Question

If p and q are two distinct irrational numbers, then which of the following is always an irrational number
(a) $\dfrac{p}{q}$
(b) $pq$
(c) ${{(p+q)}^{2}}$
(d) $\dfrac{{{p}^{2}}q+qp}{pq}$

Answer 1

Hint: To solve this problem, we must consider two distinct irrational numbers and individually check all the options. If there exist a single case for any of the options then the option will be rejected. Using this methodology, we can easily obtain the required answer.

Complete step-by-step answer:
For option (a), consider two irrational numbers $\sqrt{8}\text{ and }\sqrt{2}$. Now, performing the operation as described in option on these numbers, we get
$\begin{align}
  & \dfrac{p}{q}=\dfrac{\sqrt{8}}{\sqrt{2}}=\sqrt{4} \\
 & \dfrac{p}{q}=2 \\
\end{align}$
So, we get a rational number and hence reject this option.
For option (b), consider two irrational numbers $1+\sqrt{2}\text{ and 1}-\sqrt{2}$. Now, performing the operation as described in option on these numbers, we get
$\begin{align}
  & pq=\left( 1+\sqrt{2} \right)\left( 1-\sqrt{2} \right) \\
 & pq={{1}^{2}}-{{2}^{2}} \\
 & pq=1-4=-3 \\
\end{align}$
So, we get a rational number and hence reject this option.
For option (c), consider two irrational numbers $1+\sqrt{2}\text{ and 1}-\sqrt{2}$. Now, performing the operation as described in option on these numbers, we get
$\begin{align}
  & {{(p+q)}^{2}}={{\left( 1+\sqrt{2}+1-\sqrt{2} \right)}^{2}} \\
 & {{(p+q)}^{2}}={{2}^{2}}=4 \\
\end{align}$
So, we get a rational number and hence reject this option.
For, option (d), first we simplify the expression as,
$\begin{align}
  & \dfrac{{{p}^{2}}q+qp}{pq}=\dfrac{pq(p+1)}{pq} \\
 & \Rightarrow p+1 \\
\end{align}$
Again, consider two irrational numbers $1+\sqrt{2}\text{ and 1}-\sqrt{2}$. Now, performing the operation as described in option on these numbers, we get
$\begin{align}
  & p+1=1+1+\sqrt{2} \\
 & p+1=2+\sqrt{2} \\
\end{align}$
This option is always irrational as addition of a rational and irrational number is always an irrational number.
Therefore, option (d) is correct.

Note: The key step for solving this problem is the consideration of all the options for obtaining the correct answer. Students must consider any possible case for rejection of that particular option, so that there is no possibility of error.