1. ## Bad Math?

The following has appeared on our intranet

Is this really true?
The conjecture is that 0.9 recurring (i.e. 0.9999....9) is actually equal to 1
For this exercise I will use the notation 0.99... as notation for 0.9 recurring. This is because HTML will not let me put the little dot over the 9 without using graphics.
• Let X = 0.99...
• Then 10X = 9.99...
• Subtract X from each side
This gives us
• 9X = 9
• Divide both sides by 9
• Therefore X = 1
But hang on a moment I thought we said X was equal to 0.99...
Yes it does but from our calculations X is also equal to one. So;
• X = 0.99... = 1
Therefore 0.99... = 1
Any thoughts on this as to fatal flaws? :-?

(Note 2nd attempt at posting this)

2. This has been discussed in four or five simultaneous and very long threads. Don't say you have missed those!

3. no, 0.999... really does equal 1. And there seems to be nothing wrong with that argument.

4. lti
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hmm that proof doesnt really make any sense.

but 0.9 recurring certainly does equal 1

5. Originally Posted by lti
hmm that proof doesnt really make any sense.

but 0.9 recurring certainly does equal 1
What doesn't make sense? It's perhaps not a rigorous proof, more of an example, but it's an example that rightly convinces many people.

6. lti
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what i dont get is how people who didnt get that 0.9... equals 1 would think that 10 - 0.9... = 9

7. Originally Posted by lti
what i dont get is how people who didnt get that 0.9... equals 1 would think that 10 - 0.9... = 9

isn't it 9.999....-0.999...=9?

8. Originally Posted by lti
what i dont get is how people who didnt get that 0.9... equals 1 would think that 10 - 0.9... = 9
On the left hand side you have 10X-X, which must equal 9X, And on the right hand side you have 9.999...- 0.999... which must equal 9.

I would have thought that the step some may question is 0.999... x 10 = 9.999...

9. Here we go again. Anybody wanna lay bets on how many pages this goes on?

10. Originally Posted by worzel
Originally Posted by lti
what i dont get is how people who didnt get that 0.9... equals 1 would think that 10 - 0.9... = 9
On the left hand side you have 10X-X, which must equal 9X, And on the right hand side you have 9.999...- 0.999... which must equal 9.

I would have thought that the step some may question is 0.999... x 10 = 9.999...
that's easy, each 9 just moves one space to the left.
Each 9 becomes ten times as large.

11. Originally Posted by Frog march
Originally Posted by worzel
Originally Posted by lti
what i dont get is how people who didnt get that 0.9... equals 1 would think that 10 - 0.9... = 9
On the left hand side you have 10X-X, which must equal 9X, And on the right hand side you have 9.999...- 0.999... which must equal 9.

I would have thought that the step some may question is 0.999... x 10 = 9.999...
that's easy, each 9 just moves one space to the left.
Each 9 becomes ten times as large.
[Devil's advocate] Yeah, but, originally, you had an infinite series of nines after the decimal point, but now you have one less. So that doesn't count, does it?

12. Originally Posted by Fram
Originally Posted by Frog march
Originally Posted by worzel
Originally Posted by lti
what i dont get is how people who didnt get that 0.9... equals 1 would think that 10 - 0.9... = 9
On the left hand side you have 10X-X, which must equal 9X, And on the right hand side you have 9.999...- 0.999... which must equal 9.

I would have thought that the step some may question is 0.999... x 10 = 9.999...
that's easy, each 9 just moves one space to the left.
Each 9 becomes ten times as large.
[Devil's advocate] Yeah, but, originally, you had an infinite series of nines after the decimal point, but now you have one less. So that doesn't count, does it?
If you can say that "you had an infinite series of nines", then
you can also say that the process of moving the 9s to the left gos on for "infinity"
And at the end of the day infinity-1=infinity

13. ## Re: Bad Math?

Pi = 3, now, there's a REAL challenge!

