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Damburger
2004-Dec-06, 09:19 PM
Well, I don't believe it equals one. I'd vote for undefined, with that zero in the denominator.

Your belief doesn't alter reality. If you can't disprove any of the proofs offered in this thread that 0.999r=1, you should accept it as a fact even if you can't intuitivley understand why it is that way.

Donnie B.
2004-Dec-06, 09:45 PM
Well, I don't believe it equals one. I'd vote for undefined, with that zero in the denominator.

Your belief doesn't alter reality. If you can't disprove any of the proofs offered in this thread that 0.999r=1, you should accept it as a fact even if you can't intuitivley understand why it is that way.
I was referring to the value 1/(1-0.9...), not to 0.9... itself. The former expression is exactly equal to 1/0, hence undefined. Perhaps I misunderstood the reference in the post I was responding to.

And I couldn't agree more, my beliefs don't alter reality. Except for me, at times... :wink: 8)

Damburger
2004-Dec-06, 09:51 PM
I was referring to the value 1/(1-0.9...), not to 0.9... itself. The former expression is exactly equal to 1/0, hence undefined. Perhaps I misunderstood the reference in the post I was responding to.

And I couldn't agree more, my beliefs don't alter reality. Except for me, at times... :wink: 8)

Oops, I missed that bit.

Donnie B.
2004-Dec-06, 09:51 PM
Well, I don't believe it equals one. I'd vote for undefined, with that zero in the denominator.
Picky, picky, picky... :)
Normandy6644's "it" was "0.999~".
Ah, so I see, after Damburger's remarks caused me to take a second look. That's the danger of pronouns... :wink:

While we're being picky, why do people keep typing a bazillion nines (0.9999999999...)? Seems to me one (0.9...) is necessary and sufficient.

Disinfo Agent
2004-Dec-06, 09:53 PM
It's more expressive (though redundant). :)

Damburger
2004-Dec-06, 09:58 PM
It's more expressive (though redundant). :)

This thread is now pretty much redundant. Every way of proving it I can think of has been tried and people are still not convinced.

This does not bode well for me :(

Disinfo Agent
2004-Dec-06, 10:06 PM
This thread is now pretty much redundant. Every way of proving it I can think of has been tried and people are still not convinced.
Only a few people. And I've noticed that the latest replies to the poll have been on the yea side.

Donnie B.
2004-Dec-06, 10:15 PM
Personally, I think the simplest and most convincing proof (especially for someone who doesn't have the benefit of a college-level math education) was that 1/3 and 0.9.../3 yield the same result. It's just that some people don't accept that 1/3 = 0.3... , and that's just... perplexing. #-o

[Edited to add...] It seems to me that a more consistent position would be to argue that 0.9.../3 is not equal to 0.3..., since (by the same logic being applied to 0.9... itself) you never get to the end of the nines you're dividing into. But who am I to argue with common sense? :-?

Chuck
2004-Dec-07, 02:39 AM
I choose to interpret FWIS as "Forget What I Said". It makes debating easier.

Damburger
2004-Dec-07, 09:08 AM
I choose to interpret FWIS as "Forget What I Said". It makes debating easier.

Yeah, its hardly the best debate forum going is it?

Candy
2004-Dec-07, 10:26 AM
I choose to interpret FWIS as "Forget What I Said". It makes debating easier.

Yeah, its hardly the best debate forum going is it? Hey, Damburger. http://home.att.net/~candy.stair/wavey.gif

You caused quite a stir on FWIS. =D>

Damburger
2004-Dec-07, 10:32 AM
I choose to interpret FWIS as "Forget What I Said". It makes debating easier.

Yeah, its hardly the best debate forum going is it? Hey, Damburger. http://home.att.net/~candy.stair/wavey.gif

You caused quite a stir on FWIS. =D>

I just said what I thought, they couldn't handle it :lol:

Musashi
2004-Dec-07, 11:40 PM
Sour Grapes?

Grey
2004-Dec-08, 12:04 AM
Sour Grapes?

Hmm, Grapes can be pretty vocal when defending a position, but I don't know that I'd ever call him "sour". :D

beck0311
2004-Dec-08, 05:38 AM
At the risk of beating this thing into the ground further, it just occured to me (yes, I am a little slow) that this is a geometric series where r&lt;1?

Maybe this has been said already, I haven't read the entire thread.

Normandy6644
2004-Dec-08, 05:44 AM
At the risk of beating this thing into the ground further, it just occured to me (yes, I am a little slow) that this is a geometric series where r&lt;1?

Maybe this has been said already, I haven't read the entire thread.

Yup. Ratio is 1/10. So you get (9/10)/(9/10)=1.

Damburger
2004-Dec-08, 03:29 PM
I think the misconception that a lot of the people who believe there is a difference have is that number representations map one-to-one on to real numbers. This is not the case, and I can explain why by analogy.

Rational numbers can all be represented by fractions - but there are multiple ways of righting the same rational number using fractional notation:

4/3 = 1 1/3 = 8/6 = ....

The fractional representation system has redundancy, so there is a many-to-one mapping between fractional representations and rational numbers.

The same applies to decimal representations and real numbers. Redundancy is rarely encountered in the decimal system, but it is wrong to think it does not exist, as 0.999r=1 shows. Just as there is more than one way to write 4/3 in fractional notation, there is more than one way to write 1 in decimal notation.

Understanding that, 0.999r=1 should be self-evidently true and 0.999r!=1 should be self-evidently false.

mutineer
2004-Dec-08, 04:35 PM
Let me pose a question to Disinfo Agent in particular – but all replies welcome.

A piece of (let’s say) cheese is cut in half. One of the halves is then cut in half. One of the resultant quarters in then cut in half. With each iteration, the resulting remnants are half the size produced by the previous iteration. Most (untutored?) people would say that with infinite iterations, the remnants become infinitely small, but there is always half as much as before . . . ad infinitum. If I follow your logic, you claim that if you halve a thing a sufficient number of times it has size = 0, i.e. it vanishes. Have I got this right?

jfribrg
2004-Dec-08, 05:00 PM
Instead of Disinfo Agent, what would Zeno say?

Severian
2004-Dec-08, 05:19 PM
...a sufficient number of times...

Nope. An infinite number of times. After any finite number of times, there is still some left. Though I suppose one could argue that at some point you are dividing subatomic particles, so it isn't clear that you have cheese left at that point anyway ;) .

As I understand your question in mathematical terms, it is "Does the sequence (0.5)^n converge to zero?" The answer to that question is yes.

Normandy6644
2004-Dec-08, 05:36 PM
...a sufficient number of times...

Nope. An infinite number of times. After any finite number of times, there is still some left. Though I suppose one could argue that at some point you are dividing subatomic particles, so it isn't clear that you have cheese left at that point anyway ;) .

As I understand your question in mathematical terms, it is "Does the sequence (0.5)^n converge to zero?" The answer to that question is yes.

I agree with your first bit, but doesn't the infinite series (1/2)^n converge to 1, not 0 (starting the sum at n=1)?

mutineer
2004-Dec-08, 06:14 PM
The sequence converges to 0.
The series converges to 1.
But I am asking for the matter to be considered on a logical basis, not a mathematical one. Forget your mathematical knowledge for a moment!

Normandy6644
2004-Dec-08, 06:25 PM
The sequence converges to 0.
The series converges to 1.
But I am asking for the matter to be considered on a logical basis, not a mathematical one. Forget your mathematical knowledge for a moment!

That seems to be the downfall in this thread, people appealing to their logic rather than the math. Mathematics is logical, lest we forget that. It might not appeal to our intuition or our "common sense," but that's why math is so fun!

worzel
2004-Dec-08, 06:27 PM
The sequence converges to 0.
The series converges to 1.
But I am asking for the matter to be considered on a logical basis, not a mathematical one. Forget your mathematical knowledge for a moment!

Rigorous mathematics is pure logic. What you are asking is for them to forget logic and try to get a feel for your intuition. I think they already have, and have been trying to explain why your intuition is wrong with regards to the axiomatically (logically) well defined real number system.

Pleading to "in reality" arguments are pointless, we're talking about numbers, constructs of pure logic, you can't go out and find a couple of 0.999... to experiment on.

Damburger
2004-Dec-08, 07:41 PM
The sequence converges to 0.
The series converges to 1.
But I am asking for the matter to be considered on a logical basis, not a mathematical one. Forget your mathematical knowledge for a moment!

You are attempting to apply common sense (in this case, spatial reasoning) to a problem involving infinity, which is why you are coming up with the wrong answer.

Your brain cannot comprehend infinity, or infintesimal, as a spatial concept. If you think it can, then try imagining where two perfectly parrellel lines meet. Thats infinity.

AstroRockHunter
2004-Dec-08, 07:53 PM
Hi, everyone.

Sorry I'm so long in following up. I only get on the BABB at work and a couple of projects have gotten in the way. I hate when work gets in the way of my BABBling. :lol:

Anyway, in my last post, I asked:

Does 1/1.000...-0.999... equal

a) Undefined
b) Infinity

The answer I was looking for was b) Infinity.

Here's why.

Let's rewrite the equation like this;

1/1.000...-x=

then consider the limits:

f(x)=1/1-x
Limit f(x)
x→1-
Limit (1\1-x)
x→1-

What this asks is "What is the output of the function f(x) as the value of x approaches 1 from the left?"

The important word here is "approaches", getting as close as possible to one without ever reaching it. Or, getting as infinitly close to 1 without equalling it.

What is "infinitly close to 1 without equalling it"???

Why, that would be 0.999...

So, to answer the question "Do you think that 0.999...=1?" My answer is no.

To answer the question "Does 0.999...=1?" The answer is still no.

OK, now that I've pontificated, feel free to roast me. #-o

Bad jcsd
2004-Dec-08, 09:35 PM
Does 1/1.000...-0.999... equal

a) Undefined
b) Infinity

The answer I was looking for was b) Infinity. did you read my post after yours explaining why it is undefined?

Here's why.

Let's rewrite the equation like this;

1/1.000...-x=

D

then consider the limits:

f(x)=1/1-x
Limit f(x)
x→1-
Limit (1\1-x)
x→1-

What this asks is "What is the output of the function f(x) as the value of x approaches 1 from the left?"

The important word here is "approaches", getting as close as possible to one without ever reaching it. Or, getting as infinitly close to 1 without equalling it.

Already your going wrong, we want to know what a the expression is equal to not the limit of some function! Besides which there are so many functions that are defiend at a point which do not equal there left limit at that point!

What is "infinitly close to 1 without equalling it"???

Why, that would be 0.999...

0.99.. is equal to one though, plus there are no such things as two real numbers that are infinitely close to each other.

So, to answer the question "Do you think that 0.999...=1?" My answer is no.

To answer the question "Does 0.999...=1?" The answer is still no.

OK, now that I've pontificated, feel free to roast me. #-o

How does any of that even constitute an argument that 0.999.. is not equal to 1?

Normandy6644
2004-Dec-08, 09:46 PM
I'm sticking with the infinite series argument as being pretty solid. I agree with what Bad jcsd was saying about the limits as well, and I don't really see any solid argument against .999...=1

01101001
2004-Dec-08, 09:47 PM
Does 1/1.000...-0.999... equal

a) Undefined
b) Infinity

The answer I was looking for was b) Infinity.
Nice "proof", but I think I can shorten it for you.

What is "infinitly close to 1 without equalling it"???

Why, that would be 0.999...

0.999... is not equal to 1.

Therefore, 0.999... is not equal to 1.

Of course, that's no proof -- but it's a lot less wordy.

Frog march
2004-Dec-08, 09:51 PM
any number can be rounded off.
to round off
0.9=1
0.99=1
0.999=1
0.9999=1

in fact what ever number of decimal places you go to and round off the answer is still 1, therefore wouldn't it be safe to say that if you rounded off the number at an infinite number of decimal places the answer is also 1 or does an infinite number of decimal places sudenly constitute an exception?

Bad jcsd
2004-Dec-08, 09:57 PM
any number can be rounded off.
to round off
0.9=1
0.99=1
0.999=1
0.9999=1

in fact what ever number of decimal places you go to and round off the answer is still 1, therefore wouldn't it be safe to say that if you rounded off the number at an infinite number of decimal places the answer is also 1 or does an infinite number of decimal places sudenly constitute an exception?

We're not talking about 'rounding off', we're talking about the actual value of the number.

Frog march
2004-Dec-08, 10:03 PM
but if you round off at infinity(a number of places that doesn't actualy exist) then you HAVE the actual number.

pghnative
2004-Dec-08, 10:09 PM
So far, none of the UB's ("Unequal" Believers) have come up with an argument against what Donnie B. said several posts ago. Which was:

0.999.../3 = 0.333...
1/3 = 0.333...

When X/3 = Y/3, this means that X=Y. Therefore 0.999...=1

worzel
2004-Dec-08, 10:11 PM
The two arguments that I find the most persuasive are:

1:

1 / 3 = 0.333...

0.999... / 3 = 0.333...

0.333... = 0.333...

ego...

2:

If 0.999... doesn't equal 1 then there must be a real number in between them. How can this be? What is slightly bigger than 0.999... and slightly less than 1? No matter how many decimal places you go to (even infinite) you won't find one.

I haven't seen anything close to an argument against other than false claims of "by definition" and appeals to intuition. You can't just claim that some bit of maths is defined to be what you think it ought to be, and you can't appeal to intuition in maths, you need to be rigorous.

Bad jcsd
2004-Dec-08, 10:14 PM
Frog March - when we're talking pure math and the actual vlaues of numbers we really shouldn't talk about rounding off, besides which rounding off at inifty is not a well-define procedure.

The proof that everyone is taught at school is:

let x = 0.9999

9x = 10x - x = 9.9999... - 0.9999.. = 9 therefore x = 1.

Frog march
2004-Dec-08, 10:24 PM
neat!!!
I must have missed that day..

AstroRockHunter
2004-Dec-08, 10:42 PM
So, what you all are saying is:

0.9=1.0

Correct???

Bad jcsd
2004-Dec-08, 10:44 PM
So, what you all are saying is:

0.9=1.0

Correct???

No, I'm saying 0.99... = 1

AstroRockHunter
2004-Dec-08, 10:50 PM
You're proof of 9x=10x-x works with just 0.9.
So why would an infinite number of 9's make a difference???

edited to try to be clear. (Fat Chance!!!)

tjm220
2004-Dec-08, 10:52 PM
Why does the extra 9 make the difference???

It's not a single extra 9 but an ifinitely long string of 9's that makes the difference. It would take too long to actually type the number out hence the short-hand 0.99...

Bad jcsd
2004-Dec-08, 10:53 PM
Why does the extra 9 make the difference???

we're talking about 0.999..... i.e. an inifinite number of 9's

Normandy6644
2004-Dec-08, 10:53 PM
Frog March - when we're talking pure math and the actual vlaues of numbers we really shouldn't talk about rounding off, besides which rounding off at inifty is not a well-define procedure.

The proof that everyone is taught at school is:

let x = 0.9999

9x = 10x - x = 9.9999... - 0.9999.. = 9 therefore x = 1.

I've done that one a few times now too. :wink:

mutineer
2004-Dec-08, 10:54 PM
Earlier in this thread, martin was demanding rules.
Here is the rule for working out the value of a number.
The digit on the right is units.
The digit one to the left is tens.
The next one is hundreds.
I learned this when I was about five.
Two or three years later, I discovered that if you put in a decimal point, the first digit to the right of the point is tenths, and the next one represents one-hundredths of a unit.

This is not so much a matter of rules as a statement about the nature of decimal numbers.
If the digit to the left of the decimal point is a zero, the value of the number is less than unity, because the best successive numbers to the right can do is make up nine-tenths of the deficit. No number of added rightward digits is high enough for this principle to break down. There is no nth digit where the principle is difference than for the n-1th. The principle continues to infinity. Anyone who imagines that infinity produces a discontinuity has a faulty grasp of logic.

Please, where is the INTUITION in the previous paragraph?

It seems to me that I have outlined the defining method of understanding the value of a number, and that it can under no circumstances be abrogated by some piece of additional knowledge vouchsafed to the numerati.

