Question

Q. Prove that √7 is irrational number.
(!_!)

Answer 1

let √7 be rational in the form of p/q
so , √7=p/q (where q is not equal to zero
√7q=p
squaring both sides
(√7)²q²=p²
7q²=p² ...........(i)
7 is a factor of p²
so, 7 is a factor of p also
let p=7m
put p=7m in (i)
7q²=(7²)m²
7q²=49m²
q²=7m².......(ii)

7 is a factor of q²
so,7 is a factor of q also
7 is a factor of both p and q
so, our suppositon is wrong.
so , √2 is an irrational no.

Answer 2

Let as assume that √7 is a rational number

A rational number can be written in the form of p / q where q ≠ 0 and p & q are non negative integer

√7 = p / q ( Where p and q are co prime number )

Squaring both side

7 = p² /q²

7q² = p² ..... ( i )

p² is divisible by 7
then p is also divisible by 7

Let p = 7m ( where m is any positive integer )

Squaring both side

p² = 49m²

Putting in ( i )

7q² = 49m²

q² = 7m²

q² is divisible by 7
then q is also divisible by 7

Since
both p and q are divisible by same number 7 and they aren't co prime number

Hence our assumption is wrong p and q are not co prime
So √7 is an irrational number !