Question

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Answer 1

√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}

q

p

form Where “p, q” are integers.

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

5

=

q

p

p = \sqrt { 5 } \times qp=

5

×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”."

Answer 2

Answer:

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