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I have read the proofs about why $0.9999.... = 1$, which are satisfying. But I can't get the following argument out of my head.

Defining $0.9999....$ : Lets construct a non-terminating but recurring real number n such that all digits before decimal point are zero and all digits after decimal point are 9. Comparing $1.0000$ with $0.99999...$

Digit at ones place in $1.0$ (i.e. 1) $\ne$ Digit at ones place in $0.99999$ (i.e. 0)

Digit at tenths place in $1.0$ (i.e. 0) $\ne$ Digit at tenths place in $0.99999$ (i.e. 9). And so on....

Hence, $1.0 =0.9999...$ does not fit with our original definition of $0.9999...$ Can you find the mistake in the argument (other than saying that in-fact $1.0 = 0.9999...$)? Am I using a incorrect way to define (or perhaps compare) a number (with another)? Please help me. I am new to analysis. Thanks.

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    $\begingroup$ Downvotes are harsh here.... this is a genuine question that is trying to get to the bottom of an issue that is far from intuitive $\endgroup$
    – Mathmo123
    Jul 8, 2014 at 15:56
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    $\begingroup$ This is a question about a specific argument. The suggested duplicates are different. $\endgroup$
    – Thomas
    Jul 8, 2014 at 15:57

4 Answers 4

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The flaw with your argument is that this test you have described does not test for equality in $\mathbb{R}$.

Ultimately, $\mathbb{R}$ is all about infinite sets and limits, so intuitively it's not enough to just consider a finite comparison of digits.

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  • $\begingroup$ @vadium123 How else would you determine if two numbers in R are equal? $\endgroup$ Jul 8, 2014 at 16:03
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    $\begingroup$ @ManishKhokhar check out this Wikipedia article on the construction of real numbers from Cauchy sequences. $\endgroup$ Jul 8, 2014 at 16:11
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    $\begingroup$ @Manish: If given as their (entire) decimal representation, it turns out that decimals ending in all $9$'s and decimals ending in all $0$'s are the only special cases: so you just test if all the digits are the same, and if they're not, you check if they fit into the special case, and if they don't, they're unequal. $\endgroup$
    – user14972
    Jul 15, 2014 at 7:01
  • $\begingroup$ @Manish: And, of course, that requires being able to work with the entire decimal. If you need a "computable" algorithm that works one digit at a time, unfortunately you can prove that there is no computable algorithm to test for equality of real numbers, so no such thing exists: the best you can do is to recognize that two real numbers differ, and correctly check some cases where two numbers are given in a form that you can verify are equal. (and also that decimal notation is a rather poor choice for doing exact computation with computable numbers) $\endgroup$
    – user14972
    Jul 15, 2014 at 7:03
  • $\begingroup$ @hurkyl We can use principle of mathematical induction. No need to check all digits. $\endgroup$ Jul 15, 2014 at 7:05
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The problem is that two real numbers are not necessarily unequal just because they have a different decimal expansion. And that $1 = 0.999\dots$ is an example of that fact.

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    $\begingroup$ I think you mean, "two real numbers are not necessarily unequal just because they have a different decimal expansion." $\endgroup$
    – Théophile
    Jul 8, 2014 at 16:02
  • $\begingroup$ @Théophile: You are right. Thank. And thanks for the edit $\endgroup$
    – Thomas
    Jul 8, 2014 at 17:20
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The numerals are indeed different. However, that does not mean the numbers they represent are different.

The idea that different things can represent the same number is a familiar one: just think of $1/2$ versus $3/6$ or $1+2$ versus $3$.

I assert the main reason that things like $0.\overline{9} = 1.\overline{0}$ give people trouble is simply because every number that can be represented by a terminating decimal has a unique representation as a terminating decimal, but this property fails to hold when passing from the special case finitely long numerals to the general case infinitely long numerals, coupled with the fact that most real numbers have a unique representation in this fashion.

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  • $\begingroup$ An intuitive answer. Thanks. It helps in understanding. $\endgroup$ Jul 15, 2014 at 7:02
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The problem is indeed in your definition. There are many cases where we can write down the same thing in different ways, but they are still the same. The simplest example of this is $1+1=2$ - a seemingly obvious result, but I could very easily come along and say that there is a plus sign on the left but not on the right, so these things aren't equal.

So how do we define two numbers as being the same in $\mathbb R$? The best way is to think in terms of distance. Two numbers are the same if they have distance $0$ from each other. There are many metrics - measures of distance - on $\mathbb R$, but the most basic is:$$d(x,y) = |x-y|$$

So what is $d(0.99\ldots, 1$)? The proofs that you've already seen should convince you that the answer is indeed $0$.

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  • $\begingroup$ For any quantities a and b and any expression F(x), if a = b, then F(a) = F(b) What if we define F = [10*n], where [] means mod. Take from en.wikipedia.org/wiki/… In above function F(x) if you assume that 1.0 = 0.999... , then you will end up with 10 = 10 and hence 1.0 = 0.999.... . But that's circularity if you ask me. $\endgroup$ Jul 8, 2014 at 16:06
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    $\begingroup$ Be careful. That means that if $a=b$ then $F(a) = F(b)$... but that doesn't mean it works the other way round. $\endgroup$
    – Mathmo123
    Jul 8, 2014 at 16:11

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