The foregoing is independent of and superior to mathematics. Mathematics makes rules/axioms for operationalizing numbers; for manipulating them for purposes of computation. Its rules are of a lower order. As soon as you import any of them you import hidden assumptions that may or may not be true. For computational purposes, 0.999 recurring is sufficiently close to unity that one may put an equals sign between them. A system of rules which equates them is unlikely to pose any danger of difficulty. But one may not use the lower order rules of maths to confound what the nature of decimal numbers logically dictates.

Are my arguments logically irrefutable but mathematically wrong? Producing mathematical ‘proofs’ cuts no ice with me. Please show me the errors in my logic. Is my question about successive halving impossible to deal with on purely logical grounds?

. . . try imagining where two perfectly parrellel lines meet. Thats infinity.Please justify this statement. If you cannot, do not believe everything you have been told.

Bad jcsd
2004-Dec-08, 10:55 PM
You're proof of 9x=10x-x works with just 0.9.
So why would an infinite number of 9's make a difference???

edited to try to be clear. (Fat Chance!!!)

No it doesn't:

x = 0.9

9x = 0.81

Normandy6644
2004-Dec-08, 10:58 PM
Please, where is the INTUITION in the previous paragraph?

For the last time, it's not about intuition, it's about mathematics. If intuition had any say in how math works calculus would never have got off the ground.

Edited for emphasis.

AstroRockHunter
2004-Dec-08, 10:58 PM
Bad jcsd wrote:

x = 0.9

9x = 0.81

SORRY, 9x=8.1

Bad jcsd
2004-Dec-08, 11:06 PM
If the digit to the left of the decimal point is a zero, the value of the number is less than unity, because the best successive numbers to the right can do is make up nine-tenths of the deficit. No number of added rightward digits is high enough for this principle to break down. There is no nth digit where the principle is difference than for the n-1th. The principle continues to infinity. Anyone who imagines that infinity produces a discontinuity has a faulty grasp of logic.
ms to me that I have outlined the defining method of understanding the value of a number, and that it can under no circumstances be abrogated by some piece of additional knowledge vouchsafed to the numeration.
.

If you think the 'rules' thta you posted are even relevant think again, cos guess what they don't teach the axioms of the real numbers in the infants!

What is wrongis that you are trying to apply the principle of induction erroneously, we're not tlaking about the nth digit because infinity is not a natural number!

You could say that all decimal reprenttaions must terminate, but then you are saying that there is no decimal represnation for all real numbers or even 'most'. In order for there to be any consistency the represenation 0.999.. MUST equal 1.

Bad jcsd
2004-Dec-08, 11:08 PM
Bad jcsd wrote:

x = 0.9

9x = 0.81

SORRY, 9x=8.1

Sorry yes, but my point still stands as it doesn't prove that 0.9 = 1

as 10x - x = 9.0 -0.9 = 8.1, 8.1/9 = 0.9.

AstroRockHunter
2004-Dec-08, 11:14 PM
Bad jcsd

10x - x = 9.0 -0.9 = 8.1, 8.1/9 = 0.9.

The bolded part did not appear in your original offering of the proof. What is it doing here now???

Also, if you do apply it to your original offering, then you get:

8.999991/9=0.999999

Bad jcsd
2004-Dec-08, 11:17 PM
Bad jcsd

10x - x = 9.0 -0.9 = 8.1, 8.1/9 = 0.9.

The bolded part did not appear in your original offering of the proof. What is it doing here now???

9x/9 = x, the orignal proof showed that 9x = 9 (where x = 0.999..) and I just assumed that everyone knew that 9/9 = 1.

AstroRockHunter
2004-Dec-08, 11:19 PM
Bad jcsd

Sorry, did an edit while you were posting.

Hey, if this thread keeps going, I just might make it to apprentice!!! =D>

Frog march
2004-Dec-08, 11:22 PM
I voted YES 0.99999999999999........... times!!!!!!!! :)

Bad jcsd
2004-Dec-08, 11:23 PM
Also, if you do apply it to your original offering, then you get:

8.999991/9=0.999999

the '...' means that the decimal representation does not terminate.

so 10x = 9.99999..... , 9.99999... - 0.999999...... = 9

Chuck
2004-Dec-08, 11:30 PM
Let me pose a question to Disinfo Agent in particular – but all replies welcome.

A piece of (let’s say) cheese is cut in half. One of the halves is then cut in half. One of the resultant quarters in then cut in half. With each iteration, the resulting remnants are half the size produced by the previous iteration. Most (untutored?) people would say that with infinite iterations, the remnants become infinitely small, but there is always half as much as before . . . ad infinitum. If I follow your logic, you claim that if you halve a thing a sufficient number of times it has size = 0, i.e. it vanishes. Have I got this right?It's hard to cut cheese an infinite number of times in real life but it can be done with distance. If there's a stone wall ten meters away and I move toward it at one meter per second, I cover half the distance in 5 seconds, half of what's left in 2.5 seconds, half of what's left after that in 1.25 seconds, etc, etc. After ten seconds I've covered half the remaning distance an infinite number of times and hit the wall. There is no individual halving in the procedure in which I've eliminated the last of the distance but I'm at the wall anyway since I eliminated half the distance an infinite number of times.

Since I moved at a speed of one meter per second I must have moved ten meters after ten seconds. If this is incorrect then where will I be after ten seconds?

Normandy6644
2004-Dec-08, 11:33 PM
It's hard to cut cheese an infinite number of times in real life...

Yeah, the smell just becomes too much.

:lol:

AstroRockHunter
2004-Dec-08, 11:35 PM
Bad jcsd wrote:

the '...' means that the decimal representation does not terminate.

so 10x = 9.99999..... , 9.99999... - 0.999999...... = 9

First, I know what the '...' means.

Second, you still haven't reconciled your proof.
Putting your examples together, we get

x=0.999999...
9x=10x-x=8.99999... , 8.99999.../9=0.999999... x=0.999999...

Bad jcsd
2004-Dec-08, 11:39 PM
We're tlaking about the limit specifcally here and for an infinite decimal expansion it is equal to the limit of the series Sum a_n10^n, so arguments about pieces of cheese do not cut the mustard.

Bad jcsd
2004-Dec-08, 11:41 PM
Bad jcsd wrote:

the '...' means that the decimal representation does not terminate.

so 10x = 9.99999..... , 9.99999... - 0.999999...... = 9

First, I know what the '...' means.

Second, you still haven't reconciled your proof.
Putting your examples together, we get

x=0.999999...
9x=10x-x=8.99999... , 8.99999.../9=0.999999... x=0.999999...

Ok what does 10x equal?

edited to add and surely if 9x = 8.999... then 8x = 8.999... - 0.9999 = 8 therefore x = 1.

AstroRockHunter
2004-Dec-08, 11:52 PM
Bad jcsd wrote:

Ok what does 10x equal?

edited to add and surely if 9x = 8.999... then 8x = 8.999... - 0.9999 = 8 therefore x = 1.

10x=9.99999...

No, 8x doesn't equal 8.999...
8x equals 7.999...

This has been a fun conversation today. However, it's now 4:50pm MST and almost time for me to go home for the evening.

Hope I haven't ruffled any feathers today and I'll try to get back to the board tomorrow.

Have a great evening!!! :D

Bad jcsd
2004-Dec-08, 11:54 PM
Bad jcsd wrote:

Ok what does 10x equal?

edited to add and surely if 9x = 8.999... then 8x = 8.999... - 0.9999 = 8 therefore x = 1.

10x=9.99999...

No, 8x doesn't equal 8.999...
8x equals 7.999...

This has been a fun conversation today. However, it's now 4:50pm MST and almost time for me to go home for the evening.

Hope I haven't ruffled any feathers today and I'll try to get back to the board tomorrow.

Have a great evening!!! :D

Okay then then what does 9 + 0.99999... equal?

i.e 9 + x = 10x therefore x = 1

Severian
2004-Dec-09, 12:10 AM
No number of added rightward digits is high enough for this principle to break down.

No finite number of added digits yes...

...The principle continues to infinity. Anyone who imagines that infinity produces a discontinuity has a faulty grasp of logic.

Please, where is the INTUITION in the previous paragraph?

I put it in bold for you...

You intuition for dealing with finite decimal expansions is failing you when you inappropriately try to apply it to an infinite expansion.

worzel
2004-Dec-09, 01:14 AM
Are my arguments logically irrefutable but mathematically wrong?
No, that would be a contradiction, given that "mathematically" means "logically". You haven't made a logical argument so rather than being refutable (i.e. containing a logical flaw), your arguments are meaningless, mathematically speaking anyway (i.e. logically speaking).

Producing mathematical ‘proofs’ cuts no ice with me.
So, logic cuts no ice with you then, ok.

Is my question about successive halving impossible to deal with on purely logical grounds?
Your point was simply that continual halving will always leave some cake left. Ignoring the physical limitation of this argument, it has been pointed out many times that although that's true, the amount left is infintely small. That infintely small is not only physically impossible but also impossible to imagine should tell you something about arguments of this nature.

Please show me the errors in my logic.

Here is the rule for working out the value of a number.
The digit on the right is units.
The digit one to the left is tens.
The next one is hundreds.
I learned this when I was about five.
Two or three years later, I discovered that if you put in a decimal point, the first digit to the right of the point is tenths, and the next one represents one-hundredths of a unit.
So far so good.

This is not so much a matter of rules as a statement about the nature of decimal numbers.
You imply (or do I just infer) that our rules are just attempts to model these things we discovered called numbers. Not so, the rules are the numbers.

If the digit to the left of the decimal point is a zero, the value of the number is less than unity
This is a forgivable assumption, forgivable but wrong as shown many times on this thread. It has never been stated as a defining property of the reals, or as a consequence of their definition in any maths textbook I ever read.

because the best successive numbers to the right can do is make up nine-tenths of the deficit.
False, the best that a single number to the right can do is make up nine tenths (of unity). I'm guessing that's what you meant.

No number of added rightward digits is high enough for this principle to break down.
You have shown (or alluded to at least) that the nth place can only make up nine tenths of the (n-1)th place, you have not shown how this leads to infinite places beyond the nth place not being enough to make up the extra tenth, but just assered it as usual.

There is no nth digit where the principle is difference than for the n-1th.
Unfounded assertion number 2. But if you insist on picking a value for n then true, it alone can not make up more than nine tenths of the (n-1)th place, but you attempted to say that already.

The principle continues to infinity.
UA3 - hey, we've got a new equation labelling system for your branch of mathematics ;) Actually, there is a branch of logic called intuitionistic logic, which you could use to refute some of the proofs here (but not all of them). I don't know why you haven't googled it already 8)

Anyone who imagines that infinity produces a discontinuity has a faulty grasp of logic.
Or to paraphrase, anyone who disagrees on logical grounds with your thrice stated unfounded assertion has a faulty grasp of logic. Actually, the discontinuity you seem to be getting at is between chosing an arbitrary number of decimal places and chosing infinite. Intuitively I wouldn't expect infinity to behave like any particular choice of n - but then that's just intuition.

The foregoing is independent of and superior to mathematics.
You don't think that's just a little presumptious?

Mathematics makes rules/axioms for operationalizing numbers; for manipulating them for purposes of computation. Its rules are of a lower order.
I don't agree with that at all. Mechanical computations involve all sorts of nasty approximations in order to make computations feasible. One of the beauties of pure maths is that is not hindered so.

As soon as you import any of them you import hidden assumptions that may or may not be true.
If you really believe that you understand logic and you really believe that there is a hidden assumption in the statement that 0.999...=1 then you should be able to digest the axioms that define the real numbers and demonstrate where the hidden assumption has been introduced in every proof of this statement given in this thread.

For computational purposes, 0.999 recurring is sufficiently close to unity that one may put an equals sign between them.
Indeed, in fact so close that it is actually unity. But ironically, computatioinally this equality is often not realized and you end up getting 0.999999... (as far as your calculator/computer will go).

A system of rules which equates them is unlikely to pose any danger of difficulty. But one may not use the lower order rules of maths to confound what the nature of decimal numbers logically dictates.
There you go again. To paraphrase,

"Maths [ pure logic ] may not be used to confound what the nature of decimal numbers [ logically defined mathemetical objects ] logically dictates."

As you would say, anyone who imagines that the above is not a contradiction has a faulty grasp of logic.

mutineer
2004-Dec-09, 10:10 AM
...The principle continues to infinity. Anyone who imagines that infinity produces a discontinuity has a faulty grasp of logic.

Please, where is the INTUITION in the previous paragraph?

I put it in bold for you...

You intuition for dealing with finite decimal expansions is failing you when you inappropriately try to apply it to an infinite expansion.
The string of nines is endless. Yet you are arguing that a discontinuity exists: viz. one of the nines eliminates ALL the remaining discrepancy with unity. Please explain very carefully. Telling me (in effect) that I lack the mental capacity to understand the meaning of endless would be insufficient.

Disinfo Agent
2004-Dec-09, 12:47 PM
Let me pose a question to Disinfo Agent in particular – but all replies welcome.

A piece of (let’s say) cheese is cut in half. One of the halves is then cut in half. One of the resultant quarters in then cut in half. With each iteration, the resulting remnants are half the size produced by the previous iteration. Most (untutored?) people would say that with infinite iterations, the remnants become infinitely small, but there is always half as much as before . . . ad infinitum. If I follow your logic, you claim that if you halve a thing a sufficient number of times it has size = 0, i.e. it vanishes. Have I got this right?
I think others have already said this, but I'm going to answer anyway, since you adressed the question to me.

We have a semantic problem here with the words "always", "infinite", and "a sufficient number of times". What do you mean when you say that there will be "always half as much as before"?

Allow me to make a slight digression. Did you know that the ancient Greeks avoided the infinite like the plague? They correctly realised that the concept(s) of infinite lead to logical difficulties. Euclid, in his classical work on geometry, did not use our modern notion of a straight line as "a line that goes on forever". Instead, for him a "straight line" was what we'd call today a straight line segment. It could be prolonged into a longer segment of arbitrary but finite length. Aristotle would not say that the set of the integers was infinite, only potentially infinite. Given an integer, no matter how large, we could always construct a larger integer, but for Aristotle the actual infinite did not exist, since human beings are by their own nature finite and thus cannot contemplate the infinite. The ancient concept and the modern one are equivalent in practical terms, but there was obviously a philosophical and mathematical barrier here.

To answer your question, let me start by replacing "piece of cheese" with "piece of pie", just to make my language less awkward. I would say that a pie can potentially be divided into any finite number of smaller and smaller slices, and what remains after the final iteration will be half of the previous slice, and not zero. I wouldn't call it "infinitely small", though. It's a real number that may be very small indeed, as small as we wish, but it's still a real number, so it's not infinitely small. This is because we're really conceiving of the slicing as a sequence of actions that stop at a finite iteration (though arbitrarily large).

If, however, you take one step further and ask what would happen if you kept dividing "forever", then you're talking about an actually infinite process, where you begin, go through each finite iteration but do not stop dividing at any of those iterations. In that case there is no final iteration, so we need some rigorous way of making sense of the situation. The simplest one is to ask: what state are all those finite iterations taking you closer and closer to? In other words, we interpret the "final", infinite, iteration of the process as the limit of all finite iterations.

If the pie is approximately circular, with a circumference of x, when you slice it into smaller and smaller pieces as you've described, the circumference of the successive pieces will be x, x/2, x/4, x/8, and so on. As it turns out, the limit of this sequence is 0. In this sense, if you slice the pie "infinitely often", you should end up with a "piece" of circumference zero. This makes sense, because you know you can (theoretically) slice the pie into a piece as small as you wish, by making a large enough finite number of divisions.

In some geometrical and physical problems, it's useful to imagine that you could divide the pie into an "infinitely small" piece, with a circumference smaller than any positive real number, but still not zero. This informal, intuitive line of reasoning can be made mathematically precise by appealing to hyperreal numbers. However, when you do this you're stepping outside the real number system.

This is not so much a matter of rules as a statement about the nature of decimal numbers.
Let me stop you right there!
There is no such thing as a decimal number. Numbers are numbers. It doesn't matter what we call them! The decimal notation is just a way of denoting numbers, a numeral. But 100 is the same as 10^2 and the same as C. And 1/2 is the same as 3/6 and the same as 0.5.

