Prove that √7 is an irrational number
Answers
Lets assume that √7 is rational number. ie √7=p/q.
suppose p/q have common factor then
we divide by the common factor to get √7 = a/b were a and b are co-prime number.
that is a and b have no common factor.
√7 =a/b co- prime number
√7= a/b
a=√7b
squaring
a²=7b² .......1
a² is divisible by 7
a=7c
substituting values in 1
(7c)²=7b²
49c²=7b²
7c²=b²
b²=7c²
b² is divisible by 7
that is a and b have atleast one common factor 7. This is contradicting to the fact that a and b have no common factor.This is happen because of our wrong assumption.
√7 is irrational .
SOLUTION ☺️
let us assume that √7 be rational. then it must in the form of p / q [q ≠ 0] [p and q are co-prime] √7 = p / q => √7 x q = p squaring on both sides.=> 7q2= p2 ------> (1) p2 is divisible by 7 p is divisible by 7 p = 7c [c is a positive integer] [squaring on both sides ] p2 = 49 c2 --------- > (2) subsitute p2 in equ (1) we get, 7q2 = 49 c2 q2 =7c2 => q is divisble by 7 thus q and p have a common factor 7. Hence, There is a contradiction as our assumsion p & q are co prime but it has a common factor. so that √7 is an irrational.
Hope it help