Prove that $\sqrt 5$ is irrational

Question

I have to prove that $\sqrt 5$ is irrational.

Proceeding as in the proof of $\sqrt 2$, let us assume that $\sqrt 5$ is rational. This means for some distinct integers $p$ and $q$ having no common factor other than 1,

$$\frac{p}{q} = \sqrt5$$

$$\Rightarrow \frac{p^2}{q^2} = 5$$

$$\Rightarrow p^2 = 5 q^2$$

This means that 5 divides $p^2$. This means that 5 divides $p$ (because every factor must appear twice for the square to exist). So we have, $p = 5 r$ for some integer $r$. Extending the argument to $q$, we discover that they have a common factor of 5, which is a contradiction.

Is this proof correct?

Answer 1

It is, but I think you need to be a little bit more careful when explaining why $5$ divides $p^2$ implies $5$ divides $p$. If $4$ divides $p^2$ does $4$ necessarily divide $p$?

Answer 2

Yes, the proof is correct. Using this method you can show that $\sqrt{p}$ for any prime $p$ is irrational. During the proof you essentially use the fact that when $p|u^2$ where $p$ is a prime, then it implies that $p|u$. This is true for primes, but is not true in general. You can prove this as below Let $n|u^2,\ \gcd(n,u)=d$. Then, let $n=rd,\ u=sd $. So, $$u^2=kn \Rightarrow s^2d^2=k r d\Rightarrow s^2d=kr$$ if we have $n\not{|}\ u$, since $\gcd(s,r)=1$, we have $$r|d$$ Then, with $d>1$, $n\not{|}\ u$, but $\ n|u^2$. If $n$ is prime, then $d=1\Rightarrow r=1$ unless $\ n|u$.

Answer 3

Yes, the proof is correct. But I think you still need a lemma to reinforce your proof

Lemma: $$\text{If }P|Q^2,\text{ where P is a prime, then } P|Q$$

Proof: By the unique factorization theorem，$Q$ is able to rewrited as a product of distinct prime numbers: $$Q = P_1^{e_1}P_2^{e_2}P_3^{e_3}\ldots P_k^{e_k}\tag{1}$$ where $P_1,P_2,\ldots P_k$ are distinct prime numbers and $e_1,e_2,\ldots e_k$ are positive integers. Then: $$Q^2 = P_1^{2e_1}P_2^{2e_2}P_3^{2e_3}\ldots P_k^{2e_k}\tag{2}$$ By (1),(2), we know both $Q$ and $Q^2$ are a product of distinct prime numbers that belong to the set $\{P_1,P_2,\ldots,P_k\}$. Because $P|Q^2$ and $P$ is also a prime, It implies $P\in\{P_1,P_2,\ldots,P_k\}$. Hence, P|Q, which complete the proof.

Answer 4

The number of prime divisors of $p^2$ is even. Is that true for $5q^2$?