If the digit to the left of the decimal point is a zero, the value of the number is less than unity, because the best successive numbers to the right can do is make up nine-tenths of the deficit.
0.9 = 9 tenths
0.99 = 99 hundredths --> greater than 9 tenths!
0.999 = 999 thousandths --> greater than 99 hundredths!
...
We can make this sequence as close to 10/10 as we wish.

The foregoing is independent of and superior to mathematics. Mathematics makes rules/axioms for operationalizing numbers; for manipulating them for purposes of computation.
No, mathematics also defines numbers and notations! The decimal notation you're using was invented by Indian mathematicians in the Middle Ages. The Romans wrote V for 5, X for 10, L for 50, etc.

Are my arguments logically irrefutable but mathematically wrong? Producing mathematical ‘proofs’ cuts no ice with me. Please show me the errors in my logic.
The error in your logic, and that of others, as A Thousand Pardons wrote (http://www.badastronomy.com/phpBB/viewtopic.php?p=372194#372194) on the second page of this thread, is that you are confusing numbers with numerals, the concept with the symbol. A number is not a string of digits. That's just a representation.

Disinfo Agent
2004-Dec-09, 01:18 PM
Does 1/1.000...-0.999... equal

a) Undefined
b) Infinity

The answer I was looking for was b) Infinity.

Here's why.

Let's rewrite the equation like this;

1/1.000...-x=

then consider the limits:

f(x)=1/1-x
Limit f(x)
x?1-
Limit (1\1-x)
x?1-

What this asks is "What is the output of the function f(x) as the value of x approaches 1 from the left?"
You argue that 1/(1 - 0.999...) = infinity (don't forget the parentheses). Inverting that fraction, I get 1 - 0.999... =0, and thus 1=0.999.... :)

Damburger
2004-Dec-09, 01:38 PM
Earlier in this thread, martin was demanding rules.
Here is the rule for working out the value of a number.
The digit on the right is units.
The digit one to the left is tens.
The next one is hundreds.
I learned this when I was about five.
Two or three years later, I discovered that if you put in a decimal point, the first digit to the right of the point is tenths, and the next one represents one-hundredths of a unit.

This is not so much a matter of rules as a statement about the nature of decimal numbers.
If the digit to the left of the decimal point is a zero, the value of the number is less than unity, because the best successive numbers to the right can do is make up nine-tenths of the deficit. No number of added rightward digits is high enough for this principle to break down. There is no nth digit where the principle is difference than for the n-1th. The principle continues to infinity. Anyone who imagines that infinity produces a discontinuity has a faulty grasp of logic.

Arrogance doesn't make your ignorance sound any better. The 'rule' you stated does not in fact exist, and I challenge you to find it in any non-woowoo piece of mathematic literature.

You are essentially redefining the decimal representation system to fit your opinion. The faulty logic here is yours.

Please, where is the INTUITION in the previous paragraph?

Your intuition told you that 0.999r&lt;1 and you tried to use mathematical language to justify it.

It seems to me that I have outlined the defining method of understanding the value of a number, and that it can under no circumstances be abrogated by some piece of additional knowledge vouchsafed to the numerati.

The defining method? You are mistaking your own opinion for that of the entire maths community. If the method you describe were universal, this thread wouldn't exist.

Numerati? Is that like an evil conspiracy trying to steal real numbers from you?

The foregoing is independent of and superior to mathematics. Mathematics makes rules/axioms for operationalizing numbers; for manipulating them for purposes of computation. Its rules are of a lower order. As soon as you import any of them you import hidden assumptions that may or may not be true. For computational purposes, 0.999 recurring is sufficiently close to unity that one may put an equals sign between them. A system of rules which equates them is unlikely to pose any danger of difficulty. But one may not use the lower order rules of maths to confound what the nature of decimal numbers logically dictates.

Too good for maths? **** off then, because this thread is about maths.

Again, you are using fancy words to try and make your ignorance sound like intelligence. As I explained above, the decimal system of representing numbers has some redundancy in it, because it allows for infinite sequences of digits. You didn't listen it seems.

Are my arguments logically irrefutable but mathematically wrong? Producing mathematical ‘proofs’ cuts no ice with me. Please show me the errors in my logic. Is my question about successive halving impossible to deal with on purely logical grounds?

You arguments are pants. Your logic is based entirely on unwarranted assumptions of yours that you trumpet as irrefutable facts.

. . . try imagining where two perfectly parrellel lines meet. Thats infinity.Please justify this statement. If you cannot, do not believe everything you have been told.

The statement is intended to help you understand why intuition doesn't work with infinities.

Just try it. Close picture two parallel lines in your head, then tell yourself they meet at some point out of your field of vision.

mickal555
2004-Dec-09, 02:01 PM
*Groans* I hate maths this is so complicated......

Bad jcsd
2004-Dec-09, 02:12 PM
*Groans* I hate maths this is so complicated......#

It isn't complicated though, that's what annoys me! the informal proof I posted can be understood by an eleven year-old.

For mutineer:

Your still missing the fact that any non-terminating decimal expansion is DEFINED as the limit of a sequence of decimal fractions i.e. 0.999... is defined as the smallest number that is greater than all the numbers in the set {0.9, 0.99, 0.999, etc} which is 1.

I'll also add there is no other useful way of defining decimal expansions which is not equivalent to this defintion.

mickal555
2004-Dec-09, 02:33 PM
I understand it (I'm quite good at maths actully) but that dousn't make me like maths any more. I was going to say boring but thats harsh.

Severian
2004-Dec-09, 03:24 PM
The string of nines is endless. Yet you are arguing that a discontinuity exists: viz. one of the nines eliminates ALL the remaining discrepancy with unity. Please explain very carefully. Telling me (in effect) that I lack the mental capacity to understand the meaning of endless would be insufficient.

Nope. Not one of the nines. For any finite string of 9's, the 0.9999..99 (finite) is not 1, and adding any single 9 doesn't make it 1. But then again we aren't arguing about a finite string of 9's are we? No one is saying that you lack the capacity to understand anything, but when you say things like "after a sufficient amount of time" and "one of the nines" then you are talking about a finite process, and I can only assume that you are missing the point.

Do you disagree with the following statements:

1) A real number is either postive, negative or zero
2) Given any two real numbers, their difference is a real number
3) Given any positive real number r, there are postive rational numbers a and b with a &lt; r &lt; b

If you agree with those, answer this:
Let d = 1 - 0.99... Is d positive, negative or zero? If nonzero, give a and b as above for the absolute value of d. If zero, then do you not believe that zero is the additive identity for the reals; i.e. do you disbelieve a + 0 = a for any a?

mutineer
2004-Dec-09, 03:31 PM
Here is the rule for working out the value of a number.
The digit on the right is units.
The digit one to the left is tens.
The next one is hundreds.
I learned this when I was about five.
Two or three years later, I discovered that if you put in a decimal point, the first digit to the right of the point is tenths, and the next one represents one-hundredths of a unit.
You are essentially redefining the decimal representation system to fit your opinion.
You are telling me that is a redefinition?

Severian
2004-Dec-09, 03:39 PM
Here is the rule for working out the value of a number.
The digit on the right is units.
The digit one to the left is tens.
The next one is hundreds.
I learned this when I was about five.
Two or three years later, I discovered that if you put in a decimal point, the first digit to the right of the point is tenths, and the next one represents one-hundredths of a unit.
You are essentially redefining the decimal representation system to fit your opinion.
You are telling me that is a redefinition?

No. He is telling you that the some of the rest of your statement is the redefinition (which you amusingly cut out):

This is not so much a matter of rules as a statement about the nature of decimal numbers. If the digit to the left of the decimal point is a zero, the value of the number is less than unity, because the best successive numbers to the right can do is make up nine-tenths of the deficit. No number of added rightward digits is high enough for this principle to break down. There is no nth digit where the principle is difference than for the n-1th. The principle continues to infinity. Anyone who imagines that infinity produces a discontinuity has a faulty grasp of logic.
Interesting that you would cut this out, since it is the part that you are being told is wrong from many sides. You just thought that he was the exception?

A Thousand Pardons
2004-Dec-09, 04:15 PM
Here is the rule for working out the value of a number.
The digit on the right is units.
The digit one to the left is tens.
The next one is hundreds.
I learned this when I was about five.
Two or three years later, I discovered that if you put in a decimal point, the first digit to the right of the point is tenths, and the next one represents one-hundredths of a unit.
You are essentially redefining the decimal representation system to fit your opinion.
You are telling me that is a redefinition?
No, careful, you are misquoting.

Damburger
2004-Dec-09, 05:14 PM
Here is the rule for working out the value of a number.
The digit on the right is units.
The digit one to the left is tens.
The next one is hundreds.
I learned this when I was about five.
Two or three years later, I discovered that if you put in a decimal point, the first digit to the right of the point is tenths, and the next one represents one-hundredths of a unit.
You are essentially redefining the decimal representation system to fit your opinion.
You are telling me that is a redefinition?

I'm not going to bother slapping you down for this misquote seeing as its been done already :)

You're idea that a decimal number whos one's digit is zero has to be lower than one (honestly, calling it unity doesn't hide your lack of mathematical ability) has no grounding in maths. If you can prove me wrong, please do so.

You keep claiming that your thinking is logical. This may be so, however you seem to think that logical=correct and this is not the case. If you begin with faulty assumptions as you have, you can conclude any old crap with logic.

AstroRockHunter
2004-Dec-09, 07:00 PM
Disinfo Agent wrote:

You argue that 1/(1 - 0.999...) = infinity (don't forget the parentheses). Inverting that fraction, I get 1 - 0.999... =0, and thus 1=0.999....

Here's where we disagree. You contend that 1-0.999... = 0.

I contend that 1-0.999... ≈ 0, and thus 1 ≈ 0.999... .

BTW, I think the is a great way to spend lunch. :lol:

Disinfo Agent
2004-Dec-09, 07:06 PM
So you're saying the reciprocal of infinity is not zero after all...

A Thousand Pardons
2004-Dec-09, 07:09 PM
Here's where we disagree. You contend that 1-0.999... = 0.

I contend that 1-0.999... ≈ 0, and thus 1 ≈ 0.999... .

BTW, I think the is a great way to spend lunch. :lol:
Sure, and if only all disagreements were so agreeable :)

mutineer
2004-Dec-09, 07:13 PM
Surely what preceded my sentence "This is not so much a matter of rules as a statement about the nature of decimal numbers" was the only part of what I said that could be regarded as a "definition".

How could the sentences which followed be regarded as a "definition"?

Bad jcsd
2004-Dec-09, 07:15 PM
I've laready explained the difinition of a non-terminating decimal expansion and that's the only dfeintion we should be interested in for now.

What you posted was a brief description of some of the properties of decimal expansions not a defitnion.

Damburger
2004-Dec-09, 07:16 PM
Surely what preceded my sentence "This is not so much a matter of rules as a statement about the nature of decimal numbers" was the only part of what I said that could be regarded as a "definition".

How could the sentences which followed be regarded as a "definition"?

Stop backtracking and trying to misrepresent me. Its almost as annoying as you constantly calling the number one 'unity'

You claimed that any decimal where the first digit left of the point is 0, must be lower than 1. This is an unwarranted assumption on your part, and you've supplied no proof whatsoever.

AstroRockHunter
2004-Dec-09, 07:18 PM
Bad jcsd wrote:

Okay then then what does 9 + 0.99999... equal?

i.e 9 + x = 10x therefore x = 1

Not quite.

x=0.999…

9.000... + x = 10x

x = 10x - 9.000...

0.999… = 10 (0.999…) – 9.000...

0.999… = 0.999…

Bad jcsd
2004-Dec-09, 07:21 PM
Bad jcsd wrote:

Okay then then what does 9 + 0.99999... equal?

i.e 9 + x = 10x therefore x = 1

Not quite.

x=0.999…

9.000... + x = 10x

x = 10x - 9.000...

0.999… = 10 (0.999…) – 9.000...

0.999… = 0.999…

So let me get this straight your now saying that 9.000... is not equal to 9 !!!![-X ,

okay then what is (9 - 9.0000)/2 ?

A Thousand Pardons
2004-Dec-09, 07:23 PM
Bad jcsd wrote:

Okay then then what does 9 + 0.99999... equal?

i.e 9 + x = 10x therefore x = 1

Not quite.

x=0.999…

9.000... + x = 10x

Are you claiming that 9 + 0.999... is not 9.999...?

Can you even add 9 + 0.999... in your system?

x = 10x - 9.000...

0.999… = 10 (0.999…) – 9.000...

0.999… = 0.999…

Bad jcsd
2004-Dec-09, 07:25 PM
yep thta's a point you didn't tell us what 9 + 0.9999 was infact you delibartely avoided it!

SeanF
2004-Dec-09, 07:44 PM
Bad jcsd wrote:

Okay then then what does 9 + 0.99999... equal?

i.e 9 + x = 10x therefore x = 1

Not quite.

x=0.999…

9.000... + x = 10x

x = 10x - 9.000...

0.999… = 10 (0.999…) – 9.000...

0.999… = 0.999…

So let me get this straight your now saying that 9.000... is not equal to 9 !!!![-X ,
He didn't say that, did he? He just substituted 0.9999... for x on both sides of the equation, instead of substituting on one side and solving for the other.

Bad jcsd
2004-Dec-09, 07:51 PM
[quote="SeanF"
He didn't say that, did he? He just substituted 0.9999... for x on both sides of the equation, instead of substituting on one side and solving for the other.[/quote]

Yes I see thta now, he was jsut trying to avoid writing out the decimal expansion of 9 + 0.999...

AstroRockHunter
2004-Dec-09, 07:54 PM
Thanks, SeanF:

To respond to both Bad jcsd and ATP.

Bad jcsd used the equation 9+x = 10x to prove that x=1. This is only too true.

I, OTOH, defined x=0.999... . Then used Bad jcsd's equation.
(Bites tongue so as not to get banned from the board.)

So, to Bad jcsd and ATP, answer me this:

Given

x=?

9.000... + 0.999... = 10x

What is x equal to???

edited to add "x=?"

Bad jcsd
2004-Dec-09, 07:57 PM
x is equal to 1 as 0.999... is equal to 1 so 9 + 0.999.. = 10.

You still haven't told us the decimal expansion of 9 + 0.9999.....

A Thousand Pardons
2004-Dec-09, 07:58 PM
So, to Bad jcsd and ATP, answer me this:

First, you do agree that 9 + .999... equals 9.999... ?

Given

x=?

9.000... + 0.999... = 10x

What is x equal to???

edited to add "x=?"
Does x = 0.999...? Can we for substitude equals in that equation? To produce 9 + x = 10x

Or are you saying that 9.000... also does not equal 9?

AstroRockHunter
2004-Dec-09, 08:09 PM
Bad jcsd wrote:

x is equal to 1 as 0.999... is equal to 1 so 9 + 0.999.. = 10.

You still haven't told us the decimal expansion of 9 + 0.9999.....

and

A Thousand Pardons wrote:

Does x = 0.999...? Can we for substitude equals in that equation? To produce 9 + x = 10x

Or are you saying that 9.000... also does not equal 9?

A Thousand Pardons. No, x ≠ 0.999... . You may have missed my edit of "x=?" which means that x is unknown. So, you cannot sustitute x for 0.999...

Bad jcsd. First, 9 = 9.000... .
Second, the decimal expansion of 9 + 0.999... is 9.999... .
Third, you state that 0.999... = 1. This is a statement that I cannot agree with.

Normandy6644
2004-Dec-09, 08:12 PM
Third, you state that 0.999... = 1. This is a statement that I cannot agree with.

Well that's kind of been the whole point of the discussion...

Bad jcsd
2004-Dec-09, 08:14 PM
Bad jcsd. First, 9 = 9.000... .
Second, the decimal expansion of 9 + 0.999... is 9.999... .

That is want I wanted to hear as earlier you stated that 10*0.9999 .. = 9.999...

therfore we can say: 10x = 9 + x therefore 10x - x = 9 therfore 9x = 9 therefore x = 1.

