Numerous online resources parrot the usual proof by contradiction of the irrationality of $\sqrt{2}$. These all rely upon the assumption that the rational form (say, $a/b$) is in its simplest representation, i.e., $a$ and $b$ are co-prime, and then easily demonstrates that in for $(a/b)=\sqrt{2}$ both $a$ and $b$ must be even, which contradicts the assumption that they are co-prime.

However, in all of the renditions that I have seen, a reasoning step appears to me to be missing, or at least never stated. Specifically, the co-primacy of $a$ and $b$, and thus that $a/b$ is in simplest form, is assumed but never appears to enter into the proof. If we simply don't assume that $a$ and $b$ are co-prime, then they could easily both be even, and there would be no contradiction. And since their co-primacy never appears to enter into the proof, there's no apparent reason for making that assumption.

Of course, don't think that the proof is wrong. (I'm not insane!) I'm just looking for an explication of the missing assumption (or reasoning step?) that demands co-primacy of $a$ and $b$, or, more likely, an explanation that there's nothing missing, but instead, I'm missing something! (I.e., I'm not insane, just stupid! :-)


  • 4
    $\begingroup$ Voting to close as "needs clarity". OP keeps saying (in comments) that coprimeness is an assumption in the usual proof of irrationality of $\sqrt2$, even after it was explained several times that it is not an assumption but rather information obtained by a preliminary construction, namely reducing the fraction. $\endgroup$ Sep 19, 2020 at 16:29
  • 2
    $\begingroup$ I’m voting to close this question because OP is obstinate in comments. $\endgroup$
    – user52817
    Sep 19, 2020 at 18:53

2 Answers 2


(An answer because this is a little long for a comment) Here is a proof which does not use coprimality, but is essentially the same: Suppose that $a/b = \sqrt{2}$ where $a$ and $b$ are integers. Then $a=\sqrt{2}b$ and so $a^2 = 2 b^2$ where both sides are integers.

Let $n$ be the number of factors of 2 in $a$ and $m$ be the number of factors of two in $b$. By the fundamental theorem of arithmetic, $a^2$ and $2 b^2$ have prime factorizations and, further, they must be the same. (This uniqueness is also used in defining $n$ and $m$.) In particular, the number of factors of 2 must be equal. Thus $2n = 2m+1$, which cannot be satisfied by any integral $n$ and $m$ as it would imply that there is a number which is both even and odd. This is the desired contradiction.

Note: You don't really need coprimality, the only place that it enters in here is that it implies that at least one of $n$ or $m$ is zero.

  • $\begingroup$ Brilliant! (I don’t have enough rep to upvote.) This is way clearer (to me, anyway). Thank you! $\endgroup$ Sep 20, 2020 at 5:05
  • $\begingroup$ BTW, answers were closed and I got a private msg from god, or something, about this not being about math education. I beg to differ. It’s all about how to explain possibly the most common example of proof by contradiction, as well as possibly the most common proof of irrationality. We’re not talking about the math, but about the explanation. And bears directly on how these things are explained in classroom settings. $\endgroup$ Sep 20, 2020 at 5:14
  • $\begingroup$ @jackisquizzical I originally voted to close because the comments on your other answer seemed to make it about personal understanding. I've now voted to reopen, and maybe you could get others to vote the same if you 're-pitch' the question to be about student understanding of this theorem. $\endgroup$
    – Adam
    Sep 20, 2020 at 23:05
  • $\begingroup$ @jackisquizzical I've certainly seen students confused about the standard proof before, in a somewhat similar manner. $\endgroup$
    – Adam
    Sep 20, 2020 at 23:10
  • $\begingroup$ I've seen this doubly downvoted over a year after the fact. Not really sure why. $\endgroup$
    – Adam
    Oct 18, 2021 at 2:09

You are correct in that the coprimality of $a$ and $b$ is not used in its full strength. It is adequate to merely assume that they are not both even. But since people are so used to reducing a rational number to lowest terms, making this assumption improves the readability of the proof. The unnecessary stronger assumption of coprimality is not used, but it does no harm, since it guarantees they are not both even.

Thing of flow of proof as starting like this:

"Before we get going, let us agree that $a$ and $b$ are coprime. If they are not, then before we get started, go off and reduce them so they are. I'll wait. Done? OK, let's get going..."

Then after some thought, we end up with a conclusion that both are even, which is a contradiction.

  • 8
    $\begingroup$ @jackisquizzical I don't see why your comment isn't already answered by "If they are not, then before we get started, go off and reduce them so they are. I'll wait." The point is that you don't have to assume coprimality; you obtain it by reducing the fraction. $\endgroup$ Sep 19, 2020 at 2:08
  • 3
    $\begingroup$ This doesn't seem to be a mathematics education question but, briefly, if you drop the coprimality assumption you end up with a contradiction anyway. Once you find that $a$ and $b$ are both even you note that the argument can be repeated with the integers $a'=a/2$ and $b'=b/2$. But then the argument can be repeated with the integers $a''=a'/2=a/4$ and $b''=b'/2=b/4$. Keep going and you find that the argument can be applied to the integers $a^{(1000000)}=a/2^{1000000}$ and $b^{(1000000)}=b/2^{1000000}$. You might eventually conclude that something has gone wrong. $\endgroup$ Sep 19, 2020 at 3:09
  • 1
    $\begingroup$ Either way of arguing works. If $\sqrt{2}=a/b$, you might notice that any $a/b$ will equal some $c/d$, with at least one of $c$, $d$ odd. Then run the argument on $c/d$. If you didn't spot that, you get another chance to notice it after proving that $a$ and $b$ are both even. The infinite descent style of proof was common in Fermat's day. $\endgroup$ Sep 19, 2020 at 3:40
  • 2
    $\begingroup$ @jackisquizzical: Perhaps this will help. Assume $\sqrt{2}$ is rational. Then $\sqrt{2} = a/b$ for some integers $a$ and $b.$ Removing common factors, if any, we have $\sqrt{2} = c/d$ where $c$ and $d$ are coprime. Now continue by using $c$ and $d$ to get a contradiction, which tells us our assumption is not true. $\endgroup$ Sep 19, 2020 at 8:23
  • 4
    $\begingroup$ @jackisquizzical: There is not a missing logical step. Maybe you can link to one of the proofs you're confused about so we can all be sure we're talking about the same thing. I for one am confused about your confusion at this point! $\endgroup$
    – Thierry
    Sep 19, 2020 at 18:21

Not the answer you're looking for? Browse other questions tagged or ask your own question.