Can someone prove to me how .999...=1

When he asked for proof?

 

It comes so close to 1 that the difference doesn't even matter anymore

 

Not true. It is exactly 1.

 

The people disagreeing aren't noticing the difference between '0.999...' and '0.999'.

 

This.

1/3 = 0.333...
3/3 = 1 / 0.999...

3/3 is not equal to 0.999, however.

 

how can .999... = 1?

The limit as .999... approaches infinite decimals places out would be 1. But does it actually ever touch?

 

Yes. I already proved the convergence of the infinite series in an earlier post.

 

0.9999999 repeating I'm assuming in this case is not = 1. Here's the example I will be using to prove this. Lets take an easily understood equation.

1/(x-1) therefore we can assume x-1 =/= 0 because division by zero is impossible. Therefore x can be all values for the number system from (-&#8734;< x < 1) and (1< x < &#8734 therefore we can get infinitely close to the number 1 for example 0.999 repeating so long as we don't reach one because in this case 1 is a vertical asymptote. So in my opinion 0.99999 repeating does not equal 1.

If were referring to non repeating .9999s my post is way off its mark and for that I'm sorry.

 

We're talking about "0.999..." Not "0.999"

 

No. Several things are wrong with your argument. if f(x) = 1/x-1, f(1) exists -- it is just undefined. Second, how does your example prove that you cannot get to one? I'd just say that f(.999...) is also undefined... Last, you can still get to 1 on the x-axis, which is what you're doing...you just can't get to it if you stay on the line. But then again, you can't get to x = 0 by staying on the graph of f(x) = 1/x.

Also, this isn't an opinion. This is a fact.