Question

If p and q are any two positive integers, show that $\left| \underline{pq} \right.$ is divisible by ${{\left( \left| \underline{p} \right. \right)}^{q}}.\left| \underline{q} \right.$ and by ${{\left( \left| \underline{q} \right. \right)}^{p}}.\left| \underline{p} \right.$ .

Answer 1

Hint: We will prove the required condition by using the principle of mathematical induction. In this method, we will assume our statement to be function in any of the two variables. Here, we will take in ‘q’. Then we will prove that the given condition is true for $q=1$. After that, we will assume that the statement is true for $q=k$ where ‘k’ is a positive integer. Then, using the assumption, we will prove that the statement is true for $q=k+1$ . Once that’s proved, we can say that the given statement is true by the principal of mathematical induction.

Complete step by step answer:
We will prove the required condition by the Principle of Mathematical Induction.
Let F be statement defined as:
F(q)= If p and q are any two positive integers, then $\left| \underline{pq} \right.$ is divisible by ${{\left( \left| \underline{p} \right. \right)}^{q}}.\left| \underline{q} \right.$ and by ${{\left( \left| \underline{q} \right. \right)}^{p}}.\left| \underline{p} \right.$
We will first prove that F(q) is true for $q=1$ .
Now, let $q=1$
Then, the value of $\left| \underline{pq} \right.$will be:
$\begin{align}
  & \left| \underline{pq} \right. \\
 & \Rightarrow \left| \underline{p\left( 1 \right)} \right. \\
 & \Rightarrow \left| \underline{p} \right. \\
\end{align}$
Now, the value of ${{\left( \left| \underline{p} \right. \right)}^{q}}.\left| \underline{q} \right.$ will be:
$\begin{align}
  & {{\left( \left| \underline{p} \right. \right)}^{q}}.\left| \underline{q} \right. \\
 & \Rightarrow {{\left( \left| \underline{p} \right. \right)}^{1}}.\left| \underline{1} \right. \\
 & \Rightarrow \left| \underline{p} \right. \\
\end{align}$
Now, the value of ${{\left( \left| \underline{q} \right. \right)}^{p}}.\left| \underline{p} \right.$will be:
$\begin{align}
  & {{\left( \left| \underline{q} \right. \right)}^{p}}.\left| \underline{p} \right. \\
 & \Rightarrow {{\left( \left| \underline{1} \right. \right)}^{p}}.\left| \underline{p} \right. \\
 & \Rightarrow \left| \underline{p} \right. \\
\end{align}$
Now, we have obtained that the value of $\left| \underline{pq} \right.$is $\left| \underline{p} \right.$ , ${{\left( \left| \underline{p} \right. \right)}^{q}}.\left| \underline{q} \right.$is $\left| \underline{p} \right.$and that of ${{\left( \left| \underline{q} \right. \right)}^{p}}.\left| \underline{p} \right.$is $\left| \underline{p} \right.$.
Now, if we put in the values of $\left| \underline{pq} \right.$and ${{\left( \left| \underline{p} \right. \right)}^{q}}.\left| \underline{q} \right.$ and divide them, we will get:
$\begin{align}
  & \dfrac{\left| \underline{pq} \right.}{{{\left( \left| \underline{p} \right. \right)}^{q}}.\left| \underline{q} \right.} \\
 & \Rightarrow \dfrac{\left| \underline{p} \right.}{\left| \underline{p} \right.} \\
 & \Rightarrow 1 \\
\end{align}$
Thus, $\left| \underline{pq} \right.$is divisible by ${{\left( \left| \underline{p} \right. \right)}^{q}}.\left| \underline{q} \right.$.
Now, if we put in the values of $\left| \underline{pq} \right.$and ${{\left( \left| \underline{q} \right. \right)}^{p}}.\left| \underline{p} \right.$ and divide them, then we will get:
$\begin{align}
  & \dfrac{\left| \underline{pq} \right.}{{{\left( \left| \underline{q} \right. \right)}^{p}}.\left| \underline{p} \right.} \\
 & \Rightarrow \dfrac{\left| \underline{p} \right.}{\left| \underline{p} \right.} \\
 & \Rightarrow 1 \\
\end{align}$
Thus, $\left| \underline{pq} \right.$is divisible by ${{\left( \left| \underline{q} \right. \right)}^{p}}.\left| \underline{p} \right.$.
Thus, F(1) is true.
