Question

Use contradiction method to prove that : 

 p: √3 is irrational 

is a true statement.

Answer 1

Let us assume that √3 is a rational number. 

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

Since √3 is rational, we can write it as 

√3 = p/q

→ p/√3 = q

[ squaring both sides ] 

→ p²/3 = q²

Thus, p² must be divisible by 3. Hence p will also be divisible by 3. 

We can write p = 3c ( c is a constant ), p² = 9c² 

Putting this back in the equation

9c²/3 = q²

→ 3c² = q²

→ c² = q²/3

Thus, q² 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 d hence the condition for rationality has not been met. Thus,√3 is not rational.

∴ √3 is irrational. 

Hence, the statement p:√3 is irrational , is true.