Question

Prove that root 5 is an irrational number.

Answer 1

Let us assume that √5 is a rational number.
we know that the rational numbers are in the form of p/q form where p,q are intezers.
so, √5 = p/q
     p = √5q
we know that 'p' is a rational number. so √5 q must be rational since it equals to p
but it doesnt occurs with √5 since its not an intezer
therefore, p =/= √5q
this contradicts the fact that √5 is an irrational number
hence our assumption is wrong and √5 is an irrational number.



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Answer 2

"√5 is an “irrational number”.

Given:

√5

To prove:

√5 is a rational number

Solution:

Let us consider that √5 is a “rational number”.

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

So, $\sqrt { 5 } = \frac {p}{q}$

$p = \sqrt { 5 } \times q$

we know that 'p' is a “rational number”. So 5 \times 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 ≠ √5q

This denies that √5 is an “irrational number”

So, our consideration is false and √5 is an “irrational number”."