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Question

Question: Prove that \(3 + 2\sqrt 5 \) is an irrational number....

Prove that 3+253 + 2\sqrt 5 is an irrational number.

Explanation

Solution

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 pq\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+253 + 2\sqrt 5 is a rational number, it can be expressed as
3+253 + 2\sqrt 5 = pq\dfrac{{\text{p}}}{{\text{q}}}
5=p - 3q2q\sqrt 5 = \dfrac{{{\text{p - 3q}}}}{{{\text{2q}}}}
The RHS p - 3q2q\dfrac{{{\text{p - 3q}}}}{{{\text{2q}}}}looks like a rational number but the LHS of the equation5\sqrt 5 is not a rational number.
Hence 3+253 + 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.