Cool. I'll curl up with that tonight after supper.

Disinfo Agent
2006-Nov-03, 02:51 PM
I know I already conceded in this thread, so I hope I'm not violating any rules...

Worzel, this is exactly the thing that always bothered me in calculus as it was explained to me in high school: the whole principle of calculus is using rectangles to determine the surface of a curved plane-segment (well, at least that's how I understand it)... So being the curious, smartie-pants that I was (and not being much of a fan of maths, guilty as charged there), I deviously (and quite cleverly, I thought) asked my math teacher, "What's that funny dx doing in the equation?". Remember, this was high shool, I was about 15 years and studied the 'old school' (Latin-Greec, where the main portion of classes was languages, we only had about 2 hours maths a week), and from what I had learned from math, I got used to the idea that whenever someone presented me a problem such as '4x=y', I was supposed to enter a value for x in order to determine the value of y... So when confronted with this 'bizar' looking "...times 'd times x' (the .dx portion)", I thought I was supposed to fill in the d value too... My math teacher at the time (trust me, I don't envy him trying to teach math to a couple of youngsters who -in their minds - had better and more fun things to do then math :D) told me that the 'd' wasn't a specific value, just a very, very, very,... small number that tends to 0 as x goes to infinity. So I asked him - again, quite cleverly from where I was sitting - "But as d becomes 0, and it is part of a multiplication, doesn't the entire equation become 0? After all, multiply any number or value by 0, and you get 0, no?"...

Unfortunately, his only reply was that, given only 2 class-hours a week, he didn't have the time to go into it.

Ever since then, I've always been, well, a bit suspicious about calculus, since I was convinced he didn't know it himself :D .

But anyway, that is the point I was trying to make: in calculus, you want d to become infinitely small, since that means the (incalculated) error for the value of the surface would also become infinitely small (with 'incalculated', I mean that math 'knows' it's making an error in determining a curved surface by using smaller, non-curved surfaces, but that's the point of calculus: use enough (near-infinite) small surfaces, and the error becomes equally very, very small).

However, I still feel that calculus can't allow the d value to 'really' become exactly 0, since that would mean that dx becomes 0.
For the record, I am not the world's biggest fan of the "dx" approach to calculus, either. It puts too many hocus-pocus mystical ideas into the students' heads.

That said, I have to admit that it has a certain alluring intuitiveness and elegance, and it is the formalisation preferred by physicists, for example.

But there's one thing you did not realise. The "d" is not supposed to be a number. It's a function. And it's not "d" that is supposed to get smaller and smaller. It's the whole "dx".

When reversing the logic behind that (yes, I know, I'll regret putting it like that :p ), the difference between 1 and 0.99~ gets smaller and smaller, without becoming 0.The difference between those two things is not supposed to change. Each of them is a constant. We've been through this in this thread already.

Fazor
2006-Nov-03, 02:57 PM
Wow I can't believe this thread has gone on to nearly 50 pages :-p

The theory that .999~ == 1 is just that no matter how many sig figs you want, when you find your stopping point and round up on the last digit, that will cause the next digit and the next and the next etc. etc. to round up till you get to the end (or beginning?) and it rounds up to 1. Key there is it *rounds up to one*, but isn't 1. but it's close enough that it should work for whatever your calculations are. unless ur doing a calculation that can use .999~, in which case you don't round. Altho not being a math...smart....guy (woo simpsons) I don't know when or if you could use .999~ in a calculation without rounding... hmm.

LayMan
2006-Nov-03, 03:00 PM
For the record, I am not the world's biggest fan of the "dx" approach to calculus, either. It puts too many hocus-pocus mystical ideas into the students' heads.

That said, I have to admit that it has a certain alluring intuitiveness and elegance, and it is the formalisation preferred by physicists, for example.

But there's one thing you did not realise. The "d" is not supposed to be a number. It's a function. And it's not "d" that is supposed to get smaller and smaller. It's the whole "dx".

Yep, there's my misconception. Never looked at it that way... I guess that, whenever I see a letter in a mathematical formula, I instictively try to replace it with a number (you should see me at work with the hexadecimal system ;) )...

We've been through this in this thread already.

Sorry 'bout that...

agingjb
2006-Nov-03, 03:02 PM
I recommend "Calculus" by Michael Spivak, Cambridge University Press, (third edition recently published), ISBN: 0521867444.

Moose
2006-Nov-03, 03:06 PM
Wow I can't believe this thread has gone on to nearly 50 pages :-p

I can.

