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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.

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  • The general Euclid's Lemma is not needed to show for a specific prime $p$ that if $p$ divides $a^2$ then $p$ divides $a$. – André Nicolas Nov 12 '14 at 04:43
  • 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. – Lubin Nov 12 '14 at 04:47
  • @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. – user103816 Nov 12 '14 at 04:49
  • 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. – Lubin Nov 12 '14 at 04:52
  • @Lubin The book isn't written to be a rigorous text. It is written for first time math reader. The number of 2's, how many may be, are for sure finite in $2b^2$. I've recently noticed that FTA requires Euclid's lemma to be proved. – user103816 Nov 12 '14 at 04:54
  • @AndréNicolas Could you tell me how we can prove that "$p|a^2 \implies p|a$, $p$ is prime" without using the Euclid's lemma. By Euclid's lemma I mean this theorem: http://en.wikipedia.org/wiki/Euclid's_lemma – user103816 Nov 12 '14 at 04:56
  • @Lubin: what is actually needed if the fact that every nonzero natural number $n$ can be written in a unique way in the form $n=2^kq$ where $k\geqslant 0$ and $q$ is odd. This is way simpler than the FTA (it still require a proof). – Taladris Nov 12 '14 at 04:58
  • It might be taken as the defining properties of a prime (rather than "only divided by itself and $1$") that a number $p$ is prime iff $p>1$ and for any two integers $a$ and $b$, $p\mid ab$ implies $p\mid a$ or $p\mid b$. However this is taken (a definition or a theorem), it is used in proving the FTA, and it can be used to prove $p\mid a^2 \Rightarrow p\mid a$. FTA is a bit overkill in my opinion. – Arthur Nov 12 '14 at 05:13
  • @user31782: There may be nothing simpler than Euclid's Lemma. Luckily that has a simple proof once we know the gcd of two numbers is an integer linear combination of the numbers. However, as I pointed out earlier, we need almost nothing to prove that if $2$ divides $a^2$ then $2$ divides $a$. – André Nicolas Nov 12 '14 at 05:29
  • @AndréNicolas As I understand it to prove that "if $2$ divides $a^2$ then $2$ divides $a$" we require Fundamental Theorem of Arithmetic, which in turn uses Euclid' lemma to be proved so aren't we actually(indirectly) relying Euclid's lemma to prove that if "$2$ divides $a^2$ then $2$ divides $a$"? – user103816 Nov 12 '14 at 05:37
  • If $2$ does not divide $a$, then $a=2k+1$ for some $k$, from which we easily derive that $2$ does not divide $a^2$. No Euclid's Lemma, just primitive divisibility, non-divisibility facts, $2$ does not divide $1$. – André Nicolas Nov 12 '14 at 05:42
  • @AndréNicolas I recently noted that FTA can be proved without using Euclid's lemma, so ok I understand what did you say. – user103816 Nov 12 '14 at 05:43
  • @AndréNicolas You are proving by contradiction. $a=2k+1 \implies a^2= 2(k^2+1)+1 =$odd which is false hence a=odd is false. Ok I got it. – user103816 Nov 12 '14 at 05:51
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    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. – André Nicolas Nov 12 '14 at 05:52
  • Well, it is $(2k+1)^2=2(2k^2+2k)+1$. – André Nicolas Nov 12 '14 at 05:53
  • @AndréNicolas Could you refer me to some text(book?), which is written in easy English and covers all these kinds of topics. Oh my bad on that $a^2$ is $2(k^2+k)+1$. – user103816 Nov 12 '14 at 05:55
  • Most elementary number theory books are pretty good. If you are close to a university library I would recommend looking at the one by Silverman, part of the Friendly series. – André Nicolas Nov 12 '14 at 06:02
  • Knowledge of something like the proof in your P.P.S. is strongly hinted at in the writings of Aristotle. If there's still interest--I know this question is old, but it recently got bumped--I will try to locate it and post an excerpt of the passage I have in mind. – Will Orrick Aug 14 '15 at 07:16
  • @AndréNicolas When we say If $2$ does not divide $a$, then $a=2k+1$ for some $k$, aren't we using Euclid division algorithm? – user103816 Aug 14 '15 at 20:00
  • @WillOrrick Yes I'd appreciate it. I want to know as much as possible. I like math's history. – user103816 Aug 14 '15 at 20:01
  • @user103816: We could refer to the division algorithm, but this special case can be dealt with separately. – André Nicolas Aug 14 '15 at 20:20
  • @AndréNicolas Could you elaborate how would we show that the number $a$ is of form $2k+1$ if it is not divisble by $2$, without invoking Euclid division algo? – user103816 Aug 14 '15 at 20:22

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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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