Debunking issue in proving $ \sqrt2 $ is irrational

Question

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.

Answer 1

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.)

Answer 2

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 $\gcd(s,t) \neq 1.$