Question

Let the relation $R:\left\{ \left( a,{{a}^{3}} \right):\text{ a is a prime number less than 10} \right\}$. Find R.

Answer 1

Hint: To solve this question, we should know the definition of a prime number. A prime number is defined as the number which has factors as one and itself. We should find the prime numbers which are less than 10 to get the answer. We know that 1 is not a prime number. We should take every number from 2 to 10 and get the factors of them and find the prime numbers. We know from the question that the relation R is mapping the prime number to its cube. So, after finding the prime number, we should map it to its cube to get the element in the relation R.

Complete step-by-step answer:
A prime number is defined as the number which has factors as one and itself. We know that 1 is not a prime number. We should take every number from 2 to 10 and get the factors of them and find the prime numbers.
Let us consider the factorisation of 2.
$\begin{align}
  & 2\left| \!{\underline {\,
  2 \,}} \right. \\
 & \left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
We can write that
$2=2\times 1$
So, 2 is a prime number.
Let us consider the factorisation of 3.
$\begin{align}
  & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & \left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
We can write that
$3=3\times 1$
So, 3 is a prime number.
Let us consider the factorisation of 4.
$\begin{align}
  & 2\left| \!{\underline {\,
  4 \,}} \right. \\
 & 2\left| \!{\underline {\,
  2 \,}} \right. \\
 & \left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
We can write that
$4=2\times 2\times 1$
So, 4 is not a prime number.
Let us consider the factorisation of 5.
$\begin{align}
  & 5\left| \!{\underline {\,
  5 \,}} \right. \\
 & \left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
We can write that
$5=5\times 1$
So, 5 is a prime number.
Let us consider the factorisation of 6.
\[\begin{align}
  & 2\left| \!{\underline {\,
  6 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & \left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}\]
We can write that
$6=3\times 2\times 1$
So, 6 is not a prime number.
Let us consider the factorisation of 7.
$\begin{align}
  & 7\left| \!{\underline {\,
  7 \,}} \right. \\
 & \left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
We can write that
$7=7\times 1$
So, 7 is a prime number.
Let us consider the factorisation of 8.
$\begin{align}
  & 2\left| \!{\underline {\,
  8 \,}} \right. \\
 & 2\left| \!{\underline {\,
  4 \,}} \right. \\
 & 2\left| \!{\underline {\,
  2 \,}} \right. \\
 & \left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
We can write that
$8=2\,\times 2\times 2\times 1$
So, 8 is not a prime number.
Let us consider the factorisation of 9.
$\begin{align}
  & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & \left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
We can write that
$9=3\times 3\times 1$
So, 9 is not a prime number.
Let us consider the factorisation of 10.
$\begin{align}
  & 2\left| \!{\underline {\,
  10 \,}} \right. \\
 & 5\left| \!{\underline {\,
  5 \,}} \right. \\
 & \left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
We can write that
$10=5\times 2\times 1$
So, 10 is not a prime number.
So, we can write that the numbers 2, 3, 5, 7 are the prime numbers less than 10. Let us find the cubes of them
$\begin{align}
  & {{2}^{3}}=2\times 2\times 2=4\times 2=8 \\
 & {{3}^{3}}=3\times 3\times 3=9\times 3=27 \\
 & {{5}^{3}}=5\times 5\times 5=25\times 5=125 \\
 & {{7}^{3}}=7\times 7\times 7=49\times 7=343 \\
\end{align}$
So, we can write the elements in the set R as
$R=\left\{ \left( 2,8 \right),\left( 3,27 \right),\left( 5,125 \right),\left( 7,343 \right) \right\}$
$\therefore $The required elements in the relation R are $R=\left\{ \left( 2,8 \right),\left( 3,27 \right),\left( 5,125 \right),\left( 7,343 \right) \right\}$.

Note: Some students commit a mistake by considering 1 also as a prime number. 1 is fundamentally not a prime number because 1 doesn’t have two distinct factors. We can reduce the calculations done in the solution by applying the fact that the multiples of a number are not prime numbers. As 4, 6, 8, 10 are multiples of 2, we can conclude that they are not prime numbers. Likewise, 9 is not a prime number as it is a multiple of 3.