Question

Prove that
$\sqrt{9}$
Is irrational

Answer 1

Assume √9 is rational
Then √9=a/b (where a and b are co prime and b not equal to 0)
=√9b=a
Square both sides, we get
9b^2=a^2 —1
If 9 divides a^2, 9 can divide 'a' as well
Therefore,
9c=a
Substitute thisin eq 1
= 9b^2=(9c)^2
=9b^2=81c^2
=b^2=(81c^2)/9
=b^2=9c^2
This shows that 9 can divide b as well but a and b are co prime
This is a contradiction to the fact that a and b are co prime
This contradiction has arisen because of our incorrect assumption
Therefore √9 is irrational