Q.Write the set {x : x is a positive integer and x2 < 40} in the roster form.
Concept understanding — Set Membership
Set Membership
The idea in plain words
Every set is defined by exactly one question: "does this object belong to the set, or not?" That yes/no relationship between an object and a set is called membership. If an object is in the set, it is a member (or element) of the set; if it isn't, it simply is not.
A set is only "well-defined" if this question always has a clear answer for every possible object — that's what makes membership testable.
The notation
For a set A and an object x:
- x∈A reads "x belongs to A" or "x is an element of A" — TRUE membership.
- x∈/A reads "x does not belong to A" — FALSE membership.
Example. Let A={2,4,6,8}.
4∈A(4 is listed inside A)
5∈/A(5 is not listed inside A)
Testing membership: roster form vs. set-builder form
Roster form — just look for the object in the list.
B={1,3,5,7},3∈B,4∈/B
Set-builder form — plug the candidate into the defining rule and check if it's satisfied.
C={x∣x is a prime number less than 10}
Is 7∈C? Check: is 7 prime and less than 10? Yes → 7∈C.
Is 9∈C? Check: 9 is less than 10 but not prime (9=3×3) → 9∈/C.
Properties every student must know
- Each element is either in or out — never "partly in." Membership is binary, not a matter of degree.
- Repetition doesn't affect membership. {1,1,2}={1,2} — asking "is 1 a member?" gives the same YES either way.
- Order never affects membership. 2∈{1,2,3} is exactly the same fact as 2∈{3,2,1}.
- Nothing belongs to the empty set. For any x, x∈/∅ — there is nothing inside to belong to.
- A set can be an element of another set. If D={1,{2,3}}, then 1∈D is true, and {2,3}∈D is true, but 2∈D is false — 2 is not directly listed in D; it is only inside the set that is listed.
A common slip
Students often confuse x∈A (membership: is the object present?) with B⊆A (subset: is every element of B also in A?). The two symbols compare different kinds of things:
- ∈ / ∈/ compares an element to a set.
- ⊆ / ⊆ compares a set to a set.
So for A={1,2,3}: writing 1∈A is correct, but 1⊆A is technically wrong notation (1 is an element, not a set) — although {1}⊆A is correct, because now both sides being compared are sets.
Why it matters
Membership is the single test every other set idea is built on — union, intersection, subset, and complement are all ultimately defined by asking "x∈A?" and "x∈B?" for every candidate x. Get comfortable with ∈/∈/ first, and every later operation becomes just a combination of membership questions.
Takeaway
Membership answers one question only: is this exact object listed in this exact set? Everything else in set theory is built from asking that question, over and over, about different sets.
"Set membership symbol meaning" and "element vs subset difference class 11" are common queries around this idea, which forms the very first building block of the Sets chapter in the NCERT/CBSE Class 11 Mathematics curriculum. Getting the ∈ vs ⊆ distinction right is a small but frequent trap in board and competitive exam questions.
Why this formula?
Let's break down the definition of a set — not as a formula to memorise, but as a fundamental idea that underpins all of mathematics.
1. What is a Set? (The Core Idea)
A set is a well-defined collection of distinct objects.
The "why" here is about clarity and precision — we need to know exactly what belongs and what does not.
- Well-defined: For any object, we can say yes or no — no ambiguity.
- Distinct: No duplicates — each object appears only once.
Why? Because if we couldn't decide membership, we couldn't do any logical operations. Sets are the building blocks of all mathematical structures.
2. The Key "Formula": Set-Builder Notation
The most common way to define a set is:
S={x∣P(x)}
This reads: "S is the set of all objects x such that property P(x) is true."
Why does this work?
- x is a placeholder for any object.
- P(x) is a logical condition (a predicate) that is either true or false for each x.
- The vertical bar ∣ means "such that".
Example:
A={n∣n∈N,n is even}
Here, P(n) is "n is a natural number and n is even".
Only those n that satisfy both conditions are included.
Why this form? It avoids listing infinitely many elements. It gives a rule — a decision procedure — for membership.
3. The Two Fundamental Properties (Axioms)
Every set definition relies on two intuitive truths:
(a) Extensionality — Two sets are equal if they have the same elements.
A=B⟺(∀x)(x∈A⟺x∈B)
Why? A set is completely determined by its members. There is no other hidden property.
If you know what's inside, you know the set.
(b) Membership — The only relation is ∈ (belongs to).
x∈Sorx∈/S
Why? Because a set is just a container. The only question we can ask is: "Is this object inside?"
4. Why Can't We Just List Everything?
For small sets, listing works:
{1,2,3}
But for infinite sets (like all natural numbers), listing is impossible.
Set-builder notation solves this by giving a rule instead of a list.
Example:
N={n∣n is a positive integer}
This is not a formula to memorise — it's a definition by property.
