Question

so that 4 + under root 5 is irrational by giving reasons​

Answer 1

GIVEN:

• 4+√5

TO FIND:

• Prove that 4+√5 is an irrational number.

SOLUTION:

Let us assume to the contrary that 4+√5 is a rational number. That is, we can find co-prime integers p and q (q ≠0) such that

$\sf{\longmapsto 4+ \sqrt{5} = \dfrac{p}{q} }$

$\sf{\longmapsto \sqrt{5} = \dfrac{p}{q} - 4 }$

$\sf{\longmapsto \sqrt{5} = \dfrac{p- 4q}{q} }$

We have assumed that, 4+√5 is an irrational number.

This contradicts the fact that, √5 is an Irrational number and p-4q/q

So, 4+√5 is not a rational number

Hence, 4+√5 is an Irrational number.

______________________

Answer 2

Let us assume that ,

• 4 + √5 is rational number

Thus ,

$4 + \sqrt{5} = \frac{a}{b} \\ \\ \sqrt{5} = \frac{a - 4b}{b}$

Here , √5 is an irrational number and (a - 4b)/b is rational number

Since , irrational ≠ rational

Thus , our assumptions is wrong

$\sf \therefore\underline{ 4 + \sqrt{5} \: is \: an \: irrational \: number }$