The theory that .999~ == 1 is just that no matter how many sig figs you want, when you find your stopping point and round up on the last digit, that will cause the next digit and the next and the next etc. etc. to round up till you get to the end (or beginning?) and it rounds up to 1. Key there is it *rounds up to one*, but isn't 1. but it's close enough that it should work for whatever your calculations are. unless ur doing a calculation that can use .999~, in which case you don't round. Altho not being a math...smart....guy (woo simpsons) I don't know when or if you could use .999~ in a calculation without rounding... hmm.

Sorry, that's simply incorrect. 0.999~ is exactly equal to 1. There's no rounding involved whatsoever. Please read the first seven or so pages of this thread for the relevant explanations and proofs. If, after that, you still have questions, we'll be happier to help.

Disinfo Agent
2006-Nov-03, 03:36 PM
Yep, there's my misconception. Never looked at it that way... I guess that, whenever I see a letter in a mathematical formula, I instictively try to replace it with a number (you should see me at work with the hexadecimal system ;) )...You can think of dx exactly as a Delta-x from physics (don't know how to write the damn thing). It stands for an increase in the variable x, x1 - x0 (or more usually just written x - x0).
Now, the next idea, in calculus, is typically that you find a way of making the x's get closer and closer to each other, and try to figure out what pattern that process leads to.

Going back to 0.999...9, we can write these approximations to 0.999...~ (infinite nines) as:

x1 = 0.9
x2 = 0.99
x3 = 0.999
...
and in general
xn = 0.9 + 0.09 + ... + 9*0.1n = 9 * (1 - 0.1n)/(1 - 0.1) = 1 - 0.1n

Suppose you took two consecutive terms in this series, xn and xn+1, as your "initial" and your "final" values of x. Then you'd get dx = (1 - 0.1n+1) - (1 - 0.1n) = 0.1n - 0.1n+1 = 0.1n * (1 - 0.1) = 9 * 0.1n+1. You can make this as small as you wish, by choosing a sufficiently large value of n.

With me so far?

LayMan
2006-Nov-03, 03:57 PM
You can think of dx exactly as a Delta-x from physics (don't know how to write the damn thing). It stands for an increase in the variable x, x1 - x0 (or more usually just written x - x0).
Now, the next idea, in calculus, is typically that you find a way of making the x's get closer and closer to each other, and try to figure out what pattern that process leads to.

Going back to 0.999...9, we can write these approximations to 0.999...~ (infinite nines) as:

x1 = 0.9
x2 = 0.99
x3 = 0.999
...
and in general
xn = 0.9 + 0.09 + ... + 9*0.1n = 9 * (1 - 0.1n)/(1 - 0.1) = 1 - 0.1n

Suppose you took two consecutive terms in this series, xn and xn+1, as your "initial" and your "final" values of x. Then you'd get dx = (1 - 0.1n+1) - (1 - 0.1n) = 0.1n - 0.1n+1 = 0.1n * (1 - 0.1) = 9 * 0.1n+1. You can make this as small as you wish, by choosing a sufficiently large value of n.

With me so far?

Yes, even more then you think... Actually, it was (finally) this one that completely sold me a couple of minutes ago (in a feable attempt to actually read through the entire thread - I failed again...):

Here's a proof that shows it must equal 1. It has probably been shown before in this thread, possibly even by me, but I don't want to search through, and it is a good way of explaining it.

Lets define x as 0.999~
10x is therefore equal to 9.999~ just by shifting the decimal place.
If we subtract x (0.999~) from 10x (9.999~), we get 9. This eliminates the endless decimal. If 9x = 9, then x must necessarily equal 1. Therefore, 0.999~ can be mathematically proven to equal 1.

This probably won't end the discussion, but at least it's another way to look at it...

I didn't quite 'get that', since when I read this, another (previously) hidden misconception was clouding my judgement: it's been apparently so well engraved in the depths of my subconsious, it didn't occur to me before: I'm sooo used of thinking about "multiplying by ten means just add a 0 to the right", that I failed to see that this only applies to natural numbers, not real ones.

15 * 10 = 150, but 0.15 * 10 ≠ 0.150

It especially doesn't go for 0.99~, since there is no right to it (and even if there was, it'd be filled with 9's)...

So the proof stands. Sorry for having voted no, it should have been yes (don't wanna come off shifty here, but hey, when you're right, you're right...

