Prove that 2 \sqrt{3} is an irrational number.

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Rationalisation

Question

Prove that

2 \sqrt{3}

is an irrational number.

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Let us assume that

\sqrt{3}

is a rational number which can be expressed in the form of

\frac{p}{q}

, where

p

and

q

are integers,

q \neq 0

and

p

and

q

are co prime that is

H C F (p, q) = 1

We have,

\sqrt{3} = \frac{p}{q} \Rightarrow \sqrt{3} q = p . . . . . . (1) \Rightarrow 3 q^{2} = p^{2} (s q u a r i n g b o t h s i d e s)

\Rightarrow p^{2}

is divisible by

3

\Rightarrow p

is divisible by

3

. . . . . . (2)

Therefore, for an integer

r

p = 3 r \Rightarrow \sqrt{3} q = 3 r (f r o m (1)) \Rightarrow 3 q^{2} = 9 r^{2} (s q u a r i n g b o t h s i d e s) \Rightarrow q^{2} = \frac{9}{3} r^{2} \Rightarrow q^{2} = 3 r^{2}

\Rightarrow q^{2}

is divisible by

3

\Rightarrow q

is divisible by

3

. . . . . . (3)

From equations 2 and 3, we get that

3

is the common factor of

p

and

q

which contradicts that

p

and

q

are co prime. This means that our assumption was wrong.

Thus

\sqrt{3}

is an irrational number.

Now, since multiplication of a rational number with an irrational number is an irrational number.

Hence

2 \sqrt{3}

is an irrational number.