Let A be the set of first ten natural numbers and let R be a relation on A defined by $\left( x,y \right)\in R$. Such that $x+2y=10$ i.e. $R=\left\{ \left( x,y \right):x\in A,y\in A\,and\,x+2y=10 \right\}\,$. Express R and ${{R}^{-1}}$ as a set of ordered pairs. Determine also (i) Domain of R and ${{R}^{-1}}$ (ii) Range of R and ${{R}^{-1}}$.

Question

Let A be the set of first ten natural numbers and let R be a relation on A defined by $\left( x,y \right)\in R$. Such that $x+2y=10$ i.e. $R=\left\{ \left( x,y \right):x\in A,y\in A\,and\,x+2y=10 \right\}\,$. Express R and ${{R}^{-1}}$ as a set of ordered pairs. Determine also (i) Domain of R and ${{R}^{-1}}$ (ii) Range of R and ${{R}^{-1}}$.

Vedantu Content Team · Accepted Answer

Hint: In this question, we will first write tabular form of given relation and form that write its inverse relation. Then find domain and range.

Complete step-by-step answer:
In the given question, we have set A which is set of the first ten natural numbers. That is, $A=\left\{ 1,2,3,4,5,6,7,8,9,10 \right\}$,
Also, we have,
$R=\left\{ \left( x,y \right):x\in A,y\in A\,and \,x+2y=10 \right\}\,$
For $x+2y=10$, we have,
$y=\dfrac{10-x}{2}$
Here y will be the natural number when \[10-x\] is divisible by 2. That is when \[10-x\] is even.
For \[10-x\] to be even, x need to be even, that is \[2,4,6,8,10\]
For \[x=2\], \[y=\dfrac{10-2}{2}=4\]
For \[x=4\] , \[y=\dfrac{10-4}{2}=3\]
For \[x=6,\,y=\dfrac{10-6}{2}=2\]
For \[x=8,\,y=\dfrac{10-8}{2}=1\]
For \[x=10,\,y=\dfrac{10-10}{2}=0\], not in A.
Therefore, \[R=\left\{ \left( 2,4 \right),\left( 4,3 \right),\left( 6,2 \right),\left( 8,1 \right) \right\}\].
Now, the inverse of a relation is a set of ordered pairs obtained by interchanging the first and second term of original relation.
Therefore, inverse of R is represented as,
${{R}^{-1}}=\left\{ \left( 4,2 \right),\left( 3,4 \right),\left( 2,6 \right),\left( 1,8 \right) \right\}$
Now, the domain of relation is set of first terms of all ordered pairs in it and range of relation is set of the second term of all ordered pairs in it.
Therefore, domain of R is $\left\{ 2,4,6,8 \right\}$ and domain of ${{R}^{-1}}$ is $\left\{ 4,3,2,1 \right\}$.
And, range of R is $\left\{ 4,3,2,1 \right\}$ and range of ${{R}^{-1}}$ is $\left\{ 2,4,6,8 \right\}$.

Note: In this type of question, note that, the domain of a relation is the range of its inverse relation and range of a relation in domain of its inverse relation.