If p is a prime then √p is irrational

To prove √p is irrational, where p is a prime, we will need the following theorem:

Theorem: If $$p$$ is prime then $$p|ab,$$ then $$p|a$$ or $$p|b.$$

Proof: If $$p|a,$$ we are done. So let us assume that $$pmid a.$$ Therefore, $$gcd(p,a)=1.$$ Hence, by Euclid’s lemma, $$p|b.$$ [Euclid’s lemma: If $$a|bc,$$ with $$gcd(a,b)=1,$$ then $$a|c.$$]

Now to prove √p is irrational (where p is a prime):

Assume that √p is rational.

Then there exists two integers a, b such that $$frac{a}{b}=sqrt{p}$$ where gcd(a,b)=1 and $$bne 0.$$

Then $$frac{a^2}{b^2}=p.$$ Which implies

$$a^2=pb^2.$$ __________(1)

$$Rightarrow p$$ divides $$a^2.$$

$$Rightarrow p$$ divides $$a.$$ (By the above theorem.)

So there exists an integer $$a_1$$ such that $$a=pa_1.$$

So, from (1), we get $$(pa_1)^2=pb^2.$$

$$Rightarrow p^2a_1^2=pb^2.$$

$$Rightarrow pa_1^2=b^2.$$ (Dividing both sides by p.)

This implies p divides b. This is a contradiction since gcd(a,b)=1.

Thus, when p is a prime, √p is irrational.

Particular Case:- √2 is irrational.