Question

prove that√5 irrrational​

Answer 1

Answer:

Let us consider that \sqrt{5}5 is a “rational number”.

We were told that the rational numbers will be in the “form” of \frac {p}{q}qp form Where “p, q” are integers.

So,

\sqrt { 5 } =\frac {p}{q}5=qp

p = \sqrt { 5 } \times qp=5×q

we know that 'p' is a “rational number”.

So, 5 \times q5×q should be normal as it is equal to p

But it did not happens with √5 because it is “not an integer”

Therefore, p \neq\sqrt{5}qp=5q

This denies that \sqrt{5}5 is an “irrational number”

So, our consideration is false and \sqrt{5}5 is an “irrational number”.