Bad jcsd
2004-Dec-09, 08:15 PM
Which is my point, basic algebra cannot be consistent if you define 0.999... as anything other than 1.

pghnative
2004-Dec-09, 08:18 PM
This proof seems to have been mangled amongst the many posters. Is it possible to start from scratch? The proof goes like this:

1. Let x=0.999...
2. Therefore 10x = 9.999...
3. 10x - x = 9x (for all values of x)
4. Therefore 9.999... - 0.999... = 9*(0.999...)
5. But 9.999... - 0.999... = 9
6. Substituting equation 5 into the left hand side of equation 5 gives:
9 = 9*(0.999...)
7. Dividing both sides by 9 gives 1 = 0.999...

edited once for clarity. The challenge for any UB's would be to find fault with any of the individual steps.

Normandy6644
2004-Dec-09, 08:20 PM
I'm almost curious to count how many times that proof has been presented. I know I've done it twice, and I'm sure I'm not the only one. Maybe it needs to be repeated every so often so it's fresh in our minds. Plus the infinite series seems to give people trouble. :wink:

SeanF
2004-Dec-09, 08:24 PM
I'm almost curious to count how many times that proof has been presented. I know I've done it twice, and I'm sure I'm not the only one. Maybe it needs to be repeated every so often so it's fresh in our minds. Plus the infinite series seems to give people trouble. :wink:
We don't have to repeat the proof an infinite number of times, do we?!

Bad jcsd
2004-Dec-09, 08:29 PM
edited once for clarity. The challenge for any UB's would be to find fault with any of the individual steps.

Astrorock collector objected to step 5) claiming that 9.99.. - 0.999.. was 8.999.. (which it is, but he obviously doesn;t accept that 8.999... = 9)

Therest was just to show him that his refusal to accept that step results necessarily in a contradiction.

Normandy6644
2004-Dec-09, 08:31 PM
I'm almost curious to count how many times that proof has been presented. I know I've done it twice, and I'm sure I'm not the only one. Maybe it needs to be repeated every so often so it's fresh in our minds. Plus the infinite series seems to give people trouble. :wink:
We don't have to repeat the proof an infinite number of times, do we?!

Nope, exactly .999.... times should do. :lol:

Bad jcsd
2004-Dec-09, 08:34 PM
Here's another one to consider: what is sqrt(0.999...) is it greater than 0.999..? is it less than 1? is it equal to 0.9999..?

worzel
2004-Dec-09, 08:36 PM
edited once for clarity. The challenge for any UB's would be to find fault with any of the individual steps.

Astrorock collector objected to step 5) claiming that 9.99.. - 0.999.. was 8.999.. (which it is, but he obviously doesn;t accept that 8.999... = 9)

Therest was just to show him that his refusal to accept that step results necessarily in a contradiction.
But to say that 9.99.. - 0.99.. = 8.99.. is to say that either 0.99..=1 or 8.99.. is bigger than 9 - or, of course, 0.99.. is bigger than 1, but that's just silly :)

A Thousand Pardons
2004-Dec-09, 08:43 PM
Bad jcsd. First, 9 = 9.000... .
Second, the decimal expansion of 9 + 0.999... is 9.999... .
Third, you state that 0.999... = 1. This is a statement that I cannot agree with.
Let me make sure of your meaning. Are you agreeing with the First and Second statements? Or are you disagreeing with all three?

AstroRockHunter
2004-Dec-09, 08:54 PM
Bad jcsd wrote:

AstroRockHunter wrote:
Bad jcsd. First, 9 = 9.000... .
Second, the decimal expansion of 9 + 0.999... is 9.999... .

That is want I wanted to hear as earlier you stated that 10*0.9999 .. = 9.999...

therfore we can say: 10x = 9 + x therefore 10x - x = 9 therfore 9x = 9 therefore x = 1.

This is true IF AND ONLY IF you do not know what x equals to start. OTOH, if you start with x = 0.999... then your reasoning is flawed because you are no longer dealing with unknown variables, you're simply using variables to save from writing out the numbers.

Bad jcsd
2004-Dec-09, 08:58 PM
This is true IF AND ONLY IF you do not know what x equals to start. OTOH, if you start with x = 0.999... then your reasoning is flawed because you are no longer dealing with unknown variables, you're simply using variables to save from writing out the numbers.

So the knowledge of the person mainipulating the equations has some effect on the result even if you follow precisly the same steps!!!!!!

We are delaing with an unkown as we haven't yet decided whetehr the x is equal to 1 or not.

AstroRockHunter
2004-Dec-09, 08:58 PM
A Thousand Pardons asked:

AstroRockHunter wrote:

Bad jcsd. First, 9 = 9.000... .
Second, the decimal expansion of 9 + 0.999... is 9.999... .
Third, you state that 0.999... = 1. This is a statement that I cannot agree with.

Let me make sure of your meaning. Are you agreeing with the First and Second statements? Or are you disagreeing with all three?

I am agreeing with the First and Second statements.
I am disagreeing with the Third statement.

OALN, I only have 11.000... posts to go!!! #-o

gzhpcu
2004-Dec-09, 08:59 PM
This proof seems to have been mangled amongst the many posters. Is it possible to start from scratch? The proof goes like this:

1. Let x=0.999...
2. Therefore 10x = 9.999...
3. 10x - x = 9x (for all values of x)
4. Therefore 9.999... - 0.999... = 9*(0.999...)
5. But 9.999... - 0.999... = 9
6. Substituting equation 5 into the left hand side of equation 5 gives:
9 = 9*(0.999...)
7. Dividing both sides by 9 gives 1 = 0.999...

edited once for clarity. The challenge for any UB's would be to find fault with any of the individual steps.

I don't agree for the following reason: 0.9999..... is not really well-defined, unless you believe it is equal to one (which I don't). Step 2 tries to multiply a not-well defined number by ten, so the result is also not well-defined..And then there are further steps with undefined numbers. The results of all operations are fuzzy. Mathematicans tried to get out of this problem, by equating0.999..... with 1, but it is only to avoid problems when executing operations.

pghnative
2004-Dec-09, 09:02 PM
ARH,

Which equation in the proof is invalid?

The proof goes like this:

1. Let x=0.999...
2. Therefore 10x = 9.999...
3. 10x - x = 9x (for all values of x)
4. Therefore 9.999... - 0.999... = 9*(0.999...)
5. But 9.999... - 0.999... = 9
6. Substituting equation 5 into the left hand side of equation 5 gives:
9 = 9*(0.999...)
7. Dividing both sides by 9 gives 1 = 0.999...

edited once for clarity. The challenge for any UB's would be to find fault with any of the individual steps.

Normandy6644
2004-Dec-09, 09:02 PM
This proof seems to have been mangled amongst the many posters. Is it possible to start from scratch? The proof goes like this:

1. Let x=0.999...
2. Therefore 10x = 9.999...
3. 10x - x = 9x (for all values of x)
4. Therefore 9.999... - 0.999... = 9*(0.999...)
5. But 9.999... - 0.999... = 9
6. Substituting equation 5 into the left hand side of equation 5 gives:
9 = 9*(0.999...)
7. Dividing both sides by 9 gives 1 = 0.999...

edited once for clarity. The challenge for any UB's would be to find fault with any of the individual steps.

I don't agree for the following reason: 0.9999..... is not really well-defined, unless you believe it is equal to one (which I don't). Step 2 tries to multiply a not-well defined number by ten, so the result is also not well-defined..And then there are further steps with undefined numbers. The results of all operations are fuzzy. Mathematicans tried to get out of this problem, by equating0.999..... with 1, but it is only to avoid problems when executing operations.

Where is the ambiguity in an infinite number of 9's?

Disinfo Agent
2004-Dec-09, 09:03 PM
0.9999..... is not really well-defined, unless you believe it is equal to one (which I don't). Step 2 tries to multiply a not-well defined number by ten, so the result is also not well-defined..And then there are further steps with undefined numbers. The results of all operations are fuzzy. Mathematicans tried to get out of this problem, by equating0.999..... with 1, but it is only to avoid problems when executing operations.
It's defined the same way as all other periodical infinite decimal expansions, like 0.666~ or 0.272727~. Do you think they're ill-defined, too?

Bad jcsd
2004-Dec-09, 09:04 PM
I don't agree for the following reason: 0.9999..... is not really well-defined, unless you believe it is equal to one (which I don't). Step 2 tries to multiply a not-well defined number by ten, so the result is also not well-defined..And then there are further steps with undefined numbers. The results of all operations are fuzzy. Mathematicans tried to get out of this problem, by equating0.999..... with 1, but it is only to avoid problems when executing operations.

A couple of page sback I explained the definition of a recurring decimal like 0.99.. so it is well-defined.

It#'s just a case of choosing the mpost sensible defitnion for recurring decimals. The definition that has been choosen by those who are saying 0.999 ... is not equal to 1 is not the commonluy understood defintuion and it ensures that you are no longer talking about the real numbers or any set of numbers that are known to anyone else except the 'UB'.

Severian
2004-Dec-09, 09:05 PM
I don't agree for the following reason: 0.9999..... is not really well-defined, unless you believe it is equal to one (which I don't). Step 2 tries to multiply a not-well defined number by ten, so the result is also not well-defined..And then there are further steps with undefined numbers. The results of all operations are fuzzy. Mathematicans tried to get out of this problem, by equating0.999..... with 1, but it is only to avoid problems when executing operations.

Sure it is well defined; in fact you can find the definition several times in this thread. Did you mean to say that you don't agree because you don't know how 0.99... is defined?

pghnative
2004-Dec-09, 09:06 PM
This proof seems to have been mangled amongst the many posters. Is it possible to start from scratch? The proof goes like this:

1. Let x=0.999...
2. Therefore 10x = 9.999...
3. 10x - x = 9x (for all values of x)
4. Therefore 9.999... - 0.999... = 9*(0.999...)
5. But 9.999... - 0.999... = 9
6. Substituting equation 5 into the left hand side of equation 5 gives:
9 = 9*(0.999...)
7. Dividing both sides by 9 gives 1 = 0.999...

edited once for clarity. The challenge for any UB's would be to find fault with any of the individual steps.

I don't agree for the following reason: 0.9999..... is not really well-defined, unless you believe it is equal to one (which I don't). Step 2 tries to multiply a not-well defined number by ten, so the result is also not well-defined..And then there are further steps with undefined numbers. The results of all operations are fuzzy. Mathematicans tried to get out of this problem, by equating0.999..... with 1, but it is only to avoid problems when executing operations.

So you're saying that by your definition, 0.999...is not really well defined, but that by a mathamatician's definition, 0.999... equals 1?

Severian
2004-Dec-09, 09:10 PM
Oh, and mathematicians don't make the definition that 0.99... = 1. Mathematicians define 0.99.. to be an equivalence class of Cauchy sequences of rationals, and 1 happens to be in that class. There is a difference between these two statements.

Disinfo Agent
2004-Dec-09, 09:15 PM
This is true IF AND ONLY IF you do not know what x equals to start. OTOH, if you start with x = 0.999... then your reasoning is flawed because you are no longer dealing with unknown variables, you're simply using variables to save from writing out the numbers.

What if I rewrite pghnative's argument (http://www.badastronomy.com/phpBB/viewtopic.php?p=378296#378296) as follows?

1. Consider 0.999...
2. Therefore 10 * 0.999... = 9.999...
3. 10* 0.999... - 0.999... = 9 * 0.999...
4. Therefore 9.999... - 0.999... = 9* (0.999...)
5. But 9.999... - 0.999... = 9
6. Substituting equation 5 into the left hand side of equation 4 gives:
9 = 9*(0.999...)
7. Dividing both sides by 9 gives 1 = 0.999...
No x. Still true.

SeanF
2004-Dec-09, 09:15 PM
Here's another one to consider: what is sqrt(0.999...) is it greater than 0.999..? is it less than 1? is it equal to 0.9999..?
Oh, way to go! [-X

Square roots of numbers between 0 and 1 (non-inclusive) are larger than the number, but less than 1, right?

So the next time somebody asks a non-believer to give a number between 0.9999... and 1, all they've got to do is say sqrt(0.9999...)!

;)

VTBoy
2004-Dec-09, 09:19 PM
Astro Just wondering have you seen my proof using limits. A very valid proof , and it invloves no substitution just a direct proof.

Bad jcsd
2004-Dec-09, 09:21 PM
Here's another one to consider: what is sqrt(0.999...) is it greater than 0.999..? is it less than 1? is it equal to 0.9999..?
Oh, way to go! [-X

Square roots of numbers between 0 and 1 (non-inclusive) are larger than the number, but less than 1, right?

So the next time somebody asks a non-believer to give a number between 0.9999... and 1, all they've got to do is say sqrt(0.9999...)!

;)

Yep, usually theyof course they could come back with the answer that sqrt(0.999...) = 0.999... which is correct, but of course that means that the number is idempotent and the only two idempotent numbers in any field are 0 and 1!

SeanF
2004-Dec-09, 09:21 PM
Bad jcsd wrote:

AstroRockHunter wrote:
Bad jcsd. First, 9 = 9.000... .
Second, the decimal expansion of 9 + 0.999... is 9.999... .

That is want I wanted to hear as earlier you stated that 10*0.9999 .. = 9.999...

therfore we can say: 10x = 9 + x therefore 10x - x = 9 therfore 9x = 9 therefore x = 1.

This is true IF AND ONLY IF you do not know what x equals to start. OTOH, if you start with x = 0.999... then your reasoning is flawed because you are no longer dealing with unknown variables, you're simply using variables to save from writing out the numbers.
You're missing the point.

10x = 9+x if and only if x=1, right? Any other value of x, and 10x &lt;> 9+x.

But you've already acknowledged that:
10 * 0.9999... = 9.9999...
and that:
9 + 0.9999...=9.9999...

This means that:
10 * 0.9999... = 9 + 0.9999...

But we already know that 10x = 9+x only when x=1, so . . .

AstroRockHunter
2004-Dec-09, 09:43 PM
Disinfo Agent wrote:

AstroRockHunter wrote:
This is true IF AND ONLY IF you do not know what x equals to start. OTOH, if you start with x = 0.999... then your reasoning is flawed because you are no longer dealing with unknown variables, you're simply using variables to save from writing out the numbers.

What if I rewrite pghnative's argument as follows?

pghnative wrote:
1. Consider 0.999...
2. Therefore 10 * 0.999... = 9.999...
3. 10* 0.999... - 0.999... = 9 * 0.999...
4. Therefore 9.999... - 0.999... = 9* (0.999...)
5. But 9.999... - 0.999... = 9
6. Substituting equation 5 into the left hand side of equation 5 gives:
9 = 9*(0.999...)
7. Dividing both sides by 9 gives 1 = 0.999...

No x. Still true.

Sorry, but there seems to be a nasty little mistake between steps 2 and 3.
2 Therefore 10(0.999...) = 9.999...
3 10(0.999...) -0.999... = 9(0.999...)

For this to be true, then line 2 must be
2 Therefore 10(0.999...) = 9 + 0.999...
then line 3 becomes
3 10(0.999...) - 0.999... = 9

For you to be able to subtract the 0.999... from the left side, it must have been added to the right side. In the example used, it it is not.

Disinfo Agent
2004-Dec-09, 09:48 PM
Replace line 3 with:

10* 0.999... - 0.999... = (10 - 1) * 0.999... = 9 * 0.999...

ToSeek
2004-Dec-09, 09:49 PM
Sorry, but there seems to be a nasty little mistake between steps 2 and 3.
2 Therefore 10(0.999...) = 9.999...
3 10(0.999...) -0.999... = 9(0.999...)

Step 3 is not supposed to follow from Step 2; it's an independent statement. To return to the world of x's:

3. 10x - x = 9x

Surely a statement that does not need to be justified.

AstroRockHunter
2004-Dec-09, 09:49 PM
SeanF wrote:

You're missing the point.

10x = 9+x if and only if x=1, right? Any other value of x, and 10x &lt;> 9+x.

But you've already acknowledged that:
10 * 0.9999... = 9.9999...
and that:
9 + 0.9999...=9.9999...

This means that:
10 * 0.9999... = 9 + 0.9999...

But we already know that 10x = 9+x only when x=1, so . . .
_________________

Actually, we're saying the same thing.

In the equation 10x = 9 + x, x can only equal one value. It CANNOT equal one value on one side of the equal sign and another value on the other side.

