Q.Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = { 2, 3, 5, 7 }. Find A ∩ B and hence show that A ∩ B = B
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Set Operations
The idea in plain words
Think of your two favourite groups of friends — the ones who play cricket and the ones who play football. Some friends are in both groups, some in only one, and some in neither. Set operations are simply the mathematical ways to answer questions like: "Who's in at least one team?" or "Who's only in the cricket team?".
The whole secret? Each operation is just a different way of combining or comparing two collections — like sorting your friends into different buckets.
Why this works
Sets are just labelled buckets that hold distinct items. The universal set U is the "whole world" of things we're talking about — say, all your friends. Then each operation picks out a specific bucket:
| Operation | What it asks | Bucket contains |
|---|---|---|
| Union (A∪B) | In either? | Everything from A or B (or both) |
| Intersection (A∩B) | In both? | Only the overlap |
| Difference (A∖B) | In A but not B? | Just the part of A that doesn't touch B |
| Complement (Ac) | Not in A? | Everything outside A (inside U) |
| Symmetric Difference (A△B) | In exactly one? | The two crescent-shaped parts, excluding the overlap |
Step by step
Let's take two concrete sets so you can see each operation in action:
A={1,2,3},B={3,4,5}
Step 1: Union — gather everything from both, but don't repeat anything.
A∪B={1,2,3,4,5}
Step 2: Intersection — only what's common to both.
A∩B={3}
Step 3: Difference (A minus B) — start with A, remove anything that's also in B.
A∖B={1,2}
Step 4: Complement — needs a universal set. Let U={1,2,3,4,5}. Then:
Ac={4,5}
Step 5: Symmetric Difference — combine the two differences:
A△B=(A∖B)∪(B∖A)={1,2}∪{4,5}={1,2,4,5}
A common slip …
Concept: Set Intersection and Set Equality
The intersection A∩B consists of all elements that belong to both A and B simultaneously.
Given A={1,2,3,4,5,6,7,8,9,10} and B={2,3,5,7}.
Scanning through B: every element 2,3,5,7 is present in A. Therefore,
A∩B={2,3,5,7} …
The intersection A∩B collects all elements that belong to both sets. Since every element of B already lies in A, we get A∩B={2,3,5,7}=B.
Why intersection captures "common membership"
The intersection of two sets is the collection of elements that satisfy a double requirement: membership in the first and membership in the second. Think of it as the overlap in a Venn diagram. When we write A∩B, we are asking: which elements pass the test for both A and B?
In this problem, A is the set of the first ten natural numbers, and B is the set of prime numbers less than 10. Notice that B is already "sitting inside" A — every prime in B is automatically one of the numbers from 1 to 10. This observation will make the intersection straightforward.
Finding A∩B step by step
-
List the elements of B:
B={2,3,5,7}.
-
Check each element of B for membership in A:
- Is 2∈A? Yes, 2 is in {1,2,3,…,10}.
- Is 3∈A? Yes.
- Is 5∈A? Yes.
- Is 7∈A? Yes.
-
Collect the common elements:
Every element of B is also in A, so
A∩B={2,3,5,7}.
- Compare A∩B with B: …
1. Concept First — The Idea of Set Intersection
The intersection of two sets is one of the most fundamental ideas in set theory. When we say A ∩ B (read as "A intersection B"), we are asking:
"Which elements are common to both sets A and B?"
Think of it like a Venn diagram — two overlapping circles. The intersection is the overlapping region where both circles contain the same items.
Why does this question ask us to "show that A ∩ B = B"?
This is a special case. When every element of B is also in A, then the intersection of A and B is simply B itself. In other words:
If B is a subset of A (written B ⊆ A), then A ∩ B = B.
Here, B = {2, 3, 5, 7} — these are all prime numbers less than 10. And A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} contains all numbers from 1 to 10. So clearly, every element of B lives inside A. That's the intuition.
2. Step-by-Step Solution
Step 1: List the elements of A and B clearly
We have:
- A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
- B = {2, 3, 5, 7}
Step 2: Find the common elements (the intersection)
We go through each element of B and check if it is present in A:
| Element of B | Is it in A? | Reason |
|---|---|---|
| 2 | ✓ Yes | 2 is between 1 and 10 |
| 3 | ✓ Yes | 3 is between 1 and 10 |
| 5 | ✓ Yes | 5 is between 1 and 10 |
| 7 | ✓ Yes | 7 is between 1 and 10 |
Since all four elements of B are also in A, the intersection contains exactly these four elements.
So:
A∩B={2,3,5,7}
Step 3: Compare with B
We see that:
A∩B={2,3,5,7}andB={2,3,5,7}
They are identical sets. Therefore:
A∩B=B
3. Final Answer
A∩B={2,3,5,7} and hence A∩B=B
4. Why It Works & Exam Tip
Why it works …
Here are the common mistakes students make with set intersection problems, specifically using the sets you provided, and how to avoid each.
Mistake 1: Forgetting the Definition of "Natural Number" (N)
The Error: Students often assume natural numbers start from 0 or include negative numbers. In the Indian curriculum (NCERT/CBSE), natural numbers are defined as {1,2,3,4,...}.
How it affects the answer: If you include 0, then A∩B might incorrectly include 0. If you include negatives, the intersection with primes (D) becomes confusing.
How to Avoid: Memorize the standard definition. For Class 11 NCERT, N={1,2,3,...}. Always write this set down before solving.
Mistake 2: Confusing "Even" and "Odd" with "Prime"
The Error: Students think that because a number is prime, it cannot be even (or odd). They forget that 2 is the only even prime number.
