Question

prove that√7 is an irrational number​

Answer 1

Answer:

ANSWER

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

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.