Usually the proof of $\sqrt2$ is irrational is done by contradiction(e.g. here), but I found another similar but short proof in the book "Beginning Algebra for College Students" by Lloyd Lincoln Lowenstein.

The proof goes like this:

Suppose then that $x^2=2$ and $x=a/b$, where $a$ and $b$ are integers. Then $$\left(\dfrac ab\right)^2=2\ \ \text{or}\ \ \dfrac {a^2}{b^2}=2;$$ and $$a^2=2b^2$$ Consider the number $b^2$. If it has the factor 2, it has the factor $2$ an even number of times and $2b^2$ has the factor $2$ an odd number of times. But this says that $a^2$, the square of an integer, has the factor $2$ an odd number of times, which is impossible. We see that $\left(a/b\right)^2=2$, must be a true statement . But we know that the second statement is false, and therefore the first must be false and there is no pair of integers $a$ and $b$ such that $\left( a/b\right)^2$; or the number whose square is $2$ cannot be a rational number.

The sole idea is that the quantity on the right hand side, namely $2b^2$ does contain the factor $2$ an odd number of times and the quantity on the left hand side, namely $a^2$ will always contain $2$ an even number of times.

$a$ and $b$ are not presumed to be coprime as opposed to the case in the usual proof(e.g. here on wikipedea), nor does it use the Euclid's lemma.

The problem is that I am not able to find this proof on internet. So,

What is the name of this proof and where can I find about it in more detail? Who discovered it and how, etc

P.S: I've found the same proof and a short discussion(in the comments) in this post. It seems like that the the history of the proof is not clear, nonetheless I would like to know about it as much as it is available.

P.P.S: I've found another similar prove here on wikipedea. It is based on the fact that if $\dfrac ab$ is in its lowest terms then at least one of $a$ and $b$ should be odd. We then show that both $a$ and $b$ are even-- a contradiction.

  • $\begingroup$ The general Euclid's Lemma is not needed to show for a specific prime $p$ that if $p$ divides $a^2$ then $p$ divides $a$. $\endgroup$ Commented Nov 12, 2014 at 4:43
  • $\begingroup$ This proof uses the Fundamental Theorem of Arithmetic, which says that there is essentially only one way of expressing an integer as a product of primes; in particular, the number of factors of the prime $2$ does not depend on how you factorize your original number. The proof of the irrationality of $\sqrt2$ is implicit, once one has the FTA. But I think your text is deficient in not pointing out that this Theorem has been used: it’s not so clear at the outset that the number of $2$’s in a number is well-defined. $\endgroup$
    – Lubin
    Commented Nov 12, 2014 at 4:47
  • $\begingroup$ @Lubin I understand that it uses the "Fundamental Theorem of Arithmetic". But the book itself doesn't mention it!, perhaps the author found it too trivial to mention. $\endgroup$
    – user103816
    Commented Nov 12, 2014 at 4:49
  • $\begingroup$ Well, all the worse for the author, because the FTA does require a proof: it’s by no means trivial, as you’ll see when you go through a proof. $\endgroup$
    – Lubin
    Commented Nov 12, 2014 at 4:52
  • 1
    $\begingroup$ Actually, I believe you do not understand. We do not need FTA for the proof of the irrationality of $\sqrt{2}$. We do not need Euclid's (general) Lemma. We do not need general properties of gcd. $\endgroup$ Commented Nov 12, 2014 at 5:52

1 Answer 1


That argument can be reasonably called the 2-adic (or dyadic) argument, and that word would instantly signal to mathematicians what argument is being made. Although the proof does not elaborate on it, the argument shows not only that there are no solutions in the rational numbers $\mathbb{Q}$ but also that there are no solutions in a broader setting, the 2-adic completion $\mathbb{Q_2}$. It does this using basic facts about the "2-adic valuation" that counts the number of powers of $2$ that divide a number.

The first technique for showing a Diophantine equation has no rational solutions is to show there are no solutions mod $p^n$ for $p$ a prime. A better formulation that applies to more general sorts of algebraic situation is to take the "$p$-adic localization" of the equations you want to solve, which keeps $p$ fixed but takes account of all possible $n$ at once. The Hasse principle in number theory says essentially that this is the first thing to try when proving no solutions exist, and it is known that this principle is particularly effective for equations of degree $2$.

In the argument as presented in the question, the numerator and denominator $a$ and $b$ do not have to be handled separately from the number that hypothetically squares to $2$. Any nonzero rational number $x$ has a well-defined power of $2$ that divides it, and for $ x^2$ this power is even, which makes a rational solution of $x^2=2$ impossible.


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