Question

please explain me this question

Answer 1

Let √2 be a rational no.
them √2 =p/q where p and q are integers and co- prime no. and q is not equal to 0

Then

√2=p/q
Squaring both the sides

2 = p^2/q^2
2q^2 = p^2
This means 2 divides p^2
2 divides p

p/2 =m
p =2m

Substituting this In 1st equation

2q^2 =4m^2
q^2 =2m^2
This means 2 divides q

Bit this can only happen when they are not co-prime no.

Hence our supposition was wrong
√2 is an irrational no.

Hence proved ☺☺

Answer 2

let us assume that √2 is irrational
√2=a/b , a&b are co-prime , b≠0
squaring both side
2=a^2/b^2
2b^2=a^2
2 divides a^2 so 2 divides a. (p is a prime number.p divides a square then p divides a)

substitue a=2c
4c^2=2b^2
2c^2=b^2(the reason that given to a^2)

therefore,a&b has 2 as their factor
but a&b are co-prime.

therefore √2 is an irrational number