## 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.