# Is 0.9 recurring equal to 1?

by curleymatsuma
Tags: equal, recurring
P: 800
 Quote by 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.
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.
P: 628
 Quote by SteveL27 Otherwise, use standard terminology to avoid confusion.
So, I guess what you're saying implies 'PIN' stands for 'Personal Idetification Numerals'? Never hear that one before!
P: 15,319
 Quote by cmb So, I guess what you're saying implies 'PIN' stands for 'Personal Idetification Numerals'? Never hear that one before!
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.
P: 628
My copy of Chambers Dictionary of Science and Technology says;

 Number (Maths.). An attribute of objects or labels obtained according to a law or rule of counting.
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?
P: 12
 Quote by micromass Yes, that infinite series converges to a certain number. Namely, 1. Thus $$\sum_{n=1}^{+\infty}{\frac{9}{10^n}}=1$$
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
Mentor
P: 18,299
 Quote by 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. What solidified my issues was this proof 1/3 = 0.3333... 3(1/3) = 0.99999.... 3/3 = 1
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.
P: 12
 Quote by 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.
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.
Mentor
P: 18,299
 Quote by 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. 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.
You are correct that the sequence

$$0.9,~0.99,~0.999,...$$

will never settle. It will always come closer and closer to 1. The point is that the very definition of 0.9999.... is a limit, namely the limit of the series

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

this is the definition of 0.9999....
The above series will approach 1, and the limit of the above series will equal one. Thus 0.9999..., being defined as a limit, will equal 1.
 P: 12 I get it. 0.9999... is just short hand for a limit. It isn't a real number in the traditional sense.
Mentor
P: 18,299
 Quote by wsabol I get it. 0.9999... is just short hand for a limit. It isn't a real number in the traditional sense.
It is a real number in the traditional sense. Limits of sequences are real numbers!!!

We call a real number L the limit of $(x_n)_n$ if

$$\forall \varepsilon>0:\exists n_0:\forall n>n_0:|x_n-L|<\varepsilon$$

So you see that only real numbers can be limits, by definition.

0.9999... is simply notation for the real number defined as the limit of the series

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

This series has a limit which is a real number. We denote this real number by 0.9999...
But 1 is also a limit, thus 1=0.99999...
Mentor
P: 21,292
Quote by cmb
My copy of Chambers Dictionary of Science and Technology says;
 Number (Maths.). An attribute of objects or labels obtained according to a law or rule of counting.

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?
That definition seems very imprecise to me. We are distinguishing between "number" and "numeral" here.

For example, we can write the number 4 in a variety of ways: 4, IV (in Roman numerals), 100 (in binary). These are different numerals that all represent the same number.

In the same way, 0.999... and 1 are different representations of the same number.

I can't believe we're having this same discussion again!
P: 12
 Quote by micromass It is a real number in the traditional sense. Limits of sequences are real numbers!!! We call a real number L the limit of $(x_n)_n$ if $$\forall \varepsilon>0:\exists n_0:\forall n>n_0:|x_n-L|<\varepsilon$$ So you see that only real numbers can be limits, by definition. 0.9999... is simply notation for the real number defined as the limit of the series $$\sum_{n=1}^{\infty}{\frac{9}{10^n}}$$ This series has a limit which is a real number. We denote this real number by 0.9999... But 1 is also a limit, thus 1=0.99999...
1 is a real whole number. 0.999... is a limit. That limit is equal to 1, not the real decimal number 0.9999....(as close as you can get to infinity without getting there, because the infinite term of the sequence ever happen)...9
Mentor
P: 18,299
 Quote by wsabol 1 is a real whole number. 0.999... is a limit. That limit is equal to 1, not the real decimal number 0.9999....(as close as you can get to infinity without getting there, because the infinite term of the sequence ever happen)...9
No, 0.99999.... is a notation for a real number. The real number is defined by a limit.

Limits are real numbers.
P: 12
 Quote by micromass No, 0.99999.... is a notation for a real number. The real number is defined by a limit. Limits are real numbers.
Damn you got me. Ok.
 P: 66 There is also the infinitesimal approach. There is a hyperreal number $\epsilon$ that is smaller than the smallest real number, so we can define the following: $1 - \epsilon = .999...$. This implies that $1$ and $1 - \epsilon$ (.999...) are different numbers in the hyperreal numbering system.
Mentor
P: 18,299
 Quote by Matt Benesi There is also the infinitesimal approach. There is a hyperreal number $\epsilon$ that is smaller than the smallest real number, so we can define the following: $1 - \epsilon = .999...$. This implies that $1$ and $1 - \epsilon$ (.999...) are different numbers in the hyperreal numbering system.
I fear you have not fully understood hyperreals. In the hyperreals, the definition $1-\varepsilon=0.9999...$ is not made. Furthermore, in the hyperreals, there is no such thing as the smallest real number.
 Mentor P: 15,167 This topic has come up so often here that we have an FAQ that addresses this concept: http://www.physicsforums.com/showthread.php?t=507001. Thread closed.

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