# Validity of Statements

To find the truth value of compound statements connected with “and”, “or” and “if-then”, we follow the following rules:-

- If
*p*and*q*are statements,*p*and*q*is true if*p*is true*q*is true

*p*is true and also show that

*q*is true.

- If
*p*and*q*are statements, such that*p*or*q*is given, then- Assuming
*p*is false, show that*q*is true - Assuming
*q*is false, show that*p*is true

- Assuming
- Statement with if then
- Assuming
*p*is true, prove*q*must be true (direct method)

Or - By assuming
*q*is false, prove*p*must be false (contra positive method).

- Assuming
- Statements with “if and only if”. We need to show.
- if
*p*is true, then*q*is true.

and - if
*q*is true, then*p*is true.

- if

**Example 1:**

Check whether the given statement is true or false. If such that are even, then is even.

**Solution:**

,

*x*is even and

*y*is even

is odd.

Let us apply Rule 3 (a),

*p*is true means

*x*and

*y*are even numbers

Let

Then which is even

∴

*q*is true

∴ The statement is true.

**Example 2:**

By contradiction, check the truth value of the statement, is irrational

**Solution:**

Assume, is rational

Then (are rationals, with no common factors)

Which contradicts our assumption that there is no common factor for

*a*and

*b*.

∴ is irrational.

**Example 3:**

By giving a counter example, show that the following statement is false.

“If

*n*is a prime number, it has to be odd.”

**Solution:**

*p*:

*n*is a prime number

*q*:

*n*is odd.

Let us look for an even number which is prime. Obviously only '2' is an even prime.

∴

*p*:

*n*is a prime number

~

*q*:

*n*is even

p ⇒ ~

*q*is true

∴

*p*⇒

*q*is false.