Question

prove that √7 is irrational​

Answer 1

Answer:

Answer

Let us assume that

√7 is rational. Then, there exist co-prime positive integers a and b such that

√7=a/b

⟹√7a=b

Squaring on both sides, we get

a²=7b²

Therefore, a² is divisible by 7 and hence, a is also divisible by7

so, we can write a=7p, for some integer p.

Substituting for a, we get:

7b²=(7p)²

7b²=49a²

b²=7a²

This means, b² is also divisible by 7 and so, b is also divisible by 7.

Therefore, a and b have at least one common factor, i.e., 7.

But, this contradicts the fact that a and b are co-prime.

Thus, our supposition is wrong.

Hence,

√7 is irrational.

Answer 2

Answer:

Let us assume that ✓7 is rational. Then, there exist co-prime positive integers a and b such that

√7=a/b

a=b√7

squaring on both sides,

a^2=7b^2

Therefore, a^2 is divisible by 7 and hence, a is also divisible by7

so, we can write a=7p, for some integer p.

substituting for a we get 49p^2=7b^2

b^2=7p^2

This means, b^2 is also divisible by 7 and so, b is also divisible by 7.

Therefore, a and b have at least one common factor, i.e., 7.

But, this contradicts the fact that a and b are co-prime.

Thus, our supposition is wrong.

Hence,√7is irrational.

hope it helps you...