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# Check the validity of the statements given below by the method given against it.

(i) p : The sum of an irrational number and a rational number is irrational (by contradiction method).

(ii) q : If n is a real number with n > 3, then n² > 9 (by contradiction method).

**Solution:**

**(i)** The given statement is

p : the sum of an irrational number and a rational number is irrational.

To prove it by contradiction, assume that the given statement, p, is false. i.e., the sum of an irrational number and a rational number is rational.

Therefore, √x + y/z = a/b when √x is irrational and y, z, a, b are integers.

From this equation we get, √x = a/b - y/z ... (1)

We know that a/b - y/z is a rational number but from (1), its value is √x which is an irrational number.

This is a contradiction.

Thus, our assumption (that the given statement is false) is incorrect. Thus, it must be true.

**(ii)** The given statement, q is

If n is a real number with n > 3, then n^{2} > 9.

To prove it by contradiction, assume that n is a real number with n > 3, but n^{2} > 9 is not true i.e., n^{2} ≤ 9.

Then,

n > 3 and n is a real number.

Squaring both the sides, we obtain

n^{2} > (3)^{2}

n^{2} > 9

which is a contradiction, since we have assumed that n^{2} ≤ 9.

Thus, the given statement is true. i.e., if n is a real number with n > 3, then n^{2} > 9

NCERT Solutions Class 11 Maths Chapter 14 Exercise ME Question 6

## Check the validity of the statements given below by the method given against it. (i) p : The sum of an irrational number and a rational number is irrational (by contradiction method). (ii) q : If n is a real number with n > 3, then n² > 9 (by contradiction method)

**Summary:**

(i) We have proved statement p to be true by the contradiction method

(i) We have proved statement q to be true by the contradiction method

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