Show that $\sqrt{6}$ is irrational.

Question

Suppose $x\in \mathbb{Q}$ such that $x^{2}=6$. Since $x\in \mathbb{Q}$, there exists $m,n \in \mathbb{Z} $ where either $m$ or $n$ is odd such that $x=\frac{m}{n}$.

$\implies$ $x^2=(\frac{m}{n})^2=\frac{m^2}{n^2}=6$

$\implies$ $m^2=6n^2$, so $m^2$ is even. Hence, $m$ is even.

Since $m$ is even, $m=2k, k\in \mathbb{Z}$.

Then, $m^2=(2k)^2=4k^2=6n^2$

$\implies$ $n^2$ is even, so $n$ is even. But one of $m$ or $n$ must be odd, so $x\notin \mathbb{Q}$.

Therefore, $\sqrt{6}$ is irrational.

Does everything look alright here?

@copper.hat Well, it is enough (as stated) that they cannot both be even — Hagen von Eitzen, Oct 24 '13 at 16:29
There's a more general theorem, if $n^2\mid m^2$ then $n\mid m$. But this works fine. — Thomas Andrews, Oct 24 '13 at 16:31

score 6 · Answer 1 · answered Oct 24 '13 at 16:28

6

This is fine and in effect you show that $\sqrt{2a}$ is irrational if $a$ is odd.

answered Oct 24 '13 at 16:28

Hagen von Eitzen

374,180

score 3 · Answer 2 · answered Oct 24 '13 at 17:19

Yes, it's right. You can generalize the method; for a positive integer $z$ and a prime number $p$, denote by $\mu_p(z)$ the maximum exponent $k$ such that $p^k$ divides $z$ ($\mu_p(z)=0$ if $p$ doesn't divide $z$).

From the unique factorization, it's clear that $\mu_p(xy)=\mu_p(x)+\mu_p(y)$.

We want to prove that if $\sqrt{z}$ is rational, then $\mu_p(z)$ is even, for any prime $p$.

Suppose there exist $m$ and $n$ positive integers such that $(m/n)^2=z$. Then $$ m^2=zn^2 $$ Let $p$ be a prime; then $$ \mu_p(m^2)=2\mu_p(m)=\mu_p(zn^2)=\mu_p(z)+2\mu_p(n); $$ therefore $\mu_p(z)=2(\mu_p(m)-\mu_p(n))$ is even. In particular $z$ is a perfect square.

Since $2^2<6<3^2$, $6$ is not a perfect square, so its square root is not rational.

Show that $\sqrt{6}$ is irrational.

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