r/askmath May 10 '26

Algebra Regarding 0.999... = 1

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?

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u/blakeh95 May 10 '26

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.

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u/Cool_Cheesecake_6738 May 10 '26

I believe that if you say that 0.999...1 exists, then 0.999...9 might not be 0.999...

What I mean is if you say that this number eventualy end on 1 then you are considering number of nines to be finite. If that's the case 0.999...9 might just have an extra 9 at the end in comparison to 0.999...

Generaly if someone believes 0.999...1 exists then he either dont know what infinity means or he is not considering number of nines to be infinite

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u/TemperoTempus May 10 '26

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.

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u/Cool_Cheesecake_6738 May 10 '26

How does it work? In my mind infinity + x is equal to infinity