@Dsgrue

I think the only thing that really needs to be said is that the square root operation (like the absolute value operation) is one of those operations where two different inputs can have the same output. That's why you cannot set two values equal solely because they have the same output from these operations. Since having the same output does not imply equal input.

That is not the case with the addition, subtraction, multiplication, or division operations. Using these operations, no input will have the same output as another input. Therefore if two unknowns have the same output using these operations, then their inputs are equal. Having the same output DOES mean the inputs are the same.

The only exception is multiplying/dividing by zero, which I didn't do.

f(x,y) = x + 2

I can put an inifite number of y-values for any x to satisfy any output.

Either way x = 0.999... is not a function, as I've beaten to death already. 

I never said x = 0.9999 is a function. Not sure what the other things have to do with what I said either. Let me put it this way:

If x^2 = y^2, then we don't know if y = x. We just know y = x OR (-x). 

HOWEVER

if 10x = 10y, then we know x = y.
if 9x = 9y, then we know x = y.
if x/9 = y/9, then we know x = y.
Using these operations, we know 0.999... = 1

The above is always true. That's why the operations I used works, and the square operation would not have worked. And that's why I didn't use the square operation. So proving that the square operation brings back weird results does nothing to discount what I did, because I did multiplication and subtraction, which DO NOT give off weird results (unless 0 is used). 

Had I used the square operation, then you would have a point. But the operations I used are always fine for any number, which is what SuperMarioWorld tried (and failed) to disprove.

Again, 

Because the square function is not one-to-one, we cannot know if x=y even if x^2 = y^2.

However, if 10x = 10y or 9x = 9y or x/9 = y/9, then we do know x = y. This works all the time unless 0 is used.

None of what you said is applicable at all. Not sure why you insist that it is.

Given x = 0.999...

Show me where y is in this? 

The point is to show you that multiplying, dividing, adding, or subtracting will not give extraneous solutions like squaring will.

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But I'll play along. Do you agree with the following:?

if 10x = 10y, then x = y. ALWAYS 
if x - c = y - c, the x = y. ALWAYS
if x/9 = y/9, then x = y. ALWAYS

and vice versa

if x = y, then 10x = 10y. ALWAYS
if x = y, then x - c = y - c. ALWAYS
if x = y, then x/9 = y/9. ALWAYS

Your answer should be yes to all of the above. Now, let's go back to what I did:

x = 0.9999...

if x = y, then 10x = 10y, so we KNOW:

10x = 10(0.9999...) = 9.999...  
10x = 9.9999....

if x = y, then we know x - c = y - c. (in this case, c = x = 0.999...), so we KNOW:

10x - x = 9.9999 - x
9x = 9.99999 - 0.9999 (because x = 0.9999)
9x = 9

if x = y, then x/9 = y/9. so we KNOW:

9x/9 = 9/9
x = 1    

since x = 0999999 and (not or) x = 1, then 1 = 0.99999...

It has to be true or algebra is fundamentally flawed.

Nothing I did above is wrong. Absolutely nothing.

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You cannot do the same with squaring because if x^2 = y^2, x only = y SOMETIMES.

Fortunately, I did not square. I used operations which were ALWAYS true.

I didn't say you did anything wrong mathematically.

x = 0.999... isn't something you can solve because it already is solved. This isn't a proof was my point.

We're not solving that x = 0.999. We are showing 1 = 0.999..

...and that is not a valid way of showing it.

I've shown it was correct mathematically as you agreed was valid. Your squaring example is not applicable counter-proof. All you've proven is that sometimes things don't work out, but you've done nothing to show that what I did wouldn't work out.

You guys must know something, this kind of thread is not far from a troll =).

@Jay520: do you realize that many people in here don't even know how to construct the set of real numbers? It could close this debate once and for all, even if it wouldn't be in a satisfactory way.