Show that is irrational.

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Rational Numbers

Question

Show that $\sqrt{2}$ is irrational.

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Let

\sqrt{2}

is an rational number

So,

\sqrt{2} = \frac{p}{q}

(where p and q are co-prime number and q is not equal to 0)

Squaring both Sides,

2 = \frac{p^{2}}{q^{2}}

Cross Multiplying,

q 2 = 2p 2 \Rightarrow (1)

So,We can say that q²is divisible by 2 and q is also divisible by 2.

Let

q = 2r

Squaring both Sides, q² = 4r² ⇒ (2)

From Eq.1 and 2, 2p² = 4r²

Dividing both Sides by 2, p² = 2r²

So,We can say that p² is divisible by 2 and p is also divisible by 2.

From this, we have reached a conclusion that p and q both divisible by 2.

It contradicts the fact that p and q are co-prime numbers. So, Our Supposition is wrong.

Hence,we can say that

\sqrt{2}

is an irrational number.