(Play Salieri!)

14. Originally Posted by Fram
This has been discussed in four or five simultaneous and very long threads. Don't say you have missed those!
Sorry I seem to have missed all of them ops:

15. Originally Posted by Frog march
If you can say that "you had an infinite series of nines", then you can also say that the process of moving the 9s to the left gos on for "infinity"
And at the end of the day infinity-1=infinity
Yeah, but some people have difficulty with that. If someone thinks that 1 - 0.99... doesn't equal 0 then then they presumably think that multiplying 0.99... by 10 will inrease the difference between the 9s after the decimal point by 10 also. So taking away the original 0.99... from 10 x 0.99... should leave you with 9 + (9 times the difference between 0.99... and 1).

The demonstration only works if you think that difference is 0. But it does demostrate that it makes sense to think that, in this case.

16. Originally Posted by worzel
Originally Posted by Frog march
If you can say that "you had an infinite series of nines", then you can also say that the process of moving the 9s to the left gos on for "infinity"
And at the end of the day infinity-1=infinity
Yeah, but some people have difficulty with that. If someone thinks that 1 - 0.99... doesn't equal 0 then then they presumably think that multiplying 0.99... by 10 will inrease the difference between the 9s after the decimal point by 10 also. So taking away the original 0.99... from 10 x 0.99... should leave you with 9 + (9 times the difference between 0.99... and 1).

The demonstration only works if you think that difference is 0. But it does demostrate that it makes sense to think that, in this case.
at the start of the proof your just using the form 0.9999...., your not saying it is equal to 1 it is just a way of writing down a number.

with 9.9999....-0.9999....(expressing the numbers in the same way) doesn't it seem obvious that the 0.9999.... part of the expressions cancel out? There is a 9 in the first expression for every nine in the second which just leaves the 9 to the left of the decimal point.

17. Order of Kilopi
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Originally Posted by Frog march
doesn't it seem obvious that the 0.9999.... part of the expressions cancel out?
Just because it's obvious to you, doesn't mean it's obvious to everyone. (for the record, it is obvious to me.)

18. 9.9999.... is just 9+0.9999....

so if you put it like
9.9999....-0.9999....=
9+0.9999....-0.9999....=9

it is easier to see that the 0.9999.....s simply cancel out leaving 9.

19. Originally Posted by Sticks
Originally Posted by Fram
This has been discussed in four or five simultaneous and very long threads. Don't say you have missed those!
Sorry I seem to have missed all of them ops:
One wonders where you were then. There were polls with it in the title, etc.

Heck, I got into some serious trouble in one of them that resulted in my being temporarily banned.

Where were you?

20. Originally Posted by Tensor
Originally Posted by Frog march
doesn't it seem obvious that the 0.9999.... part of the expressions cancel out?
Just because it's obvious to you, doesn't mean it's obvious to everyone. (for the record, it is obvious to me.)
I suppose self evident is a better term.

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Do you think 0.9999999~ =1 , that is infinite 9s.
Two reasons why .999...=1
Where do you stand now?
the missing number....

By the way, Sticks, the proof in the OP is correct.

Edited to reshuffle the links.

22. in a cave ?

The phrase that comes to mind, which came from Alan Partridge is:

My God I have spawned a monster
#-o

Are we talking here about a converging series, or am I totally up the creek with out even a canoe, let alone a paddle ops:

23. Originally Posted by Sticks
Are we talking here about a converging series, or am I totally up the creek with out even a canoe, let alone a paddle ops:

it isn't a converging series, it is just a way of writing a number and it just happens to represent the same value as 1.

24. I'm okay with .99999999=1

but is 1 a prime number? is .9999999 a prime number?

those are big questions.