Now, we will assume that F(k) is true, i.e. $\left| \underline{pk} \right.$ is divisible by ${{\left( \left| \underline{p} \right. \right)}^{k}}.\left| \underline{k} \right.$ and by \[{{\left( \left| \underline{k} \right. \right)}^{p}}.\left| \underline{p} \right.\] where ‘k’ is a positive integer.
Thus, we can say that:
$\dfrac{\left| \underline{pk} \right.}{{{\left( \left| \underline{p} \right. \right)}^{k}}.\left| \underline{k} \right.}=m$
$\dfrac{\left| \underline{pk} \right.}{{{\left( \left| \underline{k} \right. \right)}^{p}}.\left| \underline{p} \right.}=n$
Where ‘m’ and ‘n’ are positive integers.
Now, we try to prove that F(k+1) is true, i.e. $\left| \underline{p\left( k+1 \right)} \right.$ is divisible by ${{\left( \left| \underline{p} \right. \right)}^{k+1}}.\left| \underline{k+1} \right.$ and by \[{{\left( \left| \underline{\left( k+1 \right)} \right. \right)}^{p}}.\left| \underline{p} \right.\] .
Now, for $\left| \underline{p\left( k+1 \right)} \right.$to be divisible by ${{\left( \left| \underline{p} \right. \right)}^{k+1}}.\left| \underline{k+1} \right.$, $\dfrac{\left| \underline{p\left( k+1 \right)} \right.}{{{\left( \left| \underline{p} \right. \right)}^{k+1}}.\left| \underline{k+1} \right.}$ should be an integer.
Now, we will try to prove the above mentioned statement.
\[\begin{align}
  & \dfrac{\left| \underline{p\left( k+1 \right)} \right.}{{{\left( \left| \underline{p} \right. \right)}^{k+1}}.\left| \underline{k+1} \right.}\div m \\
 & \Rightarrow \dfrac{\left| \underline{p\left( k+1 \right)} \right.}{{{\left( \left| \underline{p} \right. \right)}^{k+1}}.\left| \underline{k+1} \right.}\div \dfrac{\left| \underline{pk} \right.}{{{\left( \left| \underline{p} \right. \right)}^{k}}.\left| \underline{k} \right.} \\
 & \Rightarrow \dfrac{\left| \underline{p\left( k+1 \right)} \right.}{{{\left( \left| \underline{p} \right. \right)}^{k+1}}.\left| \underline{k+1} \right.}\times \dfrac{{{\left( \left| \underline{p} \right. \right)}^{k}}.\left| \underline{k} \right.}{\left| \underline{pk} \right.} \\
 & \Rightarrow \dfrac{\left| \underline{p\left( k+1 \right)} \right.}{\left| \underline{p} \right..\left( k+1 \right).\left| \underline{pk} \right.} \\
 & \Rightarrow \dfrac{\left( p\left( k+1 \right) \right).\left| \underline{\left( p\left( k+1 \right)-1 \right)} \right.}{\left| \underline{p} \right..\left( k+1 \right).\left| \underline{pk} \right.} \\
 & \Rightarrow \dfrac{\left| \underline{\left( p\left( k+1 \right)-1 \right)} \right.}{\left| \underline{p-1}.\left| \underline{pk} \right. \right.} \\
\end{align}\]
This is equal to $^{p\left( k+1 \right)-1}{{C}_{p-1}}$ which is always an integer. Therefore, \[\dfrac{\left| \underline{p\left( k+1 \right)} \right.}{{{\left( \left| \underline{p} \right. \right)}^{k+1}}.\left| \underline{k+1} \right.}\div m\]is an integer.
Since, ‘m’ is also an integer, \[\dfrac{\left| \underline{p\left( k+1 \right)} \right.}{{{\left( \left| \underline{p} \right. \right)}^{k+1}}.\left| \underline{k+1} \right.}\] is also an integer.
Therefore, $\left| \underline{p\left( k+1 \right)} \right.$ is divisible by ${{\left( \left| \underline{p} \right. \right)}^{k+1}}.\left| \underline{k+1} \right.$
Similarly, $\left| \underline{p\left( k+1 \right)} \right.$will also be divisible by ${{\left( \left| \underline{p} \right. \right)}^{k+1}}.\left| \underline{k+1} \right.$
Now, we can see that F(k+1) is true if F(k) is true and we have already established that F(1) is true. Hence, F(q) is true for all values of p and q where p and q are positive integers.

Note:
Take care while opening the values of the factorials taken and assumed. Those can be confusing and can lead to mistakes. Also, don’t forget the second step of the process, i.e. the one where we assume the statement is true for $q=k$ as it is a very small yet a very important step and many students tend to forget it.