Question

Prove that square root of 7 is irrational number.​

Answer 1

Let √7 be a rational number, then

√7=p/q ,where p, q are integers, q is not equals to zero and p, q have no common factors (except 1).

or, 7 =p^2/q^2

or, p^2=7q^2. ...(i)

As 7 divides 7q^2, so 7 divides p^2 but 7 is prime.

Let p=7m, where m is an integer

Substituting the value of p in (i) we get

(7m)^2= 7q^2

or, 49m^2= 7q^2

or, 7m^2= q^2

As 7 divides 7m^2, so 7 divides q^2 but 7 is prime or, 7 divides q

Thus p and q have a common factor 7.

This contradicts that p and q have no common factors (except 1).

Hence, √7 is an irrational number.

Answer 2

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 contridite to the fact that a and b have no common factor.This is happen because of our wrong assumption.

√7 is irrational

Thanks