Someone please help explain this to me because I cant decide whether or not .9 repeating = 1

Here is my reasoning for why it should:

1) Fraction Approach
.3 repeating = 1/3
1/3*3=1
.3 repeating*3=.9 repeating

2) Algebraic Approach
I will write .999 to represent .9 repeating to simplify things
.999=.999
x=x (substitute x for .999)
10x=10x
10x-x=9x
10(.999)-.999=9x
9.999-.999=9x
9=9x
1=x
.999=1

Here is my reasoning why it does not equal 1:

1) It does not make any sense! It should not equal 1!  

2) You could argue that 1 as a value for x in the algebraic approach is extraneous because when plugged back in, it comes out to .999=1

So thats my reasoning so far. Can someone please help to explain to me whether or not it does equal 1? Thank you

It approaches 1 as the length goes to infinity. For any value \delta more than 0 you can choose a natural number N so that the number .99999... with N digits or more will be within \delta of 1.

1) It does not make any sense! It should not equal 1! 

Do you have a specific argument against it?

I agree with the asymptote thing... It gets infinitely close to but never equals 1. That's just how it works in my head though.

This is how I think of it:
0.9 = 1 - 1/10    so with 1 digit behind the comma, it is 1/10th less than 1
0.99 = 1 - 1/100 so with 2 digits behind the comma, it's 1/100th less than 1
0.99 ... 9 (n digits behind the comma) = 1 - 1/(10^n)
and for infinity:
0.99... (∞ digits behind the comma) = 1 - 1/∞, and anything divided by infinity (except for infinity itself) is 0, so it equals 1-0=1.

If you were in a limit, 0.999... = 1^-. But since we're not in a limit, it would, for all intents and purposes, equal 1.

The explanation I have  heard is that it approaches one without reaching it, like an asymptote on a graph, it will get infinitely close, but cannot equal the value of the asymptote.

I agree.

My math teacher has this picture on a shirt  

I've had this argument with a friend. To me, a decimal number is still just that, a decimal. Nothing other than 1 can equal 1. If anything, the proof only suggests to me not that .99999 repeating equals 1, but that instead that the decimal system is a flawed way to represent partial numbers.

have you ever taken calculus? that these two things ARE equal is essential to the fundamental theorem, and, thus, most modern mathematics.
think of it this way:
.9 is close to 1, but does not equal it, .99 is closer, and, every time you add another 9, the gap between the two grows even smaller. if you add infinitely many 9s on the end, then, the gap between the two numbers will be infinitely small. an infinitely small gap is not a gap at all, and, thus, the two are equal. take a look at the concepts of limits and convergence.

EDIT: i think where people are getting tripped up is the concept of infinity.

Quote

The explanation I have  heard is that it approaches one without reaching it, like an asymptote on a graph, it will get infinitely close, but cannot equal the value of the asymptote.

I agree.

My math teacher has this picture on a shirt 

"infinity" is not a number; it's a concept. if something is "infinitely x", then it is as x as something can possibly be. if two things are "infinitely similar to one another", they are as similar as it is possible to be. the most similar two things can be to one another is to match one another in every single regard, so two things that are infinitely similar are the same thing. likewise, if something is "infinitely close to another thing", the two occupy the same space. a decimal place followed by infinitely many 9s, thus, equals one, e^(-infinity) = 0, and so on.