## 06 Apr 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.