Answer :

Let us assume that is a rational number.

For a number to be rational, it must be able to express it in the form where p and q do not have any common factor, i.e. they are co-prime in nature.

Since is rational, we can write it as

→

[ squaring both sides ]

→

Thus, p^{2} must be divisible by 3. Hence p will also be divisible by 3.

We can write p = 3c ( c is a constant ), p^{2} = 9c^{2}

Putting this back in the equation,

→

→

Thus, q^{2} must also be divisible by 3, which implies that q will also be divisible by 3.

This means that both p and q are divisible by 3 which proves that they are not co-prime and hence the condition for rationality has not been met. Thus, is not rational.

∴ is irrational.

Hence, the statement p: is irrational , is true.

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