5. The "Empty Set" — Why It Exists
The empty set ∅ (or {}) is the set with no elements.
∅={x∣x=x}
Why is this allowed?
Because the condition x=x is always false — no object satisfies it.
This is a logical necessity: if we can define a set by a property, we must allow the possibility that nothing satisfies it.
Key insight: The empty set is not "nothing" — it's a set that contains nothing. It's a mathematical object.
6. Summary: The "Why" Behind the Definition
| Concept | Why it's defined this way |
|---|---|
| Set | To have a precise, unambiguous collection — no guesswork. |
| Set-builder | To define infinite or complex sets without listing. |
| Membership (∈) | The only question that matters — is it inside or not? |
| Empty set | Logical completeness — a property may have no objects. |
Final takeaway: The definition of a set is not a formula to plug numbers into. It's a logical framework for saying: "These objects, and only these, belong here." Every formula you see later (union, intersection, complement) builds on this single idea.
We need positive integers x such that x2<40.
Testing values: 62=36<40, but 72=49 is NOT less than 40. So the largest qualifying integer is 6, and every positive integer from 1 to 6 satisfies the condition.
The set in roster form is {1,2,3,4,5,6}.
The set contains all positive integers whose square is less than 40. Listing them gives {1,2,3,4,5,6}.
Roster form means listing all the elements of a set inside curly braces, separated by commas -- no conditions, just the actual values.
The condition is: x is a positive integer and x2<40.
Step 1: Start from the smallest positive integer and test each one.
12=1<40 -- include 1.
22=4<40 -- include 2.
32=9<40 -- include 3.
42=16<40 -- include 4.
52=25<40 -- include 5.
62=36<40 -- include 6.
Step 2: Find the cutoff.
72=49, which is NOT less than 40, so 7 is excluded. Every integer larger than 7 has an even bigger square, so none of them qualify either.
A common mistake is to compare x itself to 40 instead of x2. The condition is on the SQUARE of x, not x directly.
Step 3: Collect the qualifying integers into roster form.
Equivalently, x2<40 means x<40≈6.32, so the positive integers satisfying this are 1 through 6 -- a quick mental check.
The set in roster form is {1,2,3,4,5,6}.
Concept: Interval Notation & Roster Form
Step 1: Identify the condition
We need positive integers x such that x2<40.
Step 2: Find the range of x
Since x is a positive integer, x≥1.
Solve x2<40⟹x<40≈6.324.
So x can be 1,2,3,4,5,6.
Step 3: Verify each integer
12=1, 22=4, 32=9, 42=16, 52=25, 62=36 — all are <40.
72=49>40, so stop.
Final Answer:
{1,2,3,4,5,6}
Common Mistakes in Writing Interval Notation
Mistake 1: Confusing Open and Closed Brackets
The Error:
Students often write (–4, 6] as [–4, 6] or (–4, 6), misreading the inequality signs.
Why It Happens:
<and>mean open (excluded endpoint) → use(or)≤and≥mean closed (included endpoint) → use[or]
How to Avoid:
Draw a number line for each inequality. Mark the endpoint with:
- An open circle (○) for
<or> - A filled circle (●) for
≤or≥
Then write the bracket that matches the circle.
Correct answers:
- (i) (−4,6]
- (ii) (−12,−10)
- (iii) [0,7)
- (iv) [3,4]
Mistake 2: Reversing the Order of Endpoints
The Error:
Writing (6, –4] instead of (-4, 6] — putting the larger number first.
Why It Happens:
Students copy the order from the set-builder form without thinking about the number line.
How to Avoid:
Always write the smaller number on the left. On a number line, values increase left to right. Interval notation must follow this natural order.
Memory trick:
"Left is less, right is greater — always write small → big."
Mistake 3: Forgetting the Variable is Real Numbers
The Error:
Writing (-4, 6] but thinking it only includes integers like {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6}.
Why It Happens:
Students confuse interval notation (continuous set of real numbers) with roster form (listing discrete elements).
How to Avoid:
Remember: x∈R means all real numbers between the endpoints — including fractions, decimals, and irrationals. The interval (-4, 6] contains infinitely many numbers, not just integers.
Mistake 4: Mixing Up Infinity Notation
The Error:
Writing [–∞, 6] or (–12, ∞] — using closed brackets with infinity.
Why It Happens:
Students treat infinity like a number that can be "reached" or "included."
How to Avoid:
Infinity (∞) is not a number — it's a concept meaning "unbounded." You can never include it, so always use:
(–∞, ...(open on left)..., ∞)(open on right)
Correct examples:
- (−∞,5] — includes 5, but not infinity
- (3,∞) — excludes 3, and infinity is always open
Quick Reference Table
| Inequality | Interval Notation | Bracket Rule |
|---|---|---|
| –4<x≤6 | (−4,6] | < → ( , ≤ → ] |
| –12<x<–10 | (−12,−10) | Both < → both ( |
| 0≤x<7 | [0,7) | ≤ → [ , < → ) |
| 3≤x≤4 | [3,4] | Both ≤ → both [ |
Final tip: Practice by converting back and forth:
Set-builder → Number line → Interval notation.
