Question

prove √5 is an irrational number​

Answer 1

Answer:

put 5 in place of 2

Step-by-step explanation:

please understand carefully

Answer 2

Step-by-step explanation:

$let \: us \: assume \: that \: \sqrt{5} \: is \: rational \: number$

--> There exist co-prime integers a and b (b is not equal to 0 ) such that ,

$\sqrt{5} = \frac{a}{b}$

$\sqrt{5} b = a$

$5 {b}^{2} = {a}^{2} (squaring \: on \: both \: sides \: ) \: - > 1$

$5 \: is \: factor \: of \: {a}^{2}$

$5 \: is \: factor \: of \: a \: - > 2$

Let a = 5c , where c is some integer

Substitute a = 5c in eq 1

$5 {b}^{2} = {(5c)}^{2}$

$5 {b}^{2} = 25 {c}^{2}$

${b}^{2} = 5 {c}^{2} \: (dividing \: 5 \: on \: both \: sides \: )$

$5 \: is \: a \: factor \: of \: {b}^{2}$

$5 \: is \: a \: factor \: of \: b \: - > 3$

So , 5 is common factor of a and b ( from eq 2 and 3 )

This contradicts the fact that a and b are co-prime .

So our assumption is wrong.

$hence \: \sqrt{5} \: is \: irrational \: number$