Question

prove that √5 is irrational.

Answer 1

To prove that √5 is irrational number 
Let us assume that √5 is rational 
Then √5 = $\frac{a}{b}$ 
(a and b are co primes, with only 1 common factor and b≠0) 
⇒ √5 = $\frac{a}{b}$ 
(cross multiply) 
⇒ a = √5b 
⇒ a² = 5b² -------> α
⇒ 5/a² 
(by theorem if p divides q then p can also divide q²) 
⇒ 5/a ----> 1 
⇒ a = 5c 
(squaring on both sides) 
⇒ a² = 25c² ----> β 
From equations α and β 
⇒ 5b² = 25c²
⇒ b² = 5c² 
⇒ 5/b² 
(again by theorem) 
⇒ 5/b-------> 2 

we know that a and b are co-primes having only 1 common factor but from 1 and 2 we can that it is wrong. 
This contradiction arises because we assumed that √5 is a rational number 
∴ our assumption is wrong 
∴ √5 is irrational number 

Answer 2

$\sqrt{5} \ is \ an \ irrational \ number.$

Given:

$\sqrt{5}$

To prove:

$\sqrt{5} \ is \ a \ irrational \ number$

Solution:

Let us consider that $\sqrt{5}$ is a “rational number”.

We were told that the rational numbers will be in the “form” of $\frac {p}{q}$ form Where “p, q” are integers.

So,

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

$p = \sqrt { 5 } \times q$

we know that 'p' is a “rational number”.

So, $5 \times q$ should be normal as it is equal to p

But it did not happens with √5 because it is “not an integer”

Therefore, $p \neq\sqrt{5}q$

This denies that $\sqrt{5}$ is an “irrational number”

So, our consideration is false and $\sqrt{5}$ is an “irrational number”.