This triple-check catches all four mistakes.
- COMEDK 2024Set 2024-M1 markMCQQ.Express the set A={1,7,17,31,49} in set builder form (A) {x∣x=2n2−1, where n∈N and n<5} (B) {x∣x=2n2−3, where n∈N and 2≤n≤8} (C) {x∣x=2n2+1, where n∈N and n≤7} (D) {x∣x=2n2−1, where n∈N and n≤5}
›Reveal solutionSolution
The set A={1,7,17,31,49} matches the formula x=2n2−1 for n=1,2,3,4,5, so the correct option is (D).
We need to express the given set {1,7,17,31,49} in set-builder form. The key is to find a pattern: each number is one less than twice a perfect square. Let’s check:
- 1=2(1)2−1
- 7=2(2)2−1
- 17=2(3)2−1
- 31=2(4)2−1
- 49=2(5)2−1
So the pattern is x=2n2−1 with n taking natural numbers from 1 to 5. Now we examine each option.
-
Option (A): x=2n2−1, n∈N, n<5.
This gives n=1,2,3,4 → values: 1,7,17,31. Missing 49. So incorrect.
-
Option (B): x=2n2−3, n∈N, 2≤n≤8.
For n=2: 2(4)−3=5 (not in set). So incorrect.
-
Option (C): x=2n2+1, n∈N, n≤7.
For n=1: 2+1=3 (not in set). So incorrect.
-
Option (D): x=2n2−1, n∈N, n≤5.
This gives n=1,2,3,4,5 → exactly 1,7,17,31,49. Correct.
Watch outA common mistake is to misread “n<5” as including 5 — it does not. Option (A) fails because it stops at n=4, missing 49.
TipAlways test the smallest and largest elements of the set against the formula and the range of n. Here, 49 is the key: only n=5 produces it, so the range must include 5.
✓Final answerThe correct option is (D).
ANSWER: D
- COMEDK 2023Set 2023-E1 markMCQQ.Which of the following is a singleton set? (A) {x:x2=4,x∈R} (B) {x:∣x∣<4,x∈N} (C) {x:∣x∣<−4,x∈N} (D) {x:x2=4,x∈N}
›Reveal solutionSolution
{x:x2=4, x∈N}={2} has exactly one element.
- (A) {x:x2=4, x∈R}={2,−2} — two elements.
- (B) {x:∣x∣<4, x∈N}={1,2,3} — three elements.
- (C) {x:∣x∣<−4, x∈N}=∅ — empty (no modulus is negative).
- (D) {x:x2=4, x∈N}={2} — exactly one element.
Only (D) is a singleton.
✓Final answerThe correct option is (D) — {x:x2=4, x∈N}
- KCET 2019Set A-11 markMCQQ.If U is the universal set with 100 elements; A and B are two sets such that n(A)=50, n(B)=60, n(A∩B)=20 then n(A′∩B′)= (A) 40 (B) 20 (C) 90 (D) 10
›Reveal solutionSolution
Use De Morgan’s law: A′∩B′=(A∪B)′. Find n(A∪B) via the inclusion-exclusion principle, then subtract from the total. The answer is 10.
The core idea here is that the complement of the union is exactly the region outside both sets. Instead of trying to count elements outside A and outside B separately, we use De Morgan’s law to turn the problem into one we already know how to solve: finding the size of the union.
When you see A′∩B′, always think: “this is everything that is not in A and not in B” — which is the same as “everything that is not in (A or B)”. That’s the complement of the union.
- Find n(A∪B) using the inclusion-exclusion formula. For any two sets,
n(A∪B)=n(A)+n(B)−n(A∩B)
Substitute the given values:
n(A∪B)=50+60−20=90
- Apply De Morgan’s law.
A′∩B′=(A∪B)′
This is a set identity — it always holds, no matter what the sets are.
- Find the size of the complement. The universal set U has 100 elements. The complement of any set X has size:
n(X′)=n(U)−n(X)
So:
n(A′∩B′)=n((A∪B)′)=n(U)−n(A∪B)=100−90=10
Watch outA common mistake is to think n(A′∩B′)=n(U)−n(A)−n(B). That would give 100−50−60=−10, which is impossible. The error is double-counting the intersection — elements in both A and B get subtracted twice. Always use the union first.
TipIf you prefer a visual approach: draw a Venn diagram with three regions — only A (30), only B (40), and both (20). The union fills 30+40+20=90 elements. The remaining 10 are outside both circles, which is exactly A′∩B′.
✓Final answerThe value is 10, which corresponds to option (D).
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