Question

prove that under root 3 is irrational number

Answer 1

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Answer 2

Here your answer goes

Step :- 1

Let $\sqrt{3}$ be a rational number

$\sqrt{3} = \frac{p}{q}$

Where p and q are co-prime integers and $q \neq 0$

Step :- 2

On squaring both the sides

$3 =\frac{p^2}{q^2}$

$p^2 = 3q^2$

Therefore ,

$p^2$ is divisible by 3

p is divisible by 3 ---------> (i)

Step :- 3

Let p = 2r for some integer r

$p^2 = 9r^2$

$3q^2 = 9r^2$

$q^2 = 3r^2$

Therefore ,

$q^2$ is divisible by 3

q is divisible by 3  --------> (ii)

From (i) and (ii) , p and q are divisible by 3 , which contradicts a fact that p and q are co-primes

Hence , our assumption is False

Therefore , $\sqrt{3}$ is irrational

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Thanks !