About a specific argument purporting to show $0.999\dots = 1.0$. 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.
 A: 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.
A: 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.
A: 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.
A: 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 $0$.
