Question

prove that square of root of 7 is an irrational number
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Answer 1

Answer:

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

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.