Question

State whether $2-\sqrt{5}$ is rational or irrational.

Answer 1

Hint: We recall the definition of rational and irrational numbers. We know that $\sqrt{5}$ is an irrational number, we assume $2-\sqrt{5}=\dfrac{p}{q},q\ne 0,\gcd \left( p,q \right)=1$ where $\dfrac{p}{q}$ is a rational number and proceed to prove by method of contradiction $2-\sqrt{5}$ is irrational.

Complete step-by-step answer:
We know that the rational numbers are those real numbers that can be written in the form of $\dfrac{p}{q}$ such that both p and q are integers who are relatively prime and $q0$. We will check the statement by method of contradiction with the fact that $\sqrt{5}$ is an irrational number. A number can be rational or irrational. Irrational numbers cannot be written in the same form of $\dfrac{p}{q}$
Let us assume that the given number $2-\sqrt{5}$ is rational. So we have
(a) \[\dfrac{p}{q}=2-\sqrt{5}\]
We have taken $p,q\ne 0$ here as two integers which are relatively prime which means the highest factor between $p$ and $q$ is 1. We have,
\[\begin{align}
  & \dfrac{p}{q}=2-\sqrt{5} \\
 & \Rightarrow \sqrt{5}=2-\dfrac{p}{q} \\
 & \Rightarrow \sqrt{5}=\dfrac{2q-p}{q} \\
\end{align}\]
We observe at the right hand side that $p,q$ are rational number which means $2q,-p,2q-p$ and $\dfrac{2q-p}{q}$ are also rational number because the setoff rational number excluding 0 is closed under addition, subtraction, multiplication and division. Then the number at the left hand side $\sqrt{5}$ has to be a rational number which is a contradiction. So our assumption is wrong and $2-\sqrt{5}$ is an irrational number.

Note: We can alternatively use the fact that operations like addition, subtraction, multiplication and division between a rational and an irrational number always yields an irrational number. The set of rational numbers is denoted by $Q$ and irrational numbers are denoted by ${{Q}^{'}}$ and they are disjoint whose union is the real number set.