Disinfo Agent
2006-Nov-03, 04:17 PM
Wow, you fell for it! :D ;)

I think this line in particular:

xn = 0.9 + 0.09 + ... + 9*0.1n = 9 * (1 - 0.1n)/(1 - 0.1) = 1 - 0.1nshows very explicitly how one "flips" from adding more and more nines to something which is indefinitely close to 1. (Though it should actually be 0.9 instead of 9. I made a mistake. :o)

LayMan
2006-Nov-03, 04:23 PM
Wow, you fell for it! :D ;)

True to your name, hey...? ;)

Anyway, I'm off for a weeklong holiday, catch you all later (but not in this thread, I hope) :D ...

spacemanspiff
2006-Nov-03, 07:10 PM
Spaceman Spiff
*Having scanned all the posts between his last post and this one, blinks rapidly to get the glazed look out of his eyes, wipes the drool off of his chin then picks up his copy of 'DOC' Smiths' Skylark of Space and shuffles off to the corner to read.*

Damien Evans
2006-Nov-05, 04:45 AM
Also:

1/3 = 0.3333....
0.3333.... * 3 = 0.9999....
Therefore, 0.9999.... = 1

i should have thought of that, i don't know why i voted no

Mister Earl
2006-Nov-05, 01:37 PM
0.999R < 1.
Think of it like cutting a piece of string in half over and over and over again. Your knife my approach the very end of the string, but you'll never hit it.

Thomas(believer)
2006-Nov-05, 02:10 PM
0.999R < 1.
Think of it like cutting a piece of string in half over and over and over again. Your knife my approach the very end of the string, but you'll never hit it.
Unless you can do the cutting in infinitesimal small time intervals. :think:

Chuck
2006-Nov-05, 02:24 PM
When I got up this morning I had to move 9/10 of the distance to my computer, then 9/10 of the remaining distance for a total of 99/100, then 9/10 of what remained for a total of 999/1000, etc, etc. If 0.999... were less than one I'd never have made it all the way, but here I am.

Fortis
2006-Nov-05, 02:33 PM
0.999R < 1.
Think of it like cutting a piece of string in half over and over and over again. Your knife my approach the very end of the string, but you'll never hit it.
But you can define a point arbitrarily close to the end of the string and you will always pass it in a finite time.

Moose
2006-Nov-05, 03:59 PM
0.999R < 1.
Think of it like cutting a piece of string in half over and over and over again. Your knife my approach the very end of the string, but you'll never hit it.

Middle-school physics says that if you travel a certain velocity for a certain amount of time, you'll travel a certain distance. d = vt.

Follow me so far?

Now, I want you to walk half the length of your driveway at a steady pace. It'll take you so many seconds. Call it .5t. Now walk the next quarter in half the time you took to walk half the driveway (so that you maintain the same pace throughout this experiment.) Maintain the same pace over the next eighth of the driveway, taking an eighth of the time. Etc. Keep repeating that.

Now, according to your proposition, you'll never reach the street, because you'll never pass the point where you'll pass 0.999~d and 0.999~t.

Reality, however, says you'd better be looking both ways, because you're going to get hit by traffic at time 1t. Which is to say 0.999~ driveways.

I've said it before a page or two back. Zeno's "paradox" isn't a paradox at all. Please read the first seven or so pages of this thread to see all the proofs you'll ever need as to why 0.999~ = 1, both mathematically and reinforced by simple reality.

Mister Earl
2006-Nov-05, 05:43 PM
Ah, but you're following a different set of rules that I am. Imagine that I am walking down the driveway, as you mentioned it. Each time I get 90% of the way through the remainder of the driveway, I stop. Another 90%, stop. And so on. I'll never hit the end of the driveway. I'll be close enough to be indistinguishable to the human eye, but on a mathmatical level, I'll never complete the journey.

cjl
2006-Nov-05, 05:49 PM
But why would you have to stop at each point?

Whether you stop or not, you pass the point. The stopping only adds time, it does not change the problem (except that because of the pauses, you can never make it to the end, because there are an infinite number of points to stop at).

Mister Earl
2006-Nov-05, 05:51 PM
If you walk at a constant rate, then you will reach the end of the driveway. You will reach 1, not 0.9R. The end of the driveway is the end of the driveway, not an infinitely small point before it.

cjl
2006-Nov-05, 05:57 PM
But you will also reach 0.9R

That's the thing. 0.9R is not an infinitely small point before 1, it IS 1. If you move at a constant rate, you will pass each point that is 9/10 closer to the end of the driveway, passing 0.9R only when you pass the end of the driveway (1).

