Given that \sqrt{3} is an irrational number, show that \left(\right. 5 + 2 \sqrt{3} \left.\right) is an irrational number.

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Rationalisation

Question

Given that

\sqrt{3}

is an irrational number, show that

(5 + 2 \sqrt{3})

is an irrational number.

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Given

\sqrt{3}

is an irrational number

Let

5 + 2 \sqrt{3}

is a rational number

∴

we can write

5 + 2 \sqrt{3} = \frac{p}{q}

, where

p

and

q

are integers

\Rightarrow 2 \sqrt{3} = \frac{p}{q} - 5 = \frac{p - 5 q}{q}

\sqrt{3} = \frac{p - 5 q}{2 q}

Here,

\frac{p - 5 q}{2 q}

is a rational number

So,

\sqrt{3}

is also a rational number.

But it is given that

\sqrt{3}

is irrational number.

\Rightarrow

our assumption was wrong

\Rightarrow 5 + 2 \sqrt{3}

is an irrational number.