The Consequence: For question (v) B∩D, students often write ϕ (empty set) instead of {2}.
How to Avoid: List the first few elements of each set.
- B={2,4,6,8,10,...}
- D={2,3,5,7,11,...} Now, visually scan for common elements. The only common element is 2.
Mistake 3: Assuming "Odd" and "Prime" are Mutually Exclusive
The Error: Students think that since most primes are odd, the intersection C∩D must be all odd primes. They forget that 2 is prime but not odd.
The Consequence: For question (vi) C∩D, students write {3,5,7,11,...} (all odd primes) but forget to explicitly exclude 2. While the set of odd primes is correct, the reasoning is flawed if they don't mention that 2 is excluded.
How to Avoid: Always check the boundary case (the number 2).
- C={1,3,5,7,9,...}
- D={2,3,5,7,11,...} The intersection is {3,5,7,11,...} (all odd primes). This is correct, but be explicit: "All prime numbers except 2."
Mistake 4: Writing the Answer in Roster Form Incorrectly
The Error: For infinite sets like A∩B (which is just B), students try to list all elements or write an incomplete roster like {2,4,6}.
The Consequence: Marks are deducted for not showing the pattern or using the wrong notation.
How to Avoid: Use set-builder notation for infinite answers, or use roster form with an ellipsis (...).
- Correct: A∩B={2,4,6,8,...} or A∩B={x:x is an even natural number}.
- Incorrect: A∩B={2,4,6} (this implies the set stops at 6).
Mistake 5: Misinterpreting the Intersection Symbol (∩) …
- AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQQ.If A={x∈R∣Sin−1(x2+x+1)∈[−2π,2π]} and B={y∈R∣y=Sin−1(x2+x+1),x∈A} then (A) A∩B=ϕ (B) A∩BC=[0,1] (C) AC∩B=[3π,2π] (D) A∪B=R−{[−1,0]∪[3π,2π]}
›Reveal solutionSolution
Working out A=[−1,0] and B=[π/3,π/2] shows these two sets are disjoint (one is a set of x-values near 0, the other a set of angle-values near π/2), which makes AC∩B simply equal to B itself, matching option (C).
Concept and Intuition
The trick is recognising that Sin−1(u)∈[−π/2,π/2] is automatically true for every u in the domain of Sin−1 (that's the very definition of the principal value range), so set A's defining condition reduces to just requiring x2+x+1 to be a valid arcsine input, i.e. lying in [−1,1]. Set B is then the actual set of angle outputs produced as x ranges over A — an entirely different kind of set (radians, not x-values), which is why it ends up disjoint from A.
Step-by-Step Solution
- Find A: Since x2+x+1=(x+21)2+43>0 always, x2+x+1 is real and non-negative for all real x. The condition Sin−1(x2+x+1)∈[−π/2,π/2] holds automatically whenever Sin−1 is defined, i.e. whenever x2+x+1∈[−1,1]. Since the square root is ≥0, this reduces to x2+x+1≤1⇔x2+x+1≤1⇔x2+x≤0⇔x(x+1)≤0⇔x∈[−1,0]. So A=[−1,0].
- Find the range of x2+x+1 on A: this is a upward parabola with vertex at x=−1/2, value 43; at the endpoints x=−1 and x=0, the value is 1. So on [−1,0], x2+x+1 ranges continuously over [43,1].
- Find B: x2+x+1 then ranges over [3/2,1]. Since Sin−1 is increasing, B=Sin−1([3/2,1])=[Sin−1(3/2),Sin−1(1)]=[π/3,π/2].
- Compare A and B: A=[−1,0] is a set of real numbers near the origin; B=[π/3,π/2]≈[1.047,1.571] is a set of positive numbers greater than 1. These intervals do not overlap, so A∩B=ϕ.
- Check each option: …
- CA Foundation 2025Set may-20251 markMCQQ.If A={1,2,3,4}, B={2,4,6,8} and C={3,4,5,6}, the value of A−{B∪C} is (A) {1, 2, 3} (B) {2, 3, 4, 5} (C) {1} (D) {0}
›Reveal solutionSolution
B∪C={2,3,4,5,6,8}; removing these from A leaves {1}.
Step 1 — Compute the union B∪C
{2,4,6,8}∪{3,4,5,6}={2,3,4,5,6,8}
Step 2 — Compute the difference A−(B∪C)
Keep elements of A={1,2,3,4} NOT in the union. Elements 2,3,4 are all present in the union; only 1 survives.
A−(B∪C)={1}
Why the other options are wrong: (A) {1,2,3} and (B) {2,3,4,5} keep elements that ARE in the union; (D) {0} introduces 0, which is in no set. …
- AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQQ.If A and B are any two events of a sample space, then set-theoretic description for the event: "Exactly one of the events A, B to occur" is (Here Ec denotes the compliment of the event E) (A) A∩Bc (B) (A−B)∪(A∪B) (C) (A∩Bc)∪(Ac∩B) (D) (A∩B)c∪(Ac∩Bc)
›Reveal solutionSolution
"Exactly one occurs" is the symmetric-difference event: (A but not B) union (B but not A).
Concept and Intuition
"Exactly one" excludes both the case where neither occurs and the case where both occur. It is the union of the two mutually exclusive possibilities: only A happens, or only B happens.
Step-by-Step Solution
- "A occurs, B does not" =A∩Bc.
- "B occurs, A does not" =Ac∩B.
- These two cases are disjoint and together cover "exactly one occurs", so the event is (A∩Bc)∪(Ac∩B).
Common Mistakes …
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