Zjm7891
2004-Dec-09, 09:51 PM
While .999 doesn't equal 1, for algebraic purposes it DOES come infinitly close so for algebra you just ignore the minor difference. However, in calculus the difference can be significant in determining the answer. Since with limits we get infinitly close to a value without reaching it... use the equation (1-x)/(x-1)... if you try to use the value 1 in the equation it fails sice it becomes n/0... but if you use .999.., or as x approachs 1 as you would say with the limit, it comes out as -1... explain that using fuzzy algebra. Also, there is a reason why limits have a left and a right side...... from the left .9999... from the right 1.000...1

Bad jcsd
2004-Dec-09, 09:53 PM
SeanF wrote:

You're missing the point.

10x = 9+x if and only if x=1, right? Any other value of x, and 10x &lt;> 9+x.

But you've already acknowledged that:
10 * 0.9999... = 9.9999...
and that:
9 + 0.9999...=9.9999...

This means that:
10 * 0.9999... = 9 + 0.9999...

But we already know that 10x = 9+x only when x=1, so . . .
_________________

Actually, we're saying the same thing.

In the equation 10x = 9 + x, x can only equal one value. It CANNOT equal one value on one side of the equal sign and another value on the other side.

Exactly, and simple algebra tells us that value is 1.

This is precisely the problem that people have who don'tbsee why 0.99.., = 1 they don't see that two different decimal expansions can have the same value, even thoguh they can except that the apparently different fractions can have the same value. Decimal rperesntations are not the rea numbers they are repreensations of real numbers.

Disinfo Agent
2004-Dec-09, 09:55 PM
In the equation 10x = 9 + x, x can only equal one value. It CANNOT equal one value on one side of the equal sign and another value on the other side.
It's not an other value. It's the same value. :)

use the equation (1-x)/(x-1)... if you try to use the value 1 in the equation it fails sice it becomes n/0... but if you use .999.., or as x approachs 1 as you would say with the limit, it comes out as -1...
You are mistaken. If you plug 0.999... into that expression, it also becomes 0/0. Try it.

Severian
2004-Dec-09, 09:57 PM
While .999 doesn't equal 1, for algebraic purposes it DOES come infinitly close so for algebra you just ignore the minor difference. However, in calculus the difference can be significant in determining the answer. Since with limits we get infinitly close to a value without reaching it... use the equation (1-x)/(x-1)... if you try to use the value 1 in the equation it fails sice it becomes n/0... but if you use .999.., or as x approachs 1 as you would say with the limit, it comes out as -1... explain that using fuzzy algebra
Nope. You have assumed that 0.99... is not equal to one to make this statement. To summarize your argument:
1) 0.99.. not equal to 1
2) f(x) = (1-x)/(x-1) = -1 for x not equal to 1
3) f(0.99..) = -1
4) 0.99... not equal to 1

Edit to add: If you took calculus, you should have learned that there is no number infinitely close to 1 that isn't 1. That is to say, if a number is not 1, then it is some positive distance from 1.

AstroRockHunter
2004-Dec-09, 09:57 PM
To Seek wrote

Step 3 is not supposed to follow from Step 2; it's an independent statement. To return to the world of x's:

3. 10x - x = 9x

Surely a statement that does not need to be justified.

How is it an independent statement???
If it is, how do you justify subtracting 0.999... from one side and not the other???

Then, taking your step 3;

OK, then solving for x;

10x - x =9x
9x = 9x
9x/9 = 9x/9
x = x
or
9x/x = 9x/x
9 = 9

pghnative
2004-Dec-09, 09:59 PM
SeanF wrote:

You're missing the point.

10x = 9+x if and only if x=1, right? Any other value of x, and 10x &lt;> 9+x.

But you've already acknowledged that:
10 * 0.9999... = 9.9999...
and that:
9 + 0.9999...=9.9999...

This means that:
10 * 0.9999... = 9 + 0.9999...

But we already know that 10x = 9+x only when x=1, so . . .
_________________

Actually, we're saying the same thing.

In the equation 10x = 9 + x, x can only equal one value. It CANNOT equal one value on one side of the equal sign and another value on the other side.

No one is doing that. (At least, I am not doing that.)

This, again, is the proof, with a clarifying phrase (bolded) added to step 3 and a pointed remark (also bolded) added to step 4:

1. Let x=0.999...
2. Therefore 10x = 9.999...
3. It is known that 10x - x = 9x (for all values of x)
4. Therefore 9.999... - 0.999... = 9*(0.999...) Note that 0.999... is substituted for all "x's"
5. But 9.999... - 0.999... = 9
6. Substituting equation 5 into the left hand side of equation 5 gives:
9 = 9*(0.999...)
7. Dividing both sides by 9 gives 1 = 0.999...

pghnative
2004-Dec-09, 10:02 PM
To Seek wrote

Step 3 is not supposed to follow from Step 2; it's an independent statement. To return to the world of x's:

3. 10x - x = 9x

Surely a statement that does not need to be justified.

How is it an independent statement???
If it is, how do you justify subtracting 0.999... from one side and not the other???

Then, taking your step 3;

OK, then solving for x;

10x - x =9x
9x = 9x
9x/9 = 9x/9
x = x
or
9x/x = 9x/x
9 = 9

How is it an independant statement??? :o :o :o

It just is. 10x - x = 9x, for all values of x. It stands on it's own, to be used in combination with equation 2 to generate equation 4.

Bad jcsd
2004-Dec-09, 10:03 PM
While .999 doesn't equal 1, for algebraic purposes it DOES come infinitly close so for algebra you just ignore the minor difference. However, in calculus the difference can be significant in determining the answer. Since with limits we get infinitly close to a value without reaching it... use the equation (1-x)/(x-1)... if you try to use the value 1 in the equation it fails sice it becomes n/0... but if you use .999.., or as x approachs 1 as you would say with the limit, it comes out as -1... explain that using fuzzy algebra

Ye si t does equal 1 EXACTLY and in the real numbers there is no such thing as infinntely close.

(1-x)/(x-1) is not an equation and it's not even clera thta it's a function. if your tlaking about the function f(x) = (1-x)/(x-1), then the limit does equal -1 at 1, but the relevance of limits and functions (except series) is nil in this situation. Your are simply confused about the concpet of an infinitesmial, infinitesimals are not real numbers and the decimal expansion 0.999.. iis not related to them in the slightest.

Even in sets which include the rela numbers and numbers that cna be thought of as infitely smaller than the real numbers 0.999.. is stil equal to 1!!

Severian
2004-Dec-09, 10:03 PM
How is it an independent statement???

Because it is basic algebra.

9 x + 1 x = (9 + 1) x = 10 x

Edit: heh wrote not quite the equation:
10 x - 1x = (10 - 1) x = 9 x

Disinfo Agent
2004-Dec-09, 10:04 PM
Also, there is a reason why limits have a left and a right side......
I'm curious about this, too. Do you think you could explain to me what a left-hand limit and a right-hand limit are? Because I think you have some misconceptions about those concepts.

pghnative
2004-Dec-09, 10:07 PM
ARH
I've added two more steps in an attempt to make it more clear.

1. Let x=0.999...
2. Therefore 10x = 9.999... [that is, 10*(0.999...)=(9.999...)]
3. It is known that 10x - x = 9x (for all values of x)
3b. Therefore 10*(0.999...) - (0.999...) = 9 * (0.999...)
3c. But by equation 2, 10*(0.999)=9.999... Substituting gives
4. 9.999... - 0.999... = 9*(0.999...)
5. But 9.999... - 0.999... = 9
6. Substituting equation 5 into the left hand side of equation 5 gives:
9 = 9*(0.999...)
7. Dividing both sides by 9 gives 1 = 0.999...

Severian
2004-Dec-09, 10:11 PM
Maybe you should say in step 3 10 y - y = 9 y for all values of y; in particular for y = x = 0.99...
;)

AstroRockHunter
2004-Dec-09, 10:18 PM
pghnative wrote:

This, again, is the proof, with a clarifying phrase (bolded) added to step 3 and a pointed remark (also bolded) added to step 4:

1. Let x=0.999...
2. Therefore 10x = 9.999...
3. It is known that 10x - x = 9x (for all values of x)
4. Therefore 9.999... - 0.999... = 9*(0.999...) Note that 0.999... is substituted for all "x's"
5. But 9.999... - 0.999... = 9
6. Substituting equation 5 into the left hand side of equation 5 gives:
9 = 9*(0.999...)
7. Dividing both sides by 9 gives 1 = 0.999...

Or:

1. Let x=0.999...
2. Therefore 10x = 9.999...
3. It is known that 10x - x = 9x (for all values of x)
4. Therefore 9.999... - 0.999... = 9*(0.999...) Note that 0.999... is substituted for all "x's"
5. But 9.999... - 0.999... = 9
6. Substituting equation 5 into the left hand side of equation 5 gives:
9 = 9*(0.999...)
7. But, 9*(0.999...) = 8.999...
8. Substituting equation 7 into the right hand side of equation 6 gives:
9 = 8.999...

As I learned arithmetic, longer ago that I care to admit, once you substitute for the x's, that's it. You're DONE except for getting the values on both sides of the equation to determine if they are equal.

Bad jcsd
2004-Dec-09, 10:21 PM
Your takning different steps, NOT disproving the steps that have been taken.

Zjm7891
2004-Dec-09, 10:21 PM
use the equation (1-x)/(x-1)... if you try to use the value 1 in the equation it fails sice it becomes n/0... but if you use .999.., or as x approachs 1 as you would say with the limit, it comes out as -1...
You are mistaken. If you plug 0.999... into that expression, it also becomes 0/0. Try it.

Refresh yourself on L'Hôpital's Rule

However, .99999... in the equation spits out -1

Severian
2004-Dec-09, 10:21 PM
Ok, so if we put step 3 up as step 0 would you be happy?

Bad jcsd
2004-Dec-09, 10:23 PM
use the equation (1-x)/(x-1)... if you try to use the value 1 in the equation it fails sice it becomes n/0... but if you use .999.., or as x approachs 1 as you would say with the limit, it comes out as -1...
You are mistaken. If you plug 0.999... into that expression, it also becomes 0/0. Try it.

Refresh yourself on L'Hôpital's Rule

Refresh yourself on the differnce between a function and an expression.

Severian
2004-Dec-09, 10:24 PM
Refresh yourself on L'Hôpital's Rule

Everyone agrees that the limit as x approaches 1 from the left (or hey from the right) of f(x) = (1-x)/(x-1) is -1.

However, you have the mistaken belief that there is some number not equal to 1 that is closer to one than any other number, and that you can plug that in to the equation to get the limit. That isn't how limits work (even left and right limits).

Edit - Oh, and by the way: refresh yourself on the definition of the set of real numbers.

ToSeek
2004-Dec-09, 10:30 PM
(post contents removed as superfluous)

ToSeek
2004-Dec-09, 10:32 PM
Or:

1. Let x=0.999...
2. Therefore 10x = 9.999...
3. It is known that 10x - x = 9x (for all values of x)
4. Therefore 9.999... - 0.999... = 9*(0.999...) Note that 0.999... is substituted for all "x's"
5. But 9.999... - 0.999... = 9
6. Substituting equation 5 into the left hand side of equation 5 gives:
9 = 9*(0.999...)
7. But, 9*(0.999...) = 8.999...
8. Substituting equation 7 into the right hand side of equation 6 gives:
9 = 8.999...

9. Subtracting 8 from each side:
1 = 0.999...

Normandy6644
2004-Dec-09, 10:32 PM
use the equation (1-x)/(x-1)... if you try to use the value 1 in the equation it fails sice it becomes n/0... but if you use .999.., or as x approachs 1 as you would say with the limit, it comes out as -1...
You are mistaken. If you plug 0.999... into that expression, it also becomes 0/0. Try it.

Refresh yourself on L'Hôpital's Rule

I don't see how you can used a calculus derived concept to justify a discrepancy in something that is most rigorously proved by calculus. You can't have one but not the other...

Zjm7891
2004-Dec-09, 10:36 PM
4. 9.999... - 0.999 ... = 9 (0.999...)

But obviously
5. 9.999... - 0.999 ... = 9

actually you just dropped part of the right side of the equation... the .9999999...

Severian
2004-Dec-09, 10:38 PM
4. 9.999... - 0.999 ... = 9 (0.999...)

But obviously
5. 9.999... - 0.999 ... = 9

actually you just dropped part of the right side of the equation... the .9999999...

What are you talking about?

ToSeek
2004-Dec-09, 10:42 PM
4. 9.999... - 0.999 ... = 9 (0.999...)

But obviously
5. 9.999... - 0.999 ... = 9

actually you just dropped part of the right side of the equation... the .9999999...

(This is from my superfluous post, btw, though the same reasoning is used in the remaining proofs.) Again, step 5 is independent - it's being deduced separately from step 4. As with step 3, there's no derivation to argue with. If you think the statement is false, you need to explain why.

Severian
2004-Dec-09, 10:45 PM
Ah! So he just thought you were repeating yourself incorrectly from one line to the next. And that the entire process of going from step 4 to step 5 was to incorrectly repeat yourself :P

Zjm7891
2004-Dec-09, 10:48 PM
4. 9.999... - 0.999 ... = 9 (0.999...)

But obviously
5. 9.999... - 0.999 ... = 9

actually you just dropped part of the right side of the equation... the .9999999...

What are you talking about?
.... I think I just got postings out of order... sorry I am reading this on a very small screen... 240x320 pixels

AstroRockHunter
2004-Dec-09, 10:50 PM
To Seek wrote

AstroRockHunter wrote:

Or:

1. Let x=0.999...
2. Therefore 10x = 9.999...
3. It is known that 10x - x = 9x (for all values of x)
4. Therefore 9.999... - 0.999... = 9*(0.999...) Note that 0.999... is substituted for all "x's"
5. But 9.999... - 0.999... = 9
6. Substituting equation 5 into the left hand side of equation 5 gives:
9 = 9*(0.999...)
7. But, 9*(0.999...) = 8.999...
8. Substituting equation 7 into the right hand side of equation 6 gives:
9 = 8.999...

9. Subtracting 8 from each side:
1 = 0.999...

Then, using your reasoning:

9. Subtracting 9 from each side:
0 = -0.999...

If your are correct and 1 = 0.999...

shouldn't -0.999... = -1 ???

Bad jcsd
2004-Dec-09, 10:52 PM
yes, -0.9999... does equal -1

Bad jcsd
2004-Dec-09, 10:54 PM
Any rational number can be represented as a recurring decimal.

AstroRockHunter
2004-Dec-09, 10:57 PM
Bad jcsd wrote:

yes, -0.9999... does equal -1

In reply to what I wrote:

To Seek wrote
Quote:
AstroRockHunter wrote:

Or:

1. Let x=0.999...
2. Therefore 10x = 9.999...
3. It is known that 10x - x = 9x (for all values of x)
4. Therefore 9.999... - 0.999... = 9*(0.999...) Note that 0.999... is substituted for all "x's"
5. But 9.999... - 0.999... = 9
6. Substituting equation 5 into the left hand side of equation 5 gives:
9 = 9*(0.999...)
7. But, 9*(0.999...) = 8.999...
8. Substituting equation 7 into the right hand side of equation 6 gives:
9 = 8.999...

9. Subtracting 8 from each side:
1 = 0.999...

Then, using your reasoning:

9. Subtracting 9 from each side:
0 = -0.999...

If your are correct and 1 = 0.999...

shouldn't -0.999... = -1 ???

I just showed, using some of the extremly fuzzy arithmetic rules here, that 0 = -0.999... .

How can you reconcile this???

Severian
2004-Dec-09, 10:57 PM
9. Subtracting 8 from each side:
1 = 0.999...

Then, using your reasoning:

9. Subtracting 9 from each side:
0 = -0.999...

If your are correct and 1 = 0.999...

shouldn't -0.999... = -1 ???

er, check your math on 8.99... - 9

note 8.9 - 9 = -0.1
8.99 - 9 = -0.01

mutineer
2004-Dec-09, 10:59 PM
2. Therefore 10x = 9.999... [that is, 10*(0.999...)=(9.999...)]
Trouble is, if you begin with the opinion that 0.999... represents an infinitesimal amount less than 1, you have a term which like other infinites and infinitesimals cannot be manipulated by arithmetic. This number may not be multiplied - any more than you may multiply the infinity symbol or its reciprocal. Therein lies the case for substituting it with 1, with which for most practical purposes it is equivalent, if your desire is for a usable system of numbers.

