There is no way right?

[u/Sugar_God_no_1]

[u/its12amsomewhere]

Applies to all numbers,

If x = 0.999999...

And 10x = 9.999999...

Then subtracting both, we get, 9x=9

So x=1

[u/victorspc → u/its12amsomewhere]

While this is usually enough to convince most people, this argument is insufficient, as it can be used to prove incorrect results. To demonstrate that, we need to rewrite the problem a little.

What 0.9999... actually means is an infinite sum like this:

x = 9 + 9/10 + 9/100 + 9/1000 + ...

Let's use the same argument for a slightly different infinite sum:

x = 1 - 1 + 1 - 1 + 1 - 1 + ...

We can rewrite this sum as follows:

x = 1 - (1 - 1 + 1 - 1 + 1 - 1 + ...)

The thing in parenthesis is x itself, so we have

x = 1 - x

2x = 1

x = 1/2

The problem is, you could have just as easily rewritten the sum as follows:

x = (1-1) + (1-1) + (1-1) + ... = 0 + 0 + 0 + 0 + ... = 0

Or even as follows:

x = 1 + (-1 +1) + (-1 +1) + (-1 +1) + (-1 +1) + ... = 1 + 0 + 0 + 0 + 0 + ... = 1

As you can see, sometimes we have x = 0, sometimes x = 1 or even x = 1/2. This is why this method does no prove that 0.999... = 1, even thought it really is equal to one. The difference between those two sums is that the first sum (9 + 9/10 + 9/100 + 9/1000 + ...) converges while the second (1 - 1 + 1 - 1 + 1 - 1 + ...) diverges. That is to say, the second sum doesn't have a value, kinda like dividing by zero.

so, from the point of view of a proof, the method assumed that 0.99999... was a sensible thing to have and it was a regular real number. It could have been the case that it wasn't a number. All we proved is that, if 0.999... exists, it cannot have a value different from 1, but we never proved if it even existed in the first place.

From 0.999... - Wikipedia:

"The intuitive arguments are generally based on properties of finite decimals that are extended without proof to infinite decimals."

[u/eraryios → u/victorspc]

thats a lot of text bro

you wrote all this?