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RRB Technician Grade 1 Model Paper 2

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Question : 52 of 100

Marks: +1, -0

(\sqrt{2}-\sqrt{3})^{2}

is:

a whole number
a natural number
a rational number
an irrational number

Solution:

Concept:

An irrational number cannot be expressed as a fraction

\frac{p}{q}

, where

p

and

q

are integers and

q \neq 0

.

The expansion formula used here is

(a - b)^2 = a^2 + b^2 - 2ab

.

Explanation:

We need to expand

(\sqrt{2} - \sqrt{3})^2

.

Using the identity

(a - b)^2 = a^2 + b^2 - 2ab

, substitute

a = \sqrt{2}

and

b = \sqrt{3}

.

So,

(\sqrt{2} - \sqrt{3})^2 = (\sqrt{2})^2 + (\sqrt{3})^2 - 2\sqrt{2}\sqrt{3}

.

This simplifies to

2 + 3 - 2\sqrt{6}

.

Therefore,

(\sqrt{2} - \sqrt{3})^2 = 5 - 2\sqrt{6}

.

Since

2\sqrt{6}

is irrational,

5 - 2\sqrt{6}

is also an irrational number.

Answer:

Option D — an irrational number.

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