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

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Linear Equations in One Variable

Question

Prove that

\sqrt{2}

is an irrational number.

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Let us assume on the contrary that

\sqrt{2}

is a rational number. Then, there exist positive integers

a

and

b

such that

\sqrt{2} = \frac{a}{b}

where,

a

and

b

, are co-prime i.e. their

H C F

1

\Rightarrow {(\sqrt{2},),}^{2} = {(\frac{a}{b},),}^{2}

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

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

\Rightarrow 2 ∣ a^{2} [∵ 2 ∣ 2 b^{2} a n d 2 b^{2} = a^{2}]

\Rightarrow 2 ∣ a . . . (i)

\Rightarrow a = 2 c

for some integer

c

\Rightarrow a^{2} = 4 c^{2}

\Rightarrow 2 b^{2} = {4 c}^{2} [∵ 2 b^{2} = a^{2}]

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

\Rightarrow 2 ∣ b^{2} [∵ 2 ∣ 2 c^{2}]

\Rightarrow 2 ∣ b . . . (i i)

From

(i)

and

(i i),

we obtain that

2

is a common factor of

a

and

b

. But, this contradicts the fact that

a

and

b

have no common factor other than

1

. This means that our supposition is wrong.
Hence,

\sqrt{2}

is an irrational number.