AstroRockHunter
2004-Dec-09, 11:00 PM
Severian:

You're right, my bad :oops:

ToSeek
2004-Dec-09, 11:00 PM
Then, using your reasoning:

9. Subtracting 9 from each side:
0 = -0.999...

Your subtraction is wrong. 8.999... - 9 != -0.999... but -0.000...

Look at successive approximations if you don't believe me:

8.9 - 9 = -0.1
8.99 - 9 = -0.01
8.999 - 9 = -0.001
...

EDIT: ToSeeked by Severian!!

Severian
2004-Dec-09, 11:01 PM
2. Therefore 10x = 9.999... [that is, 10*(0.999...)=(9.999...)]
Trouble is, if you begin with the opinion that 0.999... represents an infinitesimal amount less than 1, you have a term which like other infinites and infinitesimals cannot be manipulated by arithmetic. This number may not be multiplied - any more than you may multiply the infinity symbol or its reciprocal. Therein lies the case for substituting it with 1, with which for most practical purposes it is equivalent, if your desire is for a usable system of numbers.

Oh, I see. So if you begin with the assumption that 0.99.. is not equal to 1, then you run into trouble. Yes, I agree with that.

Bad jcsd
2004-Dec-09, 11:03 PM
Mutineer:

1) decimal expansions are ONLY used to represent real numbers

2) infinitesmials are NOT real numbers. not the case of the nx bets thing, it's the case theat trhe most reasonable and useful defitnion of decmal expansions means that 0.99.. = 1

AstroRockHunter
2004-Dec-09, 11:07 PM
This reminds me of the joke about the difference between a mathmatician and an engineer.

The mathmatician thinks "If I go half the distance, then half of that, then half of that ... , I'll NEVER arrive there!!!"

The engineer thinks "If I go half the distance, then half of that, then half of that ... , I'll GET CLOSE ENOUGH!!!"

And on that note, I'd better get some work done!!! :cry:

Normandy6644
2004-Dec-09, 11:16 PM
This reminds me of the joke about the difference between a mathmatician and an engineer.

The mathmatician thinks "If I go half the distance, then half of that, then half of that ... , I'll NEVER arrive there!!!"

The engineer thinks "If I go half the distance, then half of that, then half of that ... , I'll GET CLOSE ENOUGH!!!"

And on that note, I'd better get some work done!!! :cry:

Ah Zeno's paradox. Only problem is that the sum of the series (1/2)^n=1. :D

worzel
2004-Dec-09, 11:59 PM
This is true IF AND ONLY IF you do not know what x equals to start. OTOH, if you start with x = 0.999... then your reasoning is flawed because you are no longer dealing with unknown variables, you're simply using variables to save from writing out the numbers.
So let me get this straight, an equation involving x can be true for all x unless (or until ???) we choose a value for x? Or did you mean it is true for all x except 0.99..?

8. Substituting equation 7 into the right hand side of equation 6 gives:
9 = 8.999...
Ah, so you do agree after all :)

worzel
2004-Dec-10, 12:00 AM
Ah Zeno's paradox. Only problem is that the sum of the series (1/2)^n=1. :D

Or in binary:

0.111... = 1 :D

AstroRockHunter
2004-Dec-10, 12:19 AM
worzel wrote:

So let me get this straight, an equation involving x can be true for all x unless (or until ???) we choose a value for x? Or did you mean it is true for all x except 0.99..?

No. An equation involving x can be true for all x until we determine what values of x make the equation true.

However, if we define x to equal a value, then it is no longer a variable and cannot be used as one because it's value can't be solved for as it is already is defined. (How's THAT for a circular statement??? :))

Let me leave you with this:

10x = 9.000... + 0.999...

Solve for x.

VTBoy
2004-Dec-10, 12:39 AM
worzel wrote:

So let me get this straight, an equation involving x can be true for all x unless (or until ???) we choose a value for x? Or did you mean it is true for all x except 0.99..?

No. An equation involving x can be true for all x until we determine what values of x make the equation true.

Solve for x.

Incorrect. If an equation is true for all x, then no matter what x we pick it stil holds true. All possible values of x make the equation true. For 10x-x=9x this is true for all reals, including 0.9999999....

01101001
2004-Dec-10, 12:40 AM
10x = 9.000... + 0.999...

Solve for x.

multiply both sides by 10: 100x=90 + 9.999...

subtract first equation from second: 90x = 81 + 9 = 90

divide by 90: x = 90/90 = 1, also known as sum(n, 1, infinity, 9/(10^n))

worzel
2004-Dec-10, 12:43 AM
No. An equation involving x can be true for all x until we determine what values of x make the equation true.

However, if we define x to equal a value, then it is no longer a variable and cannot be used as one because it's value can't be solved for as it is already is defined. (How's THAT for a circular statement??? :))
That's not circular, that's just plain illogical, just think about what you're saying.

That it can be true for all x means we have already determined that the values of x that make the equation true are all the values x could take, including the value 0.99..

Let me leave you with this:

10x = 9.000... + 0.999...

Solve for x.
x=0.99... so what, that doesn't show that 0.99.. &lt;> 1.

If you agree that:

1. 10y-y = 9y for all y

Then you must agree that substituting in any value for y in 1. yeilds a true statement so you must be happy with substituting in y = 0.99.. which gives

2. 10 x 0.99.. - 0.99.. = 9 x 0.99..

If you also agree that

3. 10 x 0.99.. = 9.99..

then you must be happy substituting in 9.99.. for 10 x 0.99.. on the left hand side of 2. giving

4. 9.99.. - 0.99.. = 9 x 0.99..

If you aslo agree that

5. 9.99.. - 0.99.. = 9

then you must be happy substituting in 9 for 9.99.. - 0.99.. on the left hand side of 4. giving

6. 9 = 9 x 0.99..

If you agree that

7. 9z / 9 = z

Then you must be happy with dividing the both sides of 6. by 9 giving

8. 1 = 0.99..

I have, of course, assumed that you think 0.99.. is a number, and that 9/9 =1.

Which statement/s do you disagree with?

EDIT: to add, you have already agreed with 1, 3 and 5 in this thread. 2, 4 and 6 are just substitutions in an equation you agree is always true of one expression for another that you have already agreed are equal. Presumably you agree that from 6, 7 and 8 are trivial.

gzhpcu
2004-Dec-10, 02:33 AM
This proof seems to have been mangled amongst the many posters. Is it possible to start from scratch? The proof goes like this:

1. Let x=0.999...
2. Therefore 10x = 9.999...
3. 10x - x = 9x (for all values of x)
4. Therefore 9.999... - 0.999... = 9*(0.999...)
5. But 9.999... - 0.999... = 9
6. Substituting equation 5 into the left hand side of equation 5 gives:
9 = 9*(0.999...)
7. Dividing both sides by 9 gives 1 = 0.999...

edited once for clarity. The challenge for any UB's would be to find fault with any of the individual steps.

I don't agree for the following reason: 0.9999..... is not really well-defined, unless you believe it is equal to one (which I don't). Step 2 tries to multiply a not-well defined number by ten, so the result is also not well-defined..And then there are further steps with undefined numbers. The results of all operations are fuzzy. Mathematicans tried to get out of this problem, by equating0.999..... with 1, but it is only to avoid problems when executing operations.

So you're saying that by your definition, 0.999...is not really well defined, but that by a mathamatician's definition, 0.999... equals 1?

Yes, exactly. Holds for 1 divided by 3 also, for example.

gzhpcu
2004-Dec-10, 02:43 AM
IMHO: the reason for this long thread, is that some people believe that the mathematician's definitions are true, while others do not. In my view, some people are accepting that 0.99999999999....... equals 1 just based on mathematical definitions. (Sounds almost like a religion to me.. :D ) I do not accept it, because I have yet to see proof. If the proof were really there, this thread would not exist.

Normandy6644
2004-Dec-10, 02:50 AM
IMHO: the reason for this long thread, is that some people believe that the mathematician's definitions are true, while others do not. In my view, some people are accepting that 0.99999999999....... equals 1 just based on mathematical definitions. (Sounds almost like a religion to me.. :D ) I do not accept it, because I have yet to see proof. If the proof were really there, this thread would not exist.

No, we're using the tools of mathematics to show an interesting consequence. There is no ambiguity in any of it, and the proof has been shown to you many times, but you're unwilling to accept it because it defies your "intuition," something you can't use when dealing with this kind of thing. What problem do you have with the infinite series proof?

gzhpcu
2004-Dec-10, 06:37 AM
IMHO: the reason for this long thread, is that some people believe that the mathematician's definitions are true, while others do not. In my view, some people are accepting that 0.99999999999....... equals 1 just based on mathematical definitions. (Sounds almost like a religion to me.. :D ) I do not accept it, because I have yet to see proof. If the proof were really there, this thread would not exist.

No, we're using the tools of mathematics to show an interesting consequence. There is no ambiguity in any of it, and the proof has been shown to you many times, but you're unwilling to accept it because it defies your "intuition," something you can't use when dealing with this kind of thing. What problem do you have with the infinite series proof?

I guess "you can lead a horse to water but you can't make it drink", I still can not, as you correctly stated, accept because it runs against my intuition, despite all the "proofs" presented. To me, just looking at 0.99999999........ = 1, does not make sense, irregardless of all the counterarguments. Sorry.. :(

01101001
2004-Dec-10, 08:43 AM
I guess "you can lead a horse to water but you can't make it drink", I still can not, as you correctly stated, accept because it runs against my intuition, despite all the "proofs" presented. To me, just looking at 0.99999999........ = 1, does not make sense, irregardless of all the counterarguments.
Is your intuition always right? What would you accept as proof that 0.999... equals 1? Anything?

Disinfo Agent
2004-Dec-10, 09:51 AM
use the equation (1-x)/(x-1)... if you try to use the value 1 in the equation it fails sice it becomes n/0... but if you use .999.., or as x approachs 1 as you would say with the limit, it comes out as -1...
You are mistaken. If you plug 0.999... into that expression, it also becomes 0/0. Try it.
However, .99999... in the equation spits out -1
No, it does not.

1 - 0.999... =
= 1 - lim (0.9, 0.99, 0.999, ...)
= lim (1 - 0.9, 1 - 0.99, 1 - 0.999, ...)
= lim (0.1, 0.01, 0.001, ...)
= lim (0.1^n)
= 0

0.999... - 1 = - 0 =0

0/0

P.S. And how about replying to the questions posed to you?

Disinfo Agent
2004-Dec-10, 10:04 AM
To me, just looking at 0.99999999........ = 1, does not make sense, irregardless of all the counterarguments. Sorry.. :(
It may take you a while to get used to the idea, but perhaps you'll eventually see it's what makes the most sense. Meanwhile, let me just remind you that "0.99999999........" and "1" are not numbers per se; they're just symbols. Obviously, as symbols they are different, but what we're saying is that the numbers denoted by these symbols are the same.

Finite decimal expansions (in base 10) are defined like this:

"abc.def" means a*10^2 + b*10^1 + c*10^0 + d*10^(-1) + e*10^(-2) + f*10^(-2), and so on.

It's a very natural extension to say that an infinite expansion:

"abc.defg..." means a*10^2 + b*10^1 + c*10^0 + d*10^(-1) + e*10^(-2) + f*10^(-2) + g*10^(-3) + ...

In fact, it's the only natural extension (in the sense that it preserves the usual algebraic properties of the real numbers).

A Thousand Pardons
2004-Dec-10, 10:20 AM
While .999 doesn't equal 1, for algebraic purposes it DOES come infinitly close so for algebra you just ignore the minor difference.
I guess you meant ".999..."?

If so, then I would disagree. It does equal 1.

We think. :)

worzel
2004-Dec-10, 10:27 AM
IMHO: the reason for this long thread, is that some people believe that the mathematician's definitions are true, while others do not. In my view, some people are accepting that 0.99999999999....... equals 1 just based on mathematical definitions. (Sounds almost like a religion to me.. :D ) I do not accept it, because I have yet to see proof. If the proof were really there, this thread would not exist.
Mathematical truth is a completely different animal to any other sort of truth.

You speak of mathemtatical definitions and the truth of the question of this thread as if they are different things, they are not. A mathematical proof is always just an extension of this sort of reasoning:

If A implies B
and A is true
then B is true.

Pure logic, nothing more.

Take all the axioms of the real numbers, the definition of a decimal, and put them together like this

A &amp; B &amp;C &amp; &amp; E &amp; ...

If the logical consequence of all these axioms is some statement Z then you can write this as

IF (A &amp; B &amp;C &amp; &amp; E &amp; ...)
are all true
then Z is true

or

(A &amp; B &amp;C &amp; &amp; E &amp; ...) => Z

That is all that is meant by

From the axioms of the real numbers and the definition of a decimal it is true that
0.99.. = 1

As has been said many times, the only way out is to redefine the real numbers or the definition of a decimal number. Trying to talk about the objective reality of 0.99.. is meaningless, it means nothing mathematically unless well defined. If we chose to define any string of digits followed by a period followed by any string of digits (including infinitely many) as a the series given many times previously then to be consistent we have to conclude that 0.99..=1.

Disinfo Agent
2004-Dec-10, 10:33 AM
From the axioms of the real numbers and the definition of a decimal it is true that
0.99.. = 1
But I think what gzhpcu feels is that perhaps there's something wrong with the current axioms and definitions.

Disinfo Agent
2004-Dec-10, 10:38 AM
Trouble is, if you begin with the opinion that 0.999... represents an infinitesimal amount less than 1, you have a term which like other infinites and infinitesimals cannot be manipulated by arithmetic.
Not true. Infinitesimals and infinities can be algebraically manipulated.

This number may not be multiplied - any more than you may multiply the infinity symbol or its reciprocal.
The reciprocal of infinity is zero. We do manipulate zero algebraically, all the time!
And, if you've ever taken calculus, you should know that infinity can also be manipulated algebraically.

worzel
2004-Dec-10, 10:49 AM
From the axioms of the real numbers and the definition of a decimal it is true that
0.99.. = 1
But I think what gzhpcu feels is that perhaps there's something wrong with the current axioms and definitions.
But he keeps stating that they aren't really equal even though they are mathematically. That is just a pure constradiction which I think is born from a belief that these decimal representations have some sort of platonic existence outside of their mathematical description. I have come across this misunderstanding about maths quite often. I think it is because we are so used to taking all the assumptions (axioms) for granted that some people don't realize they are making them and therefore think they are talking about real objects (rather than just real numbers :) )

mutineer
2004-Dec-10, 10:52 AM
I thought I had understood the concept of a “limit” since before most BABBers were born. Reading some of the posts to this thread makes me wonder whether maths is taught differently these days. However, having had a chance to glance at a couple of textbooks, it seems that things have not changed since my own schooldays. It provides a means of proposing/asserting that the value of an expression f(x) gets close to y when x gets close to a given value, or approaches infinity.

0.9 + 0.09 + 0.009 . . . has a limit value of 1 as the number of terms approaches infinity. That means it gets close to 1.

Or in binary:
0.111... = 1
This was precisely the equation I had in mind when I asked the question about successive halving. As we add successive ones here, the difference between the left and right sides of the equation halves. But successive halving of an object never causes it to vanish (i.e. assume a value of zero). Aside from offending common sense, it would create a mathematical paradox. The equation therefore can never be true.

worzel
2004-Dec-10, 10:54 AM
But successive halving of an object never causes it to vanish (i.e. assume a value of zero). Aside from offending common sense, it would create a mathematical paradox. The equation therefore can never be true.
Can you expand on the mathematical paradox that would be created?

Disinfo Agent
2004-Dec-10, 10:56 AM
I thought I had understood the concept of a “limit” since before most BABBers were born. Reading some of the posts to this thread makes me wonder whether maths is taught differently these days. However, having had a chance to glance at a couple of textbooks, it seems that things have not changed since my own schooldays. It provides a means of proposing/asserting that the value of an expression f(x) gets close to y when x gets close to a given value, or approaches infinity.

