Question

prove tht √5 is irrational​

Answer 1

Let us assume that √5 is a rational number.

we know that the rational numbers are in the form of p/q form where p,q are intezers.

so, √5 = p/q

p = √5q

we know that 'p' is a rational number. so √5 q must be rational since it equals to p

but it doesnt occurs with √5 since its not an intezer

therefore, p =/= √5q

this contradicts the fact that √5 is an irrational number

hence our assumption is wrong and √5 is an irrational number.

hope it helped u :)

Answer 2

Step-by-step explanation:

Let us assume to the contrary that √5 is rational number.

That is we can find coprime integers a and b such that (b is not equal to zero).

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

$b \sqrt{5} = a$

squaring on both sides

$5 {b}^{2} = {a}^{2} \: \: \: .......(1)$

.°.5 divides a^2

.°.5 divides a ........(2)

$a = 5c \: \: where \: c \: is \: an \: integer \: \: .......(3)$

put (3) in (1)

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

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

${b}^{2} \: = 5 {c}^{2}$

5 divided b^2

.°. 5 divides b .........(4)

From 2 and 4 we get that 5 divides a and b.

This contradicts the fact that a and b are coprime.

Our assumption that √5 Is rational is wrong.

.•.√5 is irrational number.