Prove that $\sqrt 5$ is irrational 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?
 A: 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 $\ n|u$.
A: Let us assume $√5$ is rational.
$$√5=\frac{x}{y}$$
Square both sides of the equation above
$$5 =\frac{x^2}{y^2}$$
Multiply both sides by $y^2$
$$5  y^2 =\frac{x^2 }{y^2}$$
We get $5  y^2 = x^2$
Another important concept before we finish our proof: Prime factorization.
Key question: is the number of prime factors for a number raised to the second power an even or odd number?
For example, $6^2$, $12^2$, and $15^2$
$6^2 = 6 × 6 = 2 × 3 × 2 × 3$ ($4$ prime factors, so even number)
$12^2 = 12 × 12 = 4 × 3 × 4 × 3 = 2 × 2 × 3 × 2 × 2 × 3$ ($6$ prime factors, so even number)
$15^2 = 15 × 15 = 3 × 5 × 3 × 5$ ($4$ prime factors, so even number)
There is a solid pattern here to conclude that any number squared will have an even number of prime factors
In order words, $x^2$ has an even number of prime factors.
Let's finish the proof then!
$5  y^2 = x^2$
Since $5  y^2$ is equal to $x^2$, $5  y^2$ and $x^2$ must have the same number of prime factors.
We just showed that
$x^2$ has an even number of prime factors, $y^2$ has also an even number of prime factors.
$5  y^2$ will then have an odd number of prime factors.
The number $5$ counts as $1$ prime factor, so $1$ + an even number of prime factors is an odd number of prime factors.
$5  y^2$ is the same number as $x^2$. However, $5  y^2$ gives an odd number of prime factor while $x^2$ gives an even number of prime factors.
This is a contradiction since a number cannot have an odd number of prime factors and an even number of prime factors at the same time.
The assumption that square root of $5$ is rational is wrong. Therefore, square of $5$ is irrational.
A: 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 $p$?
A: The number of prime divisors of $p^2$ is even. Is that true for $5q^2$?
A: 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.