0.9 + 0.09 + 0.009 . . . has a limit value of 1 as the number of terms approaches infinity. That means it gets close to 1.
What's your point?

Or in binary:
0.111... = 1
This was precisely the equation I had in mind when I asked the question about successive halving. As we add successive ones here, the difference between the left and right sides of the equation halves. But successive halving of an object never causes it to vanish (i.e. assume a value of zero).
Never say never in mathematics. It does, if you extend the halving process to infinity. :)

A Thousand Pardons
2004-Dec-10, 11:22 AM
I thought I had understood the concept of a “limit” since before most BABBers were born. Reading some of the posts to this thread makes me wonder whether maths is taught differently these days.

Ad hominem.

And ironic

However, having had a chance to glance at a couple of textbooks, it seems that things have not changed since my own schooldays. It provides a means of proposing/asserting that the value of an expression f(x) gets close to y when x gets close to a given value, or approaches infinity.

0.9 + 0.09 + 0.009 . . . has a limit value of 1 as the number of terms approaches infinity. That means it gets close to 1.
And if you let the limit pass to infinity, it is 1.

The limit of (x-1)/(x-1) as x approaches 1 is equal to 1, although x can never equal 1. It doesn't matter that you can never have an infinite number of 9s--the limit does equal one.

mutineer
2004-Dec-10, 01:59 PM
Trouble is, if you begin with the opinion that 0.999... represents an infinitesimal amount less than 1, you have a term which like other infinites and infinitesimals cannot be manipulated by arithmetic.
Not true. Infinitesimals and infinities can be algebraically manipulated.
Well let's take an example. If time continues forever, there will be an infinite number of Sundays in the future. There will (in some sense) be five times as many weekdays. But one cannot multiply infinity by five. The matter lies outside the bounds of arithmetic.

worzel
2004-Dec-10, 02:03 PM
Trouble is, if you begin with the opinion that 0.999... represents an infinitesimal amount less than 1, you have a term which like other infinites and infinitesimals cannot be manipulated by arithmetic.
Not true. Infinitesimals and infinities can be algebraically manipulated.
Well let's take an example. If time continues forever, there will be an infinite number of Sundays in the future. There will (in some sense) be five times as many weekdays. But one cannot multiply infinity by five. The matter lies outside the bounds of arithmetic.

I think you can actually, infinity times 5 equals infinity.

mutineer
2004-Dec-10, 02:19 PM
But successive halving of an object never causes it to vanish (i.e. assume a value of zero). Aside from offending common sense, it would create a mathematical paradox. The equation therefore can never be true.
Can you expand on the mathematical paradox that would be created?
We begin with an object of mass = 1. We halve this. One half we discard, the other we divide in half again. Ad infinutum. All the discarded halves (those not subjected to further division) created during this process would be of a size that could be expressed as an exact ratio of the size zero "fragment" - i.e. 2:1, 4:1, 8:1, etc. So all the pieces would be of size zero!

Disinfo Agent
2004-Dec-10, 02:24 PM
What "size zero fragment"?

mutineer
2004-Dec-10, 02:26 PM
I think you can actually, infinity times 5 equals infinity.
I agree with your answer, but I am not sure whether it is arithmetic, and it is certainly not calculation - as offered in the so-called "proofs".

Disinfo Agent
2004-Dec-10, 02:27 PM
I agree with your answer, but I am not sure whether it is arithmetic [...]
Why shouldn't it be?

[...] and it is certainly not calculation - as offered in the so-called "proofs".
Define "calculation".

Severian
2004-Dec-10, 02:31 PM
I thought I had understood the concept of a “limit” since before most BABBers were born. Reading some of the posts to this thread makes me wonder whether maths is taught differently these days. However, having had a chance to glance at a couple of textbooks, it seems that things have not changed since my own schooldays. It provides a means of proposing/asserting that the value of an expression f(x) gets close to y when x gets close to a given value, or approaches infinity.

0.9 + 0.09 + 0.009 . . . has a limit value of 1 as the number of terms approaches infinity. That means it gets close to 1.

Right, and the sequence 0.9 , 0.99 , 0.999 , ... gets close to 1. No term in the sequence (that is, a finite number of nines) is equal to 1.
However, when you read those definitions, you must have noticed you were reading the definition of what it meant for the limit of a sequence to equal a value. ( lim (n-> infinity) x_n )= x means for any e > 0, there exists an N so that for n > N , |x_n-x| &lt; e.
So the limit of x_n = x means that if you can get the sequence arbitrarily close to x by going out far enough. Note that this does not say that the limit of x_n is just close to x.
So in the limit, with an infinite number of nines, is equal to 1.

Or in binary:
0.111... = 1
This was precisely the equation I had in mind when I asked the question about successive halving. As we add successive ones here, the difference between the left and right sides of the equation halves. But successive halving of an object never causes it to vanish (i.e. assume a value of zero). Aside from offending common sense, it would create a mathematical paradox. The equation therefore can never be true.

Again, any finite number of successive halving does not cause it to vanish. No one is arguing this.

mutineer
2004-Dec-10, 02:31 PM
What "size zero fragment"?I refer to worzel's claim in regard to successive halving causing an object to "vanish" . . .
It does, if you extend the halving process to infinity.

Bad jcsd
2004-Dec-10, 02:33 PM
I thought I had understood the concept of a “limit” since before most BABBers were born

Without meaning to be mean, you clearly do not understand tthe concpet of a limit otherwise you wouldn't be having this argument, look-up the formal defintion of the limit of a sequence.

Reading some of the posts to this thread makes me wonder whether maths is taught differently these days. However, having had a chance to glance at a couple of textbooks, it seems that things have not changed since my own schooldays. It provides a means of proposing/asserting that the value of an expression f(x) gets close to y when x gets close to a given value, or approaches infinity. No a limit is really just a bound of an open set i.e. a subset of the image set, a bound is just a number that is greater than or equal to the absolute value of any number in the set.

0.9 + 0.09 + 0.009 . . . has a limit value of 1 as the number of terms approaches infinity. That means it gets close to 1.

The sequence of partial sums has a limit and that limit is 1 and for that reason we define 0.999..., as 1.

Or in binary:
0.111... = 1
This was precisely the equation I had in mind when I asked the question about successive halving. As we add successive ones here, the difference between the left and right sides of the equation halves. But successive halving of an object never causes it to vanish (i.e. assume a value of zero). Aside from offending common sense, it would create a mathematical paradox. The equation therefore can never be true.[/quote]

A paradox is a contardiction, I see none.

Disinfo Agent
2004-Dec-10, 02:35 PM
What "size zero fragment"?I refer to worzel's claim in regard to successive halving causing an object to "vanish" . . .
It does, if you extend the halving process to infinity.
I'm sorry, but I do not understand your reply. Please state more clearly what is the "zero size fragment" in the pie halving example. Worzel never used that term. Which iteration of the process produces that fragment?

Severian
2004-Dec-10, 02:37 PM
But successive halving of an object never causes it to vanish (i.e. assume a value of zero). Aside from offending common sense, it would create a mathematical paradox. The equation therefore can never be true.
Can you expand on the mathematical paradox that would be created?
We begin with an object of mass = 1. We halve this. One half we discard, the other we divide in half again. Ad infinutum. All the discarded halves (those not subjected to further division) created during this process would be of a size that could be expressed as an exact ratio of the size zero "fragment" - i.e. 2:1, 4:1, 8:1, etc. So all the pieces would be of size zero!
Nope. Each discarded bit would be expressible as an exact ratio to any other discarded bit. At any finite stage of division, the discarded bits can be expressed as exact ratios to what hasn't been discarded yet. But note that the ratios of the discarded bits to what you have left change each time you subdivide.

Bad jcsd
2004-Dec-10, 02:38 PM
btw infinitesmials are really completely irrelvant, a book on analysis probably will not mention the conceppt once. Infinitesmials are rather contrived concepts that we use to manipulate equations and they do not represent actual numbers.

Damburger
2004-Dec-10, 02:38 PM
I agree with your answer, but I am not sure whether it is arithmetic, and it is certainly not calculation - as offered in the so-called "proofs".

Why do you think you have the authority to redefine arithmetic in order to fit your assumptions?

You have not been able to refute a single proof that has been offered in this thread, your only argument being that mathematical proof isn't as good as your "logic" - and that idea has been rubbished as well.

0.999r=1 and thats all there is to it. This is a nessecary consequence of the decimal system, and your inability to grasp this does not constitute a disproof of it.

Disinfo Agent
2004-Dec-10, 02:45 PM
btw infinitesmials are really completely irrelvant, a book on analysis probably will not mention the conceppt once. Infinitesmials are rather contrived concepts that we use to manipulate equations and they do not represent actual numbers.
They are irrelevant in as much as 0.999... is not an infinitesimal, or infinitely close to 1 without being equal to 1.
Unless the contrarians can show us otherwise, of course. :)

Gmann
2004-Dec-10, 03:33 PM
Let's try it this way. I don't hold that 0.999...is equal to 1 because by it's nature, 0.999...cannot be written into an equation accurately.

Bad jcsd
2004-Dec-10, 03:36 PM
Let's try it this way. I don't hold that 0.999...is equal to 1 because by it's nature, 0.999...cannot be written into an equation accurately.

Yes it can, infact you just wrote it, so clearly it can represented with absolute no ambiguity whatsoever.

Damburger
2004-Dec-10, 03:37 PM
Let's try it this way. I don't hold that 0.999...is equal to 1 because by it's nature, 0.999...cannot be written into an equation accurately.

0.999r can be written accurately - as '1'

In order to deal with an infinitely recurring number, it has to be treated as an infinite series. Because the terms (the 9s) of this series get progressively smaller, the series converges, and the value it converges on is 1.

Disinfo Agent
2004-Dec-10, 04:06 PM
Let's try it this way. I don't hold that 0.999...is equal to 1 because by it's nature, 0.999...cannot be written into an equation accurately.
You've just written it down. Why do you say it can't be written accurately?

Normandy6644
2004-Dec-10, 04:28 PM
Let's try it this way. I don't hold that 0.999...is equal to 1 because by it's nature, 0.999...cannot be written into an equation accurately.

Sum(n=1 to infinity) (9/10^n)

:)

A Thousand Pardons
2004-Dec-10, 04:59 PM
Let's try it this way. I don't hold that 0.999...is equal to 1 because by it's nature, 0.999...cannot be written into an equation accurately.

Sum(n=1 to infinity) (9/10^n)

:)
Same thing.

Normandy6644
2004-Dec-10, 05:10 PM
Let's try it this way. I don't hold that 0.999...is equal to 1 because by it's nature, 0.999...cannot be written into an equation accurately.

Sum(n=1 to infinity) (9/10^n)

:)
Same thing.

What do you mean?

Grey
2004-Dec-10, 05:14 PM
Well let's take an example. If time continues forever, there will be an infinite number of Sundays in the future. There will (in some sense) be five times as many weekdays. But one cannot multiply infinity by five. The matter lies outside the bounds of arithmetic.
Actually, this serves as an interesting side note on why our intuitive understanding of finite numbers doesn't necessarily work with infinite numbers. In any finite amount of time, there are five times as many weekdays as Sundays (well, as long as I take my time in full-week blocks, anyway). But if I look at the infinite set, I can show that I can pair up each Sunday with a weekday, and there will be no weekdays or Sundays left over. If time continues forever, there are exactly as many (infinite) weekdays as Sundays. There are other cases where I can show that some infinite values are greater or less than other infinite values, but this isn't one of them.

Bad jcsd
2004-Dec-10, 05:26 PM
Or infact I can pair up each Sunday in the set of days in any infite numebr of weeks with each day in the same set including Sundays! This is infact just the defintion of an infinite set, i.e. set for which there exists a bijection between itself and a proper subset.

mutineer
2004-Dec-10, 06:06 PM
But successive halving of an object never causes it to vanish (i.e. assume a value of zero). Aside from offending common sense, it would create a mathematical paradox. The equation therefore can never be true.
Can you expand on the mathematical paradox that would be created?
We begin with an object of mass = 1. We halve this. One half we discard, the other we divide in half again. Ad infinutum. All the discarded halves (those not subjected to further division) created during this process would be of a size that could be expressed as an exact ratio of the size zero "fragment" - i.e. 2:1, 4:1, 8:1, etc. So all the pieces would be of size zero!
Nope. Each discarded bit would be expressible as an exact ratio to any other discarded bit. At any finite stage of division, the discarded bits can be expressed as exact ratios to what hasn't been discarded yet. But note that the ratios of the discarded bits to what you have left change each time you subdivide.
Excuse my not picking up some other issues for the moment, but I want to clear up why the three sentences which follow the "Nope" (all of which I agree with) explain the "Nope".
What are the views of others on worzel's contention that infinite halving causes an object to vanish? If there is an error in the logic of the paradox I have presented, can someone make an understanding of the error accessible to me?

Disinfo Agent
2004-Dec-10, 06:22 PM
What are the views of others on worzel's contention that infinite halving causes an object to vanish?
Did worzel actually say that? I can't see where. It isn't in the text you quoted...

If there is an error in the logic of the paradox I have presented, can someone make an understanding of the error accessible to me?
The error in your alleged paradox, to me, is that I can't even understand what you're saying until you explain what you mean by "zero sized fragment". Which iterations of the halving process are supposed to produce those zero sized fragments?

Bad jcsd
2004-Dec-10, 06:28 PM
It's a matter of defintion is all, so we needn't worry about things "dividing numbers to zero", it's because we define an infinite decimal expansions as a limit of a sequence (or at leats in an equivalent way), all other arguments are just to show that this is the most consistent way to do it and the problems of definin it in any other ways.

Disinfo Agent
2004-Dec-10, 06:43 PM
What are the views of others on worzel's contention that infinite halving causes an object to vanish?
Did worzel actually say that? I can't see where. It isn't in the text you quoted...
Actually, I'm the one who said it (http://www.badastronomy.com/phpBB/viewtopic.php?p=378776#378776), and I think you confused worzel with me (http://www.badastronomy.com/phpBB/viewtopic.php?p=378847#378847).

A Thousand Pardons
2004-Dec-10, 07:02 PM
Let's try it this way. I don't hold that 0.999...is equal to 1 because by it's nature, 0.999...cannot be written into an equation accurately.

Sum(n=1 to infinity) (9/10^n)

:)
Same thing.

What do you mean?
That "0.999..." is a shorthand for "Sm(n=1 to infinity) (9/10^n)", and that neither is more or less accurate than the other.

I am not disagreeing. :)

Normandy6644
2004-Dec-10, 07:04 PM
Let's try it this way. I don't hold that 0.999...is equal to 1 because by it's nature, 0.999...cannot be written into an equation accurately.

Sum(n=1 to infinity) (9/10^n)

:)
Same thing.

What do you mean?
That "0.999..." is a shorthand for "Sm(n=1 to infinity) (9/10^n)", and that neither is more or less accurate than the other.

I am not disagreeing. :)

Oh I know you weren't, I was just thinking maybe I misunderstood something. :D

mutineer
2004-Dec-10, 09:17 PM
What are the views of others on worzel's contention that infinite halving causes an object to vanish?
Did worzel actually say that? I can't see where. It isn't in the text you quoted...
Actually, I'm the one who said it (http://www.badastronomy.com/phpBB/viewtopic.php?p=378776#378776), and I think you confused worzel with me (http://www.badastronomy.com/phpBB/viewtopic.php?p=378847#378847).
Sorry, DA. You are quite right of course. I had forgotten I was battling on so many fronts!

Gmann
2004-Dec-11, 12:00 AM
You've just written it down. Why do you say it can't be written accurately?

It can be represented in an equation, many of you have demonstrated that it is possible. The reason it can't be written accurately is because once you start, you cannot stop. You will find yourself writing "9" on the right side of the decimal point until the end of time, and still have not finished writing the number. It was mentioned before (I don't feel like looking for the author, this is a long thread) to imagine a point where 2 parallel converge. By their very definition, they can't. If there is a convergence or divergence, they cease to be parallel. No, I'm not picking on you specifically, you just happened to ask the most direct question.

Bad jcsd
2004-Dec-11, 12:27 AM
You've just written it down. Why do you say it can't be written accurately?