Mister Earl
2006-Nov-05, 06:00 PM
But you never end calculating 0.9R, (taking those 9/10th steps) because there's always another 9/10th you can shave off. It never ends. Thus, you never achieve 1.

You can never achieve 100% if you only take 90% of each remainder.

cjl
2006-Nov-05, 06:05 PM
But it will end. Although there are an infinite number of such steps you can take, each one is shorter than the last. Each therefore takes a shorter time than the last, and can be reached in a finite amount of time. If you look at it as an infinite series, this can be demonstrated.

It can be represented as the sum of 9/10^x for x = 1 to infinity. This series, though it has an infinite number of terms, sums to 1. So an infinite number of terms can easily be fit into a finite space. Basically, even though another 9/10 can always be passed, you will always reach the end.

Mister Earl
2006-Nov-05, 06:22 PM
Travelling to 1 in a finite amount of time is not the same as calculating 0.9R.
No matter how many decimal places you calculate to, there's always another 9 you can tack on, and as long as you're tacking on another 9, you never achieve 1. It's just 9's forever.

cjl
2006-Nov-05, 06:26 PM
Not true. Each 9 tacked on takes less time to travel. Although you pass another 9 and another, etc, each takes less time than the last, and you will always pass them all and reach 0.9R, which is 1.

Mister Earl
2006-Nov-05, 06:28 PM
You're still talking about someone travelling between 0-1. This isn't the same at all. I'm talking about calculating 0.9R. No matter how many times you divide the remainder by 90%, you never take the entirety of that remainder. Hence, you never achieve 1.

cjl
2006-Nov-05, 06:33 PM
But you will reach the entirety of that remainder.

As you pointed out, there is always another 9 to add between where you are and 1. Hence, as you reach 1, you have passed an infinite number of these, but only when you reach one. Before you reach it, more can be added. So, 1 is where you have reached 0.9R.

Mister Earl
2006-Nov-05, 06:34 PM
Come to think of it, if you dictate how far you move towards 1 by only going 90% of the remaining distance, you still won't achieve 1. Even if it takes less and less time to cover that remaining distance. You'll be infantesimally close, but you won't hit 1.

Mister Earl
2006-Nov-05, 06:35 PM
But you never reach 0.9R because you're always taking another 90% of the remainder. There's never an end to it.

cjl
2006-Nov-05, 06:40 PM
Come to think of it, if you dictate how far you move towards 1 by only going 90% of the remaining distance, you still won't achieve 1. Even if it takes less and less time to cover that remaining distance. You'll be infantesimally close, but you won't hit 1.

Not true. If you take less and less time for each distance, you will reach it in some cases. If you dictate that each time you move, you will take 1/n^2 time, with n increasing by one each time, you will reach it. Even with an infinite number of intervals, you will reach it in a finite time. In doing so, you will reach 0.9R at the EXACT same time that you reach 1. This is because it is one.

Mister Earl
2006-Nov-05, 06:45 PM
I respectfully disagree :D

Even with an infinite number of intervals, you will reach it in a finite time. In doing so, you will reach 0.9R at the EXACT same time that you reach 1. This is because it is one.

The problem I see with this is with an infinite number of intervals, you will never reach 0.9R. And since you cannot reach 0.9R, you'll never get past it to one.

cjl
2006-Nov-05, 06:51 PM
The definition of 0.9R is that it is an infinite number of intervals of 9/10^n summed together. So to reach it, you must travel an infinite number of such intervals, and if you do, by definition, you are at 0.9R. As long as the time between such intervals is decreasing, it can be mathematically shown to be possible to reach in a finite amount of time (as with the 1/n^2 example above). If it takes 1 second to reach the first one, 1/2 second to reach the next one, 1/4 to reach the next, etc., you will reach infinity at 2 seconds. Also, you do not need to get past 0.9R to reach one - if you reach 0.9R, you are at one.

Mister Earl
2006-Nov-05, 06:54 PM
If it takes 1 second to reach the first one, 1/2 second to reach the next one, 1/4 to reach the next, etc., you will reach infinity at 2 seconds.

What happens at 2 seconds is a failure to keep calculating 0.9R, not achieving infinity.

Also, you do not need to get past 0.9R to reach one

Why not? I'm still not convinced they are equal.

