Question

proof that root 7 is an irrational no.​

Answer 1

Step-by-step explanation:

Let us assume that

7

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

7

=

b

a

⟹a=b

7

Squaring on both sides, we get

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,

7

is irrational.

Answer 2

Answer:

Step-by-step explanation:

Let us assume that   $\sqrt{7}$ is rational. Then, there exist co-prime positive integers a and b such that

$\sqrt{7}$ =  a/b

⟹a=b  $\sqrt{7}$

Squaring on both sides, we get

$a^{2} =b^{2} 7$

Therefore, $a^{2}$ is divisible by$\sqrt{7}$  and hence, a is also divisible by$\sqrt{7}$

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

Substituting for a, we get 49$p^{2}$

$b^{2}$ =7$p^{2}$

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,   $\sqrt{7}$ is irrational.