Question

Prove that $3 + 2\sqrt 5 $ is an irrational number.

Answer 1

Hint:In order to prove that the number is irrational, we start off by considering it to be rational and disproving that it does not hold the property of a rational number i.e by contradiction method.

Complete step-by-step answer:
If a number that can be expressed in the form of $\dfrac{{\text{p}}}{{\text{q}}}$ is called rational number. Here p and q are integers,
and q is not equal to 0.
Let us assume $3 + 2\sqrt 5 $ is a rational number, it can be expressed as
$3 + 2\sqrt 5 $= $\dfrac{{\text{p}}}{{\text{q}}}$
⟹$\sqrt 5 = \dfrac{{{\text{p - 3q}}}}{{{\text{2q}}}}$
The RHS $\dfrac{{{\text{p - 3q}}}}{{{\text{2q}}}}$looks like a rational number but the LHS of the equation$\sqrt 5$ is not a rational number.
Hence $3 + 2\sqrt 5 $ is not a rational number, i.e. it is an irrational number.

Note – In order to solve questions of this type the key is to know the definitions of rational and irrational numbers. An irrational number is a number that cannot be expressed as a fraction for any integers and Irrational numbers have decimal expansions that neither terminate nor become periodic.Contradiction method is a common proof technique that is based on a very simple principle: something that leads to a contradiction can not be true, and if so, the opposite must be true.