It can be represented in an equation, many of you have demonstrated that it is possible. The reason it can't be written accurately is because once you start, you cannot stop. You will find yourself writing "9" on the right side of the decimal point until the end of time, and still have not finished writing the number. It was mentioned before (I don't feel like looking for the author, this is a long thread) to imagine a point where 2 parallel converge. By their very definition, they can't. If there is a convergence or divergence, they cease to be parallel. No, I'm not picking on you specifically, you just happened to ask the most direct question.

The represntation 0.999... is unambigious once defined (infact it's not the usually represnation, but that's just the limits of html). Whethr or not you can write an infinite amount of digits 100% irrelevant.

If 0.999... is not a 100% accurate reprenation of this number then you surely must be able to name me which other numbers it could possibly be.

Lycus
2004-Dec-11, 12:33 AM
It can be represented in an equation, many of you have demonstrated that it is possible. The reason it can't be written accurately is because once you start, you cannot stop. You will find yourself writing "9" on the right side of the decimal point until the end of time, and still have not finished writing the number. It was mentioned before (I don't feel like looking for the author, this is a long thread) to imagine a point where 2 parallel converge. By their very definition, they can't. If there is a convergence or divergence, they cease to be parallel. No, I'm not picking on you specifically, you just happened to ask the most direct question.
What is written language besides visual representations of abstract concepts? There is nothing inherent about the symbol "9" that makes it mean what it means. It translates to that amount because we designate it to. Same thing with the "...", we're designating it to represent infinite digits, so that's what it means.

01101001
2004-Dec-11, 12:47 AM
It can be represented in an equation, many of you have demonstrated that it is possible. The reason it can't be written accurately is because once you start, you cannot stop. You will find yourself writing "9" on the right side of the decimal point until the end of time, and still have not finished writing the number.

You understate it. So will I. You can write, a googol of nines per second, nay, a googolplex per femtosecond, until the end of time, and you virtually won't even have begun -- you still have a whole infinty of nines to go.

But, just because you can't write it, can't even really start to write it, because you can't discretely compute it, doesn't mean it's value is not 1. The infinte number of nines you still haven't considered exactly make up the difference between what you have so-far considered and 1.

01101001
2004-Dec-11, 01:22 AM
Have we tried a proof by contradiction, reductio ad absurdum (http://mathworld.wolfram.com/ReductioadAbsurdum.html), yet?

[Edit: Original flawed. Would also "prove", e.g., 0.999... = 2, because I didn't use the fact that b &lt; 1. Refinements in brackets.]

Assertion: 0.999... = 1.

Assumed proposition: they are unequal, that there exists a number b such that 0.999... &lt; b &lt; 1.

Since b is unequal to .999..., then for some finite number k, digit k of the decimal expansion of b will be different from the kth digit of 0.999... [Since b &lt; 1, the decimal expansion of b begins 0.{k-1 digits}{kth digit}]

The kth digit of .999... is 9 by definition.

The kth digit of b must therefore be 0, 1, 2, 3, 4, 5, 6, 7, or 8.

Thus b must be less than 0.999...

But, that contradicts the assumption that 0.999... &lt; b &lt; 1.

Therefore, the assumption is false, and the assertion 0.999... = 1 must be true.

Severian
2004-Dec-11, 04:55 AM
Excuse my not picking up some other issues for the moment, but I want to clear up why the three sentences which follow the "Nope" (all of which I agree with) explain the "Nope".
What are the views of others on worzel's contention that infinite halving causes an object to vanish? If there is an error in the logic of the paradox I have presented, can someone make an understanding of the error accessible to me?

Which piece do you think has a ratio of 2:1 with whatever you think is left?

gzhpcu
2004-Dec-11, 05:53 AM
Assumed proposition: they are unequal, that there exists a number b such that 0.999... &lt; b &lt; 1.

Disagree. IMHO the assumption is not valid. Why shouldn't 0.9999999999..... be the number closest to (but not equal to) 1?

Grey
2004-Dec-11, 06:28 AM
Why shouldn't 0.9999999999..... be the number closest to (but not equal to) 1?
Rather than addressing this rigorously, I'll try to appeal to your intuition. Don't you think you should be able to take the average of any two distinct numbers to find a number halfway between them?

VTBoy
2004-Dec-11, 07:00 AM
Assumed proposition: they are unequal, that there exists a number b such that 0.999... &lt; b &lt; 1.

Disagree. IMHO the assumption is not valid. Why shouldn't 0.9999999999..... be the number closest to (but not equal to) 1?

Well so you think the reals can be ordered, then please tell me, what is the next number after 1, and what is the number before 0.9999....

Disinfo Agent
2004-Dec-11, 01:47 PM
And the number just before 0. Or pi. :)

The reason it can't be written accurately is because once you start, you cannot stop. You will find yourself writing "9" on the right side of the decimal point until the end of time, and still have not finished writing the number.
I essentially agree with Lycus. Yes, it's true that we can't write down the numeral "0.999...", with an infinite string of nines. That would take an eternity, or an infinitely long sheet of paper to do.
But, even if we can't literally write it down, we can understand what it means (or think we do :wink:): an infinitely long, neverending string of nines. And the meaning is more important than the name. Provided we all understand which number we're talking about, it doesn't matter much that we can't write down exactly one of its representations.
Notice that even finite, but very long numerals, like a googolplex (http://www.fpx.de/fp/Fun/Googolplex/) would be difficult to write down in decimal notation in practice. But that doesn't stop us from discussing them.
Also, there are alternative notations that can be written down. For instance, "0.(9)" is sometimes used for such an infinite decimal expansion (27/99=0.272727... is denoted as "0.(27)").

[...] you just happened to ask the most direct question.
We're trying... :)

Have we tried a proof by contradiction, reductio ad absurdum (http://mathworld.wolfram.com/ReductioadAbsurdum.html), yet?

[...]
=D>
I liked it.

A Thousand Pardons
2004-Dec-11, 01:48 PM
Assumed proposition: they are unequal, that there exists a number b such that 0.999... &lt; b &lt; 1.

Disagree. IMHO the assumption is not valid. Why shouldn't 0.9999999999..... be the number closest to (but not equal to) 1?
As the others are saying, the real numbers have the property that there is always some other number between any two different numbers.

If there weren't another number, you would not be using the "real numbers", you'd be using some other system.

Damburger
2004-Dec-11, 01:57 PM
Given the number of proofs that have failed to convince people, I don't think proof of any kind is going to work.

Why don't you accept that 0.999r=1?

Is it because you don't believe a number whos significant digits are all after the decimal point can be equal to 1?

Is it because you don't believe two decimal representations can correspond to the same number?

I'm giving you the benefit of the doubt here and assuming the doubters aren't just trolling/refusing to admit they are wrong.

Bad jcsd
2004-Dec-11, 02:30 PM
It follows on from the archmedean property that every interval of real numbers contains infinitely many real numbers, yet if 0.999... is not equal to 1 and they are 'infinitely close togehter' then (0.999..., 1) = {} is an interval of real numbers. Given that the reals are THE complete archimedean ordered field it would be completely contradictory if 0.999.. and 1 were 'infinitely close'.

Gmann
2004-Dec-11, 02:36 PM
Excellent responses from all. I understand the mathematical necessity to assign and end value to a repeating number. Without that, any computation would crash as soon as the person or device doing the computation encountered one. Classic example would be the 1/3+1/3+1/3=3/3=1 expressed decimally .333...+.333...+.333...=.999=1 :o

VTBoy brings up a point that I had not considered before. What is the next number in this type of sequence? Good question. Maybe it's 1.000...with another 1 hiding inside something somewhere. On second thought that is a stupid answer. Although I always knew that .999...was considered as equal to 1, it never made any sense. I had never considered what came after 1 or before .999...

You have to admit, this is a tough concept, and not easy to wrap your head around. Logic dictates that 0.999...does not equal 1, but logic cannot resolve the number before or after .999... In this case, math wizards 1 Logicians 0 =D>

Bad jcsd
2004-Dec-11, 02:46 PM
The fault in the logic is that people assume that real numbers ARE their decimal expansions, this is not the case as the real numbers are abstract entites whose properties are independent of the representaion used. As they've assumed that real numbers are their decmal expansions the mistake is made that every unique decimal expansion is a unique real number, which again is not the case (every rational number that can be represented as a decimal fraction has two decimal expansions, one finite, one recurring).

Damburger
2004-Dec-11, 02:51 PM
The fault in the logic is that people assume that real numbers ARE their decimal expansions, this is not the case as the real numbers are abstract entites whose properties are independent of the representaion used.

As I tried to illustrate by comparing this redundancy to the redundancy in rational numbers. Apparantly that didn't convince them, so I changed tactic.

As they've assumed that real numbers are their decmal expansions the mistake is made that every unique decimal expansion is a unique real number, which again is not the case (every rational number that can be represented as a decimal fraction has two decimal expansions, one finite, one recurring).

Are you sure about that?

I can see how it would be the case for 1/4 expanding as 0.25 or 0.24999r, but is there a finite expansion for 1/3?

Bad jcsd
2004-Dec-11, 02:53 PM
As I tried to illustrate by comparing this redundancy to the redundancy in rational numbers. Apparantly that didn't convince them, so I changed tactic.

Yep people don't seem to bothered that rational numbers are not ordered pairs of integers, rather equivalency classes of ordered pairs of integers,

Are you sure about that?

I can see how it would be the case for 1/4 expanding as 0.25 or 0.24999r, but is there a finite expansion for 1/3?

1/3 can't be represented as a decimal fraction therefore it only has one decimal expansion.

Disinfo Agent
2004-Dec-11, 03:01 PM
You have to admit, this is a tough concept, and not easy to wrap your head around. Logic dictates that 0.999...does not equal 1, but logic cannot resolve the number before or after .999... In this case, math wizards 1 Logicians 0 =D>
I don't think it's so much a logical difficulty, as it is a misunderstanding about the decimal notation. We all know that there is not a one-to-one correspondence between fractions and rational numbers, since every rational number can be written as an infinite number of fractions: (*)

1/2=2/4=3/6=4/8=...
1=1/1=2/2=3/3=...

But the decimal notation has an almost one-to-one correspondence with the real numbers, and perhaps this is what confuses some people. The only exceptions are decimals ending in "d999...", in base 10 (and "d111..." in base 2, etc.), which redundantly represent the same number as "(d+1)000..."

(Sometimes, in mathematical proofs, we "outlaw" one of those expansions, so as to get a one-to-one correspondence. For instance, by making the convention that numbers ending in "d999..." shall always be denoted with "(d+1)000..."; but usually we prefer to allow both expansions to be used.)

(*) This is not right. See below.

Bad jcsd
2004-Dec-11, 03:17 PM
Careful! there is a one-to-one correpsdnace between the rationals and ZxZ (i.e. the set of fraction if you like) and there is also a one-to-one corresponance between the set of unique decimal expansions and the set of real numbers [-X :)

Disinfo Agent
2004-Dec-11, 03:21 PM
Yes, you're right! And now I don't know how to rephrase my argument... :-? #-o

worzel
2004-Dec-11, 03:31 PM
Careful! there is a one-to-one correpsdnace between the rationals and ZxZ (i.e. the set of fraction if you like) and there is also a one-to-one corresponance between the set of unique decimal expansions and the set of real numbers [-X :)
Does "unique decimal expansions" rule out recurring 9s?

Bad jcsd
2004-Dec-11, 03:33 PM
Does "unique decimal expansions" rule out recurring 9s?

No, what I mean is all different possible decimal expansons including those whiuich may represent the same real number.

worzel
2004-Dec-11, 03:38 PM
Does "unique decimal expansions" rule out recurring 9s?

No, what I mean is all different possible decimal expansons including those whiuich may represent the same real number.

I must be a bit slow today then, because 1 and 0.99.. are unique decimal expansions right? And they both map to the same real, so

and there is also a one-to-one corresponance between the set of unique decimal expansions and the set of real numbers

Isn't true, is it?

Bad jcsd
2004-Dec-11, 03:43 PM
]I must be a bit slow today then, because 1 and 0.99.. are unique decimal expansions right? And they both map to the same real, so

Yep, but that's just one map.

Isn't true, is it?

When we say there is a one-to-one correspondance between two sets we are saying that at least one map out of all possible maps is bijective.

Disinfo Agent
2004-Dec-11, 03:48 PM
We don't want just any one-to-one correspondence; we want one that preserves algebraic properties (and so we're back to that...), i.e. an isomorphism.

There are one-to-one mappings between N, Z, Z*Z, and Q, so obviously there is a one-to-one mapping between the set of all formal fractions "a/b" (which we can identify with Z*Z minus the elements of the form (a, 0)) and Q. However, this mapping will not correspond to our idea of "fraction". For instance, it will make "1/2" and "2/4" correspond to two different elements of Q... The property a/b = (a:c)/(b:c) fails.

worzel
2004-Dec-11, 04:10 PM
Myabe I'm just too hung over today, but I think I get it. You're saying that the set of decimal expansions and reals are both of the same order of infinity so there must be a one to one correspondence between them, just not necessarilly one that actually maps decimal expansions to the reals they represent?

EDITED for grammar

Candy
2004-Dec-11, 04:14 PM
So does 1=0.9999...., too? Just wondering. :lol:

A Thousand Pardons
2004-Dec-11, 04:16 PM
Logic dictates that 0.999...does not equal 1, but logic cannot resolve the number before or after .999... In this case, math wizards 1 Logicians 0 =D>
Logic does not dictate that. It is only your intuition.

worzel
2004-Dec-11, 04:19 PM
So does 1=0.9999...., too? Just wondering. :lol:
Not in base 12 :)

Candy
2004-Dec-11, 04:24 PM
So does 1=0.9999...., too? Just wondering. :lol:
Not in base 12 :)
Not in any base, I think. 8-[

Disinfo Agent
2004-Dec-11, 04:26 PM
What if we decided to use base 0.999...? :D

Candy
2004-Dec-11, 04:29 PM
What if we decided to use base 0.999...? :D
But would 1=0.9999....? If 1 doesn't equal 0.9999...., then why would 0.9999....=1? Just a thought. 8-[

worzel
2004-Dec-11, 04:30 PM
What if we decided to use base 0.999...? :D
Back to counting beans!

Candy
2004-Dec-11, 04:33 PM
What if we decided to use base 0.999...? :D
Back to counting beans! Are you ducking my question?

Disinfo Agent
2004-Dec-11, 04:37 PM
But would 1=0.9999....? If 1 doesn't equal 0.9999...., then why would 0.9999....=1? Just a thought. 8-[
1 (unit) = 0.999... :)

Though, to be honest, b in base "b" is denoted by "10", not "1".

worzel
2004-Dec-11, 04:40 PM
What if we decided to use base 0.999...? :D
Back to counting beans! Are you ducking my question?
No, I didn't see your question when I posted. But I am wondering what the equivalent question would be in base 1 (base 0.999..)

0.000...=1 ? :-?

Guess you can't represent fraction in unary anyway.

Candy
2004-Dec-11, 04:44 PM
What if we decided to use base 0.999...? :D
Back to counting beans! Are you ducking my question?
No, I didn't see your question when I posted. But I am wondering what the equivalent question would be in base 1 (base 0.999..)

0.000...=1 ? :-?

Guess you can't represent fraction in unary anyway.
No bases. Does 1=.9999....? Simple question, no fluffing. Use the same process as you did with .9999....=1. Don't use any other mathematic formulas, please. YES OR NO.

worzel
2004-Dec-11, 04:44 PM
But would 1=0.9999....? If 1 doesn't equal 0.9999...., then why would 0.9999....=1? Just a thought. 8-[
1 (unit) = 0.999... :)

Though, to be honest, b in base "b" is denoted by "10", not "1".

But in unary every postion is 1^x so 1 in unary could be 10, 100, 1000, etc.

Candy
2004-Dec-11, 04:47 PM
But would 1=0.9999....? If 1 doesn't equal 0.9999...., then why would 0.9999....=1? Just a thought. 8-[
1 (unit) = 0.999... :)

Though, to be honest, b in base "b" is denoted by "10", not "1".

But in unary every postion is 1^x so 1 in unary could be 10, 100, 1000, etc.
Is that a YES or NO to my question?