Define a relation $R$ on the set $N$ of natural numbers by $R = \{(x, y) : y = x + 5, x \text{ is a natural number less than } 4; x, y \in N\}$. Depict this relationship using roster form. Write down the domain and the range.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The relation is defined as $R = \{(x, y) : y = x + 5, x \in \{1, 2, 3\}, x, y \in N\}$.
Since $x$ is a natural number less than $4$,the possible values for $x$ are $1, 2,$ and $3$.
For $x = 1, y = 1 + 5 = 6$.
For $x = 2, y = 2 + 5 = 7$.
For $x = 3, y = 3 + 5 = 8$.
Thus,in roster form,$R = \{(1, 6), (2, 7), (3, 8)\}$.
The domain is the set of all first elements of the ordered pairs: $\text{Domain} = \{1, 2, 3\}$.
The range is the set of all second elements of the ordered pairs: $\text{Range} = \{6, 7, 8\}$.

Explore More

Similar Questions

Given $A = \{1, 2, 3, 4, 5\}$ and $B = \{1, 4, 5\}$. If $R$ is a relation from $A$ to $B$ such that $(x, y) \in R$ with $x > y$,then the range of $R$ is:

Let $A = \{1, 2, 3, \ldots, 14\}$. Define a relation $R$ from $A$ to $A$ by $R = \{(x, y) : 3x - y = 0, \text{ where } x, y \in A\}$. Write down its domain,codomain,and range.

Two finite sets $A$ and $B$ are such that $n(A) = 2$ and $n(B) = 3$. Then the total number of relations from $A$ to $B$ is:

Let $A = \{1, 2, 3, 4\}$,$B = \{1, 5, 9, 11, 15, 16\}$,and $f = \{(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)\}$. Is $f$ a relation from $A$ to $B$? Justify your answer.

If $A = \{x, y, z\}$ and $B = \{1, 2\}$,then the total number of relations from set $A$ to set $B$ is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo