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).

Asked by Pragya Singh | 11 months ago |  125

1 Answer

Solution :-

(i) The given statement is as follows

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

Let us assume that the statement p is false. That is,

The sum of an irrational number and a rational number is rational.

Hence,

where

  

\( \sqrt{a}\) is irrational and b, c, d, e are integers.

\( \dfrac{d}{e}-\dfrac{b}{c}=\sqrt{a}\)

But here, \( \dfrac{d}{e}-\dfrac{b}{c}\) is a rational number and \( \sqrt{a}\) is an irrational number

This is a contradiction. Hence, our assumption is false.

The sum of an irrational number and a rational number is rational.

Hence, the given statement is true.

 

(ii) The given statement q is as follows

If n is a real number with n > 3, then n2 > 9

Let us assume that n is a real number with n > 3, but n2 > 9 is not true

i.e. n2 < 9

So, n > 3 and n is a real number

By squaring both sides, we get

n2 > (3)2

This implies that n2 > 9 which is a contradiction, since we have assumed that n2 < 9

Therefore, the given statement is true i.e., if n is a real number with n > 3, then n2 > 9.

Answered by Abhisek | 11 months ago

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