# Is 0.9 recurring equal to 1?

1. Oct 19, 2011

### curleymatsuma

Hi there, this is my first post here so I'm sorry if this is in the wrong place, asked before, standard newbie apologies :P

So I am well aware of the proof that 0.9 recurring =1

x = 0.99999999...

10 x = 9.99999999...

10x - x = 9.999999... - 0.99999...

9x = 9

x = 1 = 0.999999...

However my question is this. Is 0.9 recurring accepted in mathematics as being equal to 1, or are they considered distinct numbers?

I also understand that there are an infinite number of nines and in pretty much ANY situation theoretical or otherwise the difference is completely negligible. I would just like to clarify in my mind whether this number is thought of as being equal to 1 or not?

Thanks,
Matt

2. Oct 19, 2011

### gb7nash

Yes. They're two different representations of the same number.

3. Oct 19, 2011

### curleymatsuma

Thanks! this has been bugging me a bit recently :P

4. Oct 19, 2011

### DaveC426913

See the FAQ:

Last edited by a moderator: May 5, 2017
5. Oct 20, 2011

### SteveL27

1 and .999... are two different representations of the same number, as others have noted.

But the "proof" above, which is often shown to people, is actually a fake proof. That's because without carefully defining the real numbers and proving some theorems about infinite series, you don't actually know for sure that you can multiply an infinitely-long decimal such as .999... by 10 term-by-term; and you don't know that you can subtract one infinitely-long decimal from another. It turns out that you can; but only after you prove you can.

6. Oct 20, 2011

### SW VandeCarr

I think it's more correct to say that 0.9999999....... approaches 1 as a limit as the approximation is extended. It seems clear that there will always be a non zero difference between 0.9999999..... and 1.

7. Oct 20, 2011

### jgens

But 0.999... is the limit. So the difference between 0.999... and 1 is always 0.

Unless of course you mean 0.999... = 0.999...9, in which case there is definitely a non-zero difference. But this is irrelevant to the OP since 0.999... where the decimal expansion never terminates is equal to 1.

8. Oct 20, 2011

### micromass

This is not true. It makes no sense to say that 0.99999.... "approaches" 1. It makes as much sense as saying that 2 approaches 3. There is no approaching going on. 0.9999... is just a number.

9. Oct 20, 2011

### wisvuze

Obviously, if you stop writing 9's and have a finite expansion 0.9999, it will never be 1. But you are writing 0.9999..... ( with the dots ) , so you are implicitly talking about the "limiting decimal expansion"

10. Oct 20, 2011

### cmb

I'd suggest, possibly pendantic but perhaps also clarification of mathematics:

0.9999... and 1 are the same, and equal, value, but they are different numbers.

A number is a thing we have invented to represent a value. Therefore there may be a many-to-one correspondence, depending on what and why we invented the numbers.

11. Oct 20, 2011

### micromass

They have the same value and they are the same number. 1 and 0.999... are absolutely equal in all respects.

12. Oct 20, 2011

### SW VandeCarr

Can we consider the limit of the sum:

$lim \sum_{n=1}^\infty (1 -(1/10)^{n}) = 1$

?

13. Oct 20, 2011

### micromass

Limit with respect to which variable????

14. Oct 20, 2011

### SW VandeCarr

I corrected the formula which was only up for about 1 minute. The variable is the exponent of ten ( a positive integer).

In other words I'm talking about an infinite series 9/10 + 9/100 + 9/1000..............

Last edited: Oct 20, 2011
15. Oct 20, 2011

### Fredrik

Staff Emeritus
It's more accurate to say that they are different "expressions", or different "strings of text", that represent the same number.

16. Oct 20, 2011

### micromass

Yes, that infinite series converges to a certain number. Namely, 1. Thus

$$\sum_{n=1}^{+\infty}{\frac{9}{10^n}}=1$$

17. Oct 20, 2011

### Fredrik

Staff Emeritus
That series would be written as $$\sum_{k=1}^\infty \frac{9}{10^k}=\lim_n\sum_{k=1}^n\frac{9}{10^k}$$ The sequence $\langle\sum_{k=1}^n 9/10^k\rangle_{n=1}^\infty$ of partial sums is convergent. Its limit is 1.

The string of text "0.99..." represents "the limit of the sequence 0.9+0.09+0.009+...". Since that limit is 1, the two strings of text "1" and "0.99..." represent the same number.

Edit: Lol...it's not the first time that Micromass said what I was going to say while I was typing it.

18. Oct 20, 2011

### cmb

I can assure you that I am pressing a different number button to type '0.999..' compared with '1' on my number pad.

Like I say, I'd not say it is necessarily more than pedantery, but my point is I would define numbers as human inventions to represent values, whereas values are values. This is a Big Deal for little kids when they start learning maths, and in some cases until someone points out that numbers are just things we've invented to describe the world around us, they often don't get arithmetic because they think there is something magical and God-given, thus unintelligible, about the numbers themselves.

19. Oct 20, 2011

### SteveL27

What you are calling numbers, the rest of us call numerals.

What you are calling values, the rest of us call numbers.

If your objections go beyond these syntactic substitutions, please advise. Otherwise, use standard terminology to avoid confusion.

20. Oct 20, 2011

### cmb

So, I guess what you're saying implies 'PIN' stands for 'Personal Idetification Numerals'? Never hear that one before!

21. Oct 20, 2011

### DaveC426913

This is a math forum.

If this were commerce forum where discussions of PINs were common and there were some ambiguity in terminology, yes, you'd be asked to use that industry's terminology.

22. Oct 20, 2011

### cmb

My copy of Chambers Dictionary of Science and Technology says;

I don't see that this is describing a 'value' of a quantity or size. Sounds more like my definition.

Can we clarify what/whose definitions we're using here, then?

23. Oct 20, 2011

### wsabol

I agree, but infinity isn't a real number. Therefore for 9/10^∞ isn't real. Just like saying the function y = 1/x is never equal zero. Is is "at" infinity, but where is infinity? its not real, so zero is not part of the set of values for y.

In my mind comparing 0.9999... and 1 is apples to oranges: one is real, one isn't. The sum mention above approaches 1 as the number of terms approaches infinity, yes. But you won't EVER have that infinite term, so they won't ever be equal as long as both numbers are real. Once you hit that infinite term, 0.99... becomes non-real.

1/3 = 0.3333...

3(1/3) = 0.99999....
3/3 = 1

24. Oct 20, 2011

### micromass

You don't have a term $9/10^\infty$. Such a term simply does not occur. The notation $\sum_{i=1}^{\infty}{9/10^n}$ is well-defined without using infinity. Indeed, it can be defined as the limit of the partial sums. You don't need to work with infinity to have that.

0.9999... is a real number and it is equal to 1. 0.9999... is as real as 1.00000.... There is no difference between recurring 9's and recurring 0's.

25. Oct 20, 2011

### wsabol

I feel like you just made my point... The infinite term simply does not occur, therefore, is will always be 0.999999.....9 which is not equal to one. The limit equaling one is exactly that, a limit.

0 is not a member of the set of values in the sequence 9/10^n for n[1,Inf). This sequence is always changing; the derivative is also nonzero. Therefore its sum will never settle at any number, let alone 1. The sum will approach 1, get closer and closer, but never get there, unless you are at Infinity, which doesn't exist.

I don't understand the break down in this logic.