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Solution :

Let us assume ,that √ 3 is rational

i.e√ 3=x/y (where,x and y are co-primes)

y√ 3=x

Squaring both sides

We get,(y√3)2=x2

3y2=x2……..(1)

x2 is divisible by 3

So, x is also divisible by 3

 therefore we can write x=3k (for some values of k)

Substituting ,x=3k in eqn 1

3y2=(3k)2 

y2=3k2

y2 is divisible by 3 it means y is divisible by 3

 therefore x and y are co-primes.

Since ,our assumption about √ 3 is rational is incorrect .

Hence, √ 3 is irrational number.

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