Question

Prove that
$(2 + \sqrt{5)}$
is an irrational number,given that
$\sqrt{5}$
is a irrational number​

Answer 1

Let us assume  $\bf{2+\sqrt{5} }$  is a rational number.

$\bf{2+\sqrt{5} } = \dfrac{p}{q}$ where p,q∈ z,q ≠ 0

$\bf{2-\dfrac{p}{q} = -\sqrt{5}$

$\bf{\dfrac{2q-p}{q} = -\sqrt{5}$

$\longrightarrow \bf{-\sqrt{5}$ is a rational number

$\bf{\dfrac{2q-p}{q}$ is a rational number

But $-\sqrt{5}$ is not a rational number.

Our supposition $\bf{2+\sqrt{5}}$ is a rational number is wrong.

⇒ $\bf{2+\sqrt{5}}$ is an irrational number.

Answer 2

Answer

To prove :-

2+√5 is a irrational number

Given :-

√5 is a irrational number

Solution:

Let us assume that 2+√5 is a rational number and a and b are co prime number , b≠0

now,

2+√5 = a/b

√5 = a+2b/b

wkt

a+2b/b is a rational number but √5 is a irrational number....

so, our assumption is wrong

2+√5 is a irrational number