Question

Prove that there is no rational number whose square is 5

Answer 1

acc to question
x^2=5
x=√5
we know that √5 is irrational no 
then x is also irrational no 
hence  there is no rational number whose square is 5

Answer 2

Therefore, $5$ is irrational.

Step-by-step explanation:

Given,

Prove that there is no rational number whose square is $5$

Let's assume that $5=(\frac{p}{q} )$, where $p,q \epsilon \mathbb{R}$ and $p$ and $q$are co-prime.

Then we have,

$(\frac{p}{q}) ^{2} =5^{2} =25$

So,

$p^{2} =25\times q^{2}$

⇒ $p^{2} =25\times q^{2}$

This implies that $p$ is even.

Then,

$p=k$

So,

$k^{2} =25\times q^{2}$

⇒ $k^{2} =25\times q^{2}$

⇒$k^{2} =25\times q^{2}$

Thus,

⇒$k^{2} =25\times q^{2}$

⇒$k=5q$

Then,

∴ $(\frac{p}{q})=(\frac{5q}{p})$

which contradicts $p$ and $q$ being co-prime.

Therefore, $5$ is irrational.