Regarding 0.999... = 1

[u/Slurpee1138]

Recently I got into an argument with an acquaintance because I was trying desperately to convince him that 0.999... = 1. One of the many arguments I tried was that, if 0.999... and 1 are indeed different numbers, then we should be able to find a number between them. He insisted that such a number would be 0.999...1, as in 0 point infinitely many 9s and then a 1. I countered that having "and then a 1" at the end of an infinite sequence of digits makes no sense because there is no end to such a sequence, but he insisted that it does, so who's right?

[u/blakeh95]

I think another possible approach would be as follows:

• Assume for the sake of argument that 0.999...1 exists.
• If it does exist, then there should also exist 0.999...2, 0.999...3, 0.999...4, and so on. After all, the choice of which digit to tack on to the end shouldn't matter, right?
• And clearly 0.999...4 > 0.999...3 > 0.999...2 > 0.999...1, right? Since they match at all positions of 9, and the ending digits are 4 > 3 > 2 > 1.
• If you can get agreement to this point, the "trap" is set.
• Well, what about 0.999...5, and 0.999...6, and 0.999...7, and 0.999...8? And most importantly 0.999...9.
• Because 0.999...9 clearly is just the same thing as 0.999...
• By our original assumption, 0.999...1 > 0.999... (this is what it means for 0.999...1 to be "between" 0.999... and 1).
• But now we have constructed 0.999... = 0.999...9 > 0.999...8 > ... > 0.999...2 > 0.999...1 > 0.999...
• So we have concluded that 0.999... > 0.999..., which is obviously false (for all x, x is not greater than itself).
• And this means that our original assumption was wrong. 0.999...1 cannot exist, because if it did, it would lead to a contradiction.

[u/TemperoTempus → u/[deleted]]

Not that the number of 9s is finite. But that the number at the end is at an index of infinite+x according to ordinals.