Write the relation $R = \{(x, x^3) : x\text{ is a prime number less than }10\}$ in roster form.

Write the relation R = \(x, x³) : x is a prime number less than 10\ in roster form.

Exercise 2.2 · Q7

R = \{(x, x^3) : x\text{ is a prime number less than }10\}

Manipur CohsemTextbookSubjective· 2mImportance★★★★★est

23% · 23/100 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Relation Arrow Diagram

Relation Arrow Diagram

Imagine two groups of people at a party: one group of hosts, one group of guests. If you drew a string from every host to every guest they personally invited, you'd get a tangle of strings connecting the two groups. A relation arrow diagram does exactly this for sets — it draws an arrow from each element that is "related to" an element in the other set, so you can see a relation instead of just listing ordered pairs.

The Intuition First

A relation pairs up elements of one set with elements of another. Suppose:

Set $A = \{\text{Ravi}, \text{Anu}, \text{Kiran}\}$ (students)
Set $B = \{\text{Maths}, \text{Physics}, \text{Chemistry}\}$ (subjects)

and the relation is "studies": Ravi studies Maths and Physics; Anu studies only Chemistry; Kiran studies all three.

To draw the arrow diagram, list the elements of $A$ in one oval on the left, the elements of $B$ in another oval on the right, and draw one arrow for every pairing:

Ravi $\rightarrow$ Maths
Ravi $\rightarrow$ Physics
Anu $\rightarrow$ Chemistry
Kiran $\rightarrow$ Maths
Kiran $\rightarrow$ Physics
Kiran $\rightarrow$ Chemistry

Count the arrows leaving each name in $A$ : Ravi has 2, Anu has 1, Kiran has 3 — six arrows in total, one for each ordered pair in the relation. That is the whole diagram: two ovals of labelled points, joined by one arrow per related pair.

Note

The name comes from the arrows, not the ovals. What matters is: which element points to which, and in which direction.

The Precise Definition

A

Every arrow corresponds to exactly one ordered pair in $R$ . If a diagram has $k$ arrows, the relation has exactly $k$ ordered pairs — nothing more, nothing less.

Key Points to Remember

Direction matters. An arrow always starts at an element of $A$ and ends at an element of $B$ . An arrow "Ravi $\to$ Maths" is not the same statement as "Maths $\to$ Ravi" — the first element of the ordered pair is always where the arrow starts.
An element of $A$ can send multiple arrows. Kiran, above, sends three — one element can be related to many elements of $B$ .
An element of $A$ can send zero arrows. Nothing requires every element of $A$ to be related to something in $B$ . If a student studies none of the listed subjects, no arrow leaves their name.
An element of $B$ can receive multiple arrows. Both Ravi and Kiran point to Maths — an element of $B$ can be related to many elements of $A$ .

Watch out

A common mistake: assuming every element of $A$ must have at least one outgoing arrow, or that every element of $B$ must receive one. Neither is required. A relation can leave elements on either side completely unconnected.

A Relation on a Single Set

When a relation goes from a set to itself, both ovals contain the same elements, so they are usually drawn as one oval with arrows looping back into it. …

Start your 14-day free trial to unlock the full solution →

The relation $R$ pairs each prime number less than 10 with its cube. The primes are 2, 3, 5, 7; their cubes are 8, 27, 125, 343. So $R = \{(2,8), (3,27), (5,125), (7,343)\}$ .

The key idea here is simple: a relation in roster form is just a set of ordered pairs. The rule tells you exactly how to build each pair — take a prime number less than 10, and pair it with its cube.

Let’s first list the primes less than 10. A prime number has exactly two distinct positive divisors: 1 and itself. Checking numbers from 2 to 9:

2 is prime (divisors: 1, 2)
3 is prime (divisors: 1, 3)
4 is not prime (divisors: 1, 2, 4)
5 is prime (divisors: 1, 5)
6 is not prime (divisors: 1, 2, 3, 6)
7 is prime (divisors: 1, 7)
8 is not prime (divisors: 1, 2, 4, 8)
9 is not prime (divisors: 1, 3, 9)

So the primes are: $2, 3, 5, 7$ .

Now, for each prime $x$ , compute $x^3$ :

$x = 2 \implies x^3 = 2^3 = 8$ → ordered pair $(2, 8)$
$x = 3 \implies x^3 = 3^3 = 27$ → ordered pair $(3, 27)$
$x = 5 \implies x^3 = 5^3 = 125$ → ordered pair $(5, 125)$
$x = 7 \implies x^3 = 7^3 = 343$ → ordered pair $(7, 343)$ …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2022Set ANNUAL2 marks
$R = \{(x, x+5) : x \in \{0,1,2,3,4,5\}\}$
›Reveal solutionSolution

Substituting each $x\in\{0,1,2,3,4,5\}$ into $x+5$ gives the relation's pairs directly, from which the domain and range are read off.

$R=\{(x,x+5): x\in\{0,1,2,3,4,5\}\}$

Computing each pair:

$(0,5), (1,6), (2,7), (3,8), (4,9), (5,10)$

The domain is the set of all first components (the $x$ -values):

$\{0,1,2,3,4,5\}$

…

Start your 14-day free trial to see the full solution →
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2021Set ANNUAL2 marks
$A = \{1, 2, 3, ..., 14\}$
›Reveal solutionSolution

⚠️ This question's printed relation definition is genuinely unclear in the scan — it shows " $(x,-1)$ " as the pair template alongside the rule " $3x-y=0$ ", which is inconsistent (that rule needs a variable $y$ as the second coordinate, not the fixed value $-1$ ). Read honestly, the legible and mathematically sensible part is the rule $3x-y=0$ (i.e. $y=3x$ ) applied over $A=\{1,2,\dots,14\}$ — the standard textbook form of this exact problem — so that is what is solved here, restricted to pairs where both $x$ and $y$ lie in $A$ .

Note on the ambiguity: the printed stem reads " $R=\{(x,-1):3x-y=0,\text{ where }x,y\in R\}$ ", which cannot be read literally as written (a fixed " $-1$ " second coordinate contradicts a rule that defines $y$ via $3x-y=0$ , and " $x,y\in R$ " should say " $x,y\in A$ " since $R$ is the relation being defined, not a domain). The clearly legible and internally consistent core of the question is the rule $3x-y=0$ , i.e. $y=3x$ , with $x,y$ drawn from $A=\{1,2,\dots,14\}$ — this matches the standard NCERT-style problem this question is based on, so it is solved on that basis rather than guessed at literally.

…

Start your 14-day free trial to see the full solution →

←Previous question · 2.3Q6 — Determine the domain and range of the relation R defined by R = \(x, x…→Next question · 2.3Q8 — Let A = \x, y, z\ and B = \1, 2\. Find the number of relations from A …

Q.Write the relation R={(x,x3):x is a prime number less than 10}R = \{(x, x^3) : x\text{ is a prime number less than }10\}R={(x,x3):x is a prime number less than 10} in roster form.

🎓Unlock everything free for 14 days

Q.Write the relation $R = \{(x, x^3) : x\text{ is a prime number less than }10\}$ in roster form.

Unlock everything free for 14 days