Proving $\sqrt{2}$ is irrational: $ \sqrt{2}= \frac{a}{b} $ in lowest terms, thus $ {2}= \frac{a^2}{b^2} $ which leads to $ {2}= \frac{2{b^2}}{b^2} $, so $ {a} $ must be even, as even times even returns an even result (odd times odd returns an odd result). Thus, $ {a} $ can be represented as $ {2k} $. Now, $ {b^2}= \frac{a^2}{2} $ which evaluates to ${b^2}= \frac{(2k)^2}{2}$ resulting in $b^2=2k^2$. As such, $b^2$ is even, therefore $b$ must be even. This concludes that the fraction $\frac{a}{b}$ of $\sqrt2$ can never be in lowest terms, and as such it is irrational. However, I could not wrap my head around this because from my understanding $b = \pm\sqrt{2k^2}$ so that $$b =\sqrt{2}\cdot k$$ So, if $b$ is to be even or nonzero, then whatever value $k$ holds, it must multiply with the assumed irrational number $\sqrt{2}$ to give an even rational number. However, that is not possible because for all positive real numbers multiplication of an irrational number and a rational number always outputs an irrational result. Futhermore, since $a=\sqrt{2b^2}$ leading to $a=2k$, which results into the following: $$ \frac{2k}{\sqrt{2}k} = \frac{2}{\sqrt{2}} = \sqrt2$$ And when squared it reveals exactly why we consider it irrational: $$\sqrt2 = \frac{2}{\sqrt2}\rightarrow 2 = \frac{4}{2} \rightarrow \frac{2}{1}$$ So I would also like to know why we base this irrationality of $\sqrt2$ off of $\frac{4}{2}$ which can be simplified. Please indicate any wrong doings in my evaluation or if I missing some big picture here.

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    $\begingroup$ Your $b=\sqrt2k$ starts from the assumption that $b$ is a non-zero integer such that there is another integer $a$ such that $a^2/b^2=2$. If you work around the conclusions coming from said FALSE assumption, you'll end up proving all kinds of false statement (in fact, you might prove any statement, be it false, true or undecidable). $\endgroup$
    – user239203
    Apr 5, 2020 at 21:48
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    $\begingroup$ $b = \sqrt2 k$ is perfectly possible... if $\sqrt 2$ were a rational number, which we don't know yet in the proof if it is or not. $\endgroup$
    – Milten
    Apr 5, 2020 at 21:57
  • $\begingroup$ In fact, a funny thing I used to do in set theory exercises was to dissect proofs by contradiction as such: if the goal was to prove the predicate $P$, I would first write a proof of $P$ assuming $\neg P$, and then I would basically said that $$\text{True}\equiv P\lor\neg P\implies P\lor P\equiv P$$ $\endgroup$
    – user239203
    Apr 5, 2020 at 22:01

2 Answers 2


Keep in mind that $a$ and $b$ don't actually exist - this is a proof by contradiction. The proof should really begin "Suppose $\sqrt{2}={a\over b}$ with ${a\over b}$ in lowest terms." That "suppose" may seem like a small point, but it's really the key here: we're not outright asserting that such $a$ and $b$ exist, we're considering what would happen if they did - and we're going to show that they can't.

In a bit more detail, we begin by supposing that there are such $a$ and $b$. We then deduce something impossible - namely, that both $a$ and $b$ are even. At that point it's clear that our initial assumption was wrong: that is, $\sqrt{2}$ cannot be written as a fraction in lowest terms. Since every rational number can be written in lowest terms, this means that $\sqrt{2}$ is irrational.

So the fact that "$b=\sqrt{2}k$" doesn't make any sense isn't a problem: the whole point is that we've figured out that such $a$ and $b$ are impossible in the first place.

(Meanwhile your second question - about ${4\over 2}$ vs. ${1\over 2}$ - is a bit unclear to me. But I think once you understand the above, this will clear up too.)

  • $\begingroup$ matter of taste, I like the quadratic forms style approach. There is no need to introduce a square root symbol, and the coprime assumption and the `both even' conclusion/contradiction are fairly well separated. $\endgroup$
    – Will Jagy
    Apr 5, 2020 at 23:55

I prefer the version with quadratic forms, the relevant word is anisotropic. I think the "infinite descent" makes people nervous. I do all the descent first.

The form is going to be $x^2 - 2 y^2$ and we are to show it cannot be zero with rational and nonzero $x,y.$ First, ASSUME we have such rationals, multiply them both by the least common multiple of the numerators. We still get zero, but now we have $u^2 - 2 v^2 = 0$ with $u,v$ nonzero integers. Next, find $\gcd(u,v)$ and divide both $u,v$ by that. We still get zero, but now we have COPRIME nonzero integers with $s^2 - 2 t^2 = 0.$ That is the current state of the ASSUMPTION. As Samuel L. Jackson said in a movie "Long Kiss Goodnight" with Geena Davis, "When you make an assumption, you make an ass out of you and umption."

LEMMA: if $$ m^2 - 2 n^2 \equiv 0 \pmod 4 \; , \; $$ then both $m,n$ are even. Proof: well, $m$ must be even, so now $2 n^2 \equiv 0 \pmod 4.$ But that requires $n^2$ even, therefore $n$ is even.

Our assumption became that we had coprime nonzero integers $s,t$ with $s^2 - 2 t^2 = 0.$ Well, this implies $s^2 - 2 t^2 \equiv 0 \pmod 4,$ so both $s,t$ are EVEN, and $\gcd(s,t) \neq 1.$


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