>>11560822

>.00...1 = 0

I've thought about this. So basically the issue here is whether any numbers come before, but not after the ellipses, or bar if you're using bar notation. Earlier I thought that by showing that

0.999... = 0.9 + 0.0999...

0.9 = 0.8999...

0.999... = 0.8999... + 0.0999..

0.999...8

That this would show the equivalence to be wrong. But when you actually try adding this number 0.999...8 to anything, the 8 has no effect, because there is always a 9 preceding it. Likewise, in the case of 0.000...1, there is always a 0 preceding, thus 0.999...8 is effectively the same as 0.999... and 0.000...1 is effectively the same as 0.

Any arithmetic operation on 0.999... produces the same effects as if it were 1, but with a different decimal representation.

for example:

2 * 1 = 2

2 * 0.999... = 1.999...8

You might again say that, well 1.999...8 is not the same as 2, but, once again, if you realize that anything after the ellipses has no effect, then you are left with 1.999... which means that

0.999... + 0.999... = 1.999...

Which could only be true if 0.999... were equal to 1.

So, since 0.999... has the same arithmetic effect as one, it must be equal to 1.

It's interesting this debate is actually is a good segue into abstract algebra because it forces you to determine whether two numbers are equal based on their effects on other numbers rather than just what they "look like," i.e. this is a unit because it behaves as a unit, rather than, this is a unit because it looks like one.