What you are trying to do is negate infinity.

cjl
2006-Nov-05, 06:56 PM
Not quite.

How is it possible to reach 2 seconds, using the formula I gave, to not reach infinity? Not only is infinity reachable, it is unavoidable in this situation.

agingjb
2006-Nov-05, 06:58 PM
Or you can avoid "infinity" by speaking of limits; the limit of the series denoted by ".9 recurring" is 1, and the reals are complete, so that the notation clearly denotes 1.

However it is obvious that rather a large proportion of presumably mathematically educated people (they post here, don't they) do not accept, and will never accept, this conclusion.

Is there an analogue of Phil Plait called the "Bad Mathematician"?

Mister Earl
2006-Nov-05, 07:03 PM
CJL, let's conduct your experiment mentally. In order to physically conduct your test, the person or machine conducting this experiment would have to calculate 90% of the remainder, and move there, at ever-increasing speeds. Very quickly the calculations would have to occur so quickly to achieve your two seconds that it becomes impossible. You would either halt or take that last millionth step over the line, breaking the calculation and invalidating the experiment. There is no way to calculate this formulae at infinite speeds. Infinity is preserved no matter what happens, because this formula cannot be calculated at infinite speeds. And as you can never breach infinity, you cannot achieve 1, as long as you adhere to the formula and the rules we've implied.

Mister Earl
2006-Nov-05, 07:20 PM
We need to get Stephen Hawking involved in this :D

hhEb09'1
2006-Nov-05, 08:08 PM
CJL, let's conduct your experiment mentally. In order to physically conduct your test, the person or machine conducting this experiment would have to calculate 90% of the remainder, and move there, at ever-increasing speeds. Very quickly the calculations would have to occur so quickly to achieve your two seconds that it becomes impossible. You would either halt or take that last millionth step over the line, breaking the calculation and invalidating the experiment. There is no way to calculate this formulae at infinite speeds. Infinity is preserved no matter what happens, because this formula cannot be calculated at infinite speeds. And as you can never breach infinity, you cannot achieve 1, as long as you adhere to the formula and the rules we've implied.There are such things as analog computers :)

Or there used to be. In this case, our test body is the memory storage and data readout. At any point, the computation is the distance it has traveled.

If it is moving at a constant speed from left to right, say 1 meter per second, then after .9 seconds it will have traveled .9 meters. After .99 seconds, .99 meters. It will cover the infinite number of such intervals, doing all the calculations, in one second. No problem.

Mister Earl
2006-Nov-05, 08:54 PM
Show me a computer that can do an infinite number of calculations in a finite amount of time.

Fortis
2006-Nov-05, 08:57 PM
If people want to imagine that they are writing down the number 0.9~ 1 digit at a time, say one every second, then can they find a constant number, less than one, that cannot be reached in a finite time?

Chuck
2006-Nov-05, 09:08 PM
You don't sum an infinite series by adding one term at a time. You add all of the terms at the same time. If you do that then you won't have another 9 to add because you're already done.

hhEb09'1
2006-Nov-05, 10:08 PM
Show me a computer that can do an infinite number of calculations in a finite amount of time.That's what I just described.

Donnie B.
2006-Nov-05, 10:32 PM
Mister Earl, you seem to have the same misconception as many other "doubters": namely, that the notation 0.99~ (or 0.99... or 0.99R) represents a process. It doesn't. It represents a number, one with an infinite number of 9s to the right of the decimal.

All your arguments about continually adding ever-smaller fractions are moot. Those nines are already there, represented by the '~' or '...' or 'R'.

If you're so certain that 0.99~ is unequal to 1, you have to do only one thing to convince us: specify a number n such that 0.99~ < n < 1. If you can do this, you will have revolutionized mathematics (or at least redefined some fundamental elements of mathematics).

If you cannot, then by the definition of the real numbers, 0.99~ must be equal to 1. (That's because if two reals are different, there's an infinite number of reals between them. If the latter is not true, then the original numbers are equal.)

Fortis
2006-Nov-05, 11:37 PM
Show me a computer that can do an infinite number of calculations in a finite amount of time.
What is the integral of x^2 between 0 and 1?

Moose
2006-Nov-05, 11:48 PM
If you can do this, you will have revolutionized mathematics (or at least redefined some fundamental elements of mathematics).

Are you kidding? Doing that would blow a hole clean through limit theory, and all calculus that depends on it. Bye-bye, engineering. So long, stock market modelling. It's been nice knowing you, advanced computing.