PRIME NUMBER. Is 1 a prime number? Most textbooks today call it neither prime nor composite, but older texts generally considered it to be prime. In 1859, Lebesgue stated explicitly that 1 is prime in Exercices d'analyse numérique. It is prime in Primary Elements of Algebra for Common Schools and Academies (1866) by Joseph Ray and Standard Arithmetic (1892) by William J. Milne. A list of primes to 10,006,721 published in 1914 by D. N. Lehmer includes 1. [David Cantrell suggests this is more a historical issue and is inappropriate for this page. He writes, "No high-school students (unless they're doing historical research) are going to encounter a legitimate definition of prime which would include 1." This entry may be removed.]
http://members.aol.com/jeff570/ambiguities.html

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Originally Posted by Frog march
Originally Posted by Sticks
Are we talking here about a converging series, or am I totally up the creek with out even a canoe, let alone a paddle ops:
it isn't a converging series, it is just a way of writing a number and it just happens to represent the same value as 1.
Sorry, Frog March. It is a converging series, too:

0.999... = 0.9 + 0.09 + 0.009 + ...

26. Yes,ok.

but when people start talking about converging series they seem to become more confusesed.

writing 0.9999..... on its own is already a complete number. So is [sumof] 9/10+9/100...... but it doesn't seem like that to some, more like a process than a complete number.

27. Originally Posted by Frog march
Originally Posted by worzel
Originally Posted by Frog march
If you can say that "you had an infinite series of nines", then you can also say that the process of moving the 9s to the left gos on for "infinity"
And at the end of the day infinity-1=infinity
Yeah, but some people have difficulty with that. If someone thinks that 1 - 0.99... doesn't equal 0 then then they presumably think that multiplying 0.99... by 10 will inrease the difference between the 9s after the decimal point by 10 also. So taking away the original 0.99... from 10 x 0.99... should leave you with 9 + (9 times the difference between 0.99... and 1).

The demonstration only works if you think that difference is 0. But it does demostrate that it makes sense to think that, in this case.
at the start of the proof your just using the form 0.9999...., your not saying it is equal to 1 it is just a way of writing down a number.

with 9.9999....-0.9999....(expressing the numbers in the same way) doesn't it seem obvious that the 0.9999.... part of the expressions cancel out? There is a 9 in the first expression for every nine in the second which just leaves the 9 to the left of the decimal point.
Worzel's concern is not that one may doubt that

9.999... - 9 = 0.999...

but rather that one may doubt that

0.999... x 10 = 9.999...

And it's a valid concern with this particular proof. Proving the latter would essentially be proving that 0.999... = 1, which means that assuming the later is assuming your conclusion.

28. 0.9999....=

9/10+9/100+9/1000+9/10000.......

all you do if you multiply by 10 is turn that into

9+9/10+9/100+9/1000...... (all the denominators loose a zero)

ie 9.999..... you don't have to already assume that 0.9999....=1, just keep the number in the same format.

29. ## Re: Bad Math?

It's easy to summarize this thread in non-mathematical terms.

J.J. McClure: When you don't want him he's around! When you want him he's not around! I'm gonna go get a beer!
Captain Chaos: DA-DA-DUM!
(Captain Chaos disappears into the brawling mob)
J.J. McClure: I hate it when he does that!

If'n yer math's bad, then your deliverance is, ya skweel like a piggy!

30. Originally Posted by Frog march
0.9999....=

9/10+9/100+9/1000+9/10000.......

all you do if you multiply by 10 is turn that into

9+9/10+9/100+9/1000...... (all the denominators loose a zero)

ie 9.999..... you don't have to already assume that 0.9999....=1, just keep the number in the same format.
You're missing the point.

If 0.999... &lt;> 1, then 1-0.999... &lt;> 0. In other words, there's a difference between 1 and 0.999... Let's call that difference d.

If the difference between 1 and 0.999... is d, then the difference between 10*1 and 10*0.999... is 10d, but the difference between 9+1 and 9+0.999... is d.

And if that's true, and obviously 9+1 = 10*1, then that means that 9+0.999... &lt;> 10*0.999...

Thus the proof fails.

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