Question

Prove that 5 is irrational.​

Answer 1

Answer:

$\large{\bf{\purple{proof : }}} \:$

Let us assume that square root 5 is rational. where p, q are the integers, and q is not equal to 0. The integers p and q are coprime numbers. thus,

$\sqrt{5} = \large{ \frac{p}{q}}$

$\implies {\sqrt{5}\times \: q} \: \: \: \: \: \: \: \: \: \: \large {\rightarrow} (1)$

On squaring both sides we get,

$\implies p^2 = 5 \times q^2 \:$

$\implies \frac{ p^2}{5} = q^2 \rightarrow (2)$

Assuming if p was a prime number and p divides a2, then p divides a, where a is any positive integer.

Hence, 5 is a factor of p2.

This implies that 5 is a factor of p.

Thus we can write p = 5a

Substituting p = 5a in (2), we get

$\frac{(5a) ^2 }{5}= q^2 \:$

$\implies \frac{25a^2} {5} = q^2 \:$

$\implies 5a^2 = q^2 \:$

$\implies a^2 = \frac{q^2)}{5} \rightarrow (3) \:$

Hence 5 is a factor of q (from 3)

(2) indicates that 5 is a factor of p and (3) indicates that 5 is a factor of q. This contradicts our assumption that √5 = p/q.

Therefore, the square root of 5 is irrational.