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

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

zeros of a polynomial

Question

Prove that

\sqrt{3}

is an irrational number.

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Let us assume on the contrary that

\sqrt{3}

is a rational number.

Then, there exist positive integers

a

and

b

such that

\sqrt{3} = \frac{a}{b}

where,

a

and

b

, are co-prime i.e. their

H C F

1

Now,

\sqrt{3} = \frac{a}{b}

\Rightarrow 3 = \frac{a^{2}}{b^{2}}

\Rightarrow 3 b^{2} = a^{2}

\Rightarrow 3

divides

a^{2} [∵ 3 divides 3 b^{2}]

\Rightarrow 3

divides

a . . . (i)

\Rightarrow a = 3 c

for some integer

c

\Rightarrow a^{2} = 9 c^{2}

\Rightarrow 3 b^{2} = {9 c}^{2} [∵ a^{2} = 3 b^{2}]

\Rightarrow b^{2} = {3 c}^{2}

\Rightarrow 3

divides

b^{2} [∵ 3 divides 3 c^{2}]

\Rightarrow 3

divides

b . . . (i i)

From

(i)

and

(i i),

we observe that

a

and

b

have at least

3

as a common factor. But, this contradicts the fact that

a

and

b

are co-prime. This means that our assumption is not correct.

Hence,

\sqrt{3}

is an irrational number.