Question

Prove that √2 is irrational no.​

Answer 1

√2 is irrational number

__________ [TO PROVE]

Let us assume that, √2 is a rational number

√2 = $\dfrac{a}{b}$

Here, a and b are co-prime numbers.

• Squaring on both sides, we get

=> (√2)² = $( { \dfrac{a}{b}) }^{2}$

=> 2 = $\dfrac{ {a}^{2} }{ {b}^{2} }$

=> 2b² = a² ________( eq 1)

Clearly;

a² is divisible by 2.

So, a is also divisible by 2.

Now, let integer be c.

=> a = 2c

• Squaring on both sides.

=> a² = 4c²

=> 2b² = 4c² [From (eq 1)]

=> b² = 2c² _______(eq 2)

This means that, 2 divides b², and so 2 divides b also.

2 divides b² and 2 divide a² also.

So, our assumption is wrong.

√2 is irrational number.

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