How can I prove $\sqrt{5}$ is irrational using: for $n \in \mathbb{Z}$, $n^2 = 5a$ for some integer $a$ if and only if $n = 5b$ for some integer b$

Question

This looks to be a proof by contradiction. So, I have to assume $\sqrt{5}$ is rational ($= \frac{c}{d}; c,d \in \mathbb{Z}; d \neq 0$). I guess I should square both sides which brings me to: $$5 = \frac{c^2}{d^2}$$

From there I can multiply both sides by $d^2$ to get $c^2 = 5d^2$. I see that this is simillar to the fact given, but how should I continue from here?

Answer 1

Guide:

Impose the condition thta $gcd(c,d)=1$, that is $c$ and $d$ have no common divisor.

Once you get $$c^2=5d^2, \tag{1}$$ we know that $c$ is a multiple of $5$, write $c=5k$, $k\in \mathbb{Z}$.

Try to substitute that back into $(1)$ and look for the contradiction with the condition that $gcd(c,d)=1$.

Answer 2

Used in Song's guide:

$c^2 = 5d^2$, I.e. $5$ divides $c^2.$

Theorem of Number Theory:

If $p$ is prime, and $p$ divides $ab$,

$\rightarrow $:

$p$ divides $a$, or $p$ divides $b$, where $a,b$ are integers.

Hence : $5$ divides $c$.