Q.Let V = { a, e, i, o, u } and B = { a, i, k, u}. Find V – B and B – V
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Set Difference
Set Difference
The idea in plain words
Imagine two groups of students: those who play cricket (A) and those who play football (B). The set difference A−B (also written A∖B) answers one specific question: "Who plays cricket but NOT football?" You start with everything in A, then remove whatever also happens to be in B.
Set difference is a one-way street: A−B keeps only what's uniquely in A. It has nothing to do with what's uniquely in B.
The precise definition
For two sets A and B:
A−B={x∣x∈A and x∈/B}
Read as: "the set of all x such that x is in A but x is not in B."
Worked example
Let:
A={1,2,3,4,5},B={3,4,5,6,7}
Step 1: Go through each element of A.
Step 2: Keep it only if it is NOT also in B.
- 1∈A, 1∈/B → keep
- 2∈A, 2∈/B → keep
- 3∈A, 3∈B → remove
- 4∈A, 4∈B → remove
- 5∈A, 5∈B → remove
A−B={1,2}
Now compute the other direction:
B−A={6,7}
Notice A−B=B−A — set difference is not commutative.
Key properties
| Property | Statement |
|---|---|
| Not commutative | A−B=B−A in general |
| Difference with itself | A−A=∅ |
| Difference with empty set | A−∅=A, and ∅−A=∅ |
| Difference with universal set | U−A=Ac (the complement of A) |
| Disjoint sets | If A∩B=∅, then A−B=A |
The last property is worth pausing on: if two sets share nothing in common, subtracting one from the other changes nothing — there was nothing to remove.
Set difference vs. complement — the classic mix-up
Students frequently confuse A−B with Ac (complement of A). The difference is what you're comparing against:
- Complement Ac is always relative to the universal set U: everything outside A.
- Difference A−B is relative to whatever second set you name: everything in A that isn't in B.
In fact, complement is just a special case: Ac=U−A.
Set difference vs. symmetric difference …
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. …
The set difference A−B contains all elements that belong to A but do not belong to B.
For V−B, we start with V={a,e,i,o,u} and remove every element that appears in B={a,i,k,u}. The common elements are a, i, and u. Removing these leaves e and o. …
Set difference removes elements of the second set from the first. V−B={e,o} and B−V={k}.
The set difference operation A−B (also written A∖B) gives you all elements that belong to A but do not belong to B. Think of it as filtering: you start with everything in A, then throw out anything that also appears in B.
This is fundamentally different from intersection (which keeps only common elements) or union (which combines everything). Set difference is about exclusion—what remains in the first set after removing any overlap with the second.
Let's identify what we have:
- V={a,e,i,o,u} — the five vowels
- B={a,i,k,u} — a mix of vowels and the consonant k
Finding V−B
-
Start with all elements of V: We have a,e,i,o,u.
-
Identify which elements of V also appear in B: Looking at B={a,i,k,u}, we see that a, i, and u are common to both sets.
-
Remove the common elements from V: After removing a, i, and u, we're left with e and o.
Therefore, V−B={e,o}.
The elements e and o are vowels that appear in V but not in B. The consonant k from B is irrelevant here—we only care about what's in V.
Finding B−V …
Method: Direct Set Difference (Element Comparison)
Concept: The set difference A−B (also written A∖B) contains all elements that are in A but not in B.
Steps
-
List the elements of the first set
V={a,e,i,o,u}
-
For each element, check if it is present in the second set (B={a,i,k,u})
-
Keep only those elements that are NOT in B
Finding V−B
| Element in V | In B? | Keep in V−B? |
|---|---|---|
| a | Yes | No |
| e | No | Yes |
| i | Yes | No |
| o | No | Yes |
| u | Yes | No |
Result:
V−B={e,o}
Finding B−V
| Element in B | In V? | Keep in B−V? | …
Common Mistakes in Set Difference (V – B and B – V)
Mistake 1: Confusing the Order (Thinking V – B = B – V)
The error: Students often assume set difference is commutative — that V−B gives the same result as B−V.
Why it happens: Subtraction of numbers is not commutative (5−3=3−5), but students sometimes forget this applies to sets too.
How to avoid: Always read the operation as "elements in the first set that are NOT in the second set."
- V−B = elements in V that are not in B
- B−V = elements in B that are not in V
Correct results:
- V−B={e,o} (V has a, e, i, o, u; remove a, i, u which are in B)
- B−V={k} (B has a, i, k, u; remove a, i, u which are in V)
Mistake 2: Removing All Common Elements from Both Sets
The error: Students list V−B as {e,o,k} — removing the intersection from both sets.
Why it happens: Confusing set difference with symmetric difference (V△B), which removes common elements from both.
How to avoid: Remember the "only from the first set" rule. Draw two overlapping circles. For V−B, shade only the part of V that does not touch B.
Mistake 3: Including Elements That Are Not in the First Set
The error: Writing V−B={e,o,k} — including k which is not in V at all.
Why it happens: Students think "remove common elements" means remove intersection from the union of both sets.
How to avoid: Check: Can the answer contain an element not present in the first set? No. For V−B, every element must belong to V.
Mistake 4: Forgetting to List All Elements of the First Set
The error: Writing V−B={e} — missing 'o'. …
Showing the 12 most recent of 17 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.If A={2,3,4}, B={4,5,6}, then the value of A−B is —(a) {3,4}(b) {2,3}(c) {5,6}(d) {3,4,5}
›Reveal solutionSolution
A−B={2,3}, option (b).
The set difference A−B consists of all elements that belong to A but do NOT belong to B.
Here A={2,3,4} and B={4,5,6}. Check each element of A:
- 2∈A, 2∈/B → keep …
- CBSE 2025Set ANNUAL1 markMCQQ.Given U={1,2,…,15}, A={1,2,3,5,15}, B={2,4,6,8,10,12,14}, C={2,3,5,7,11,13}. A−B=(a) {1,2,3,5}(b) {1,3,5,15}(c) {2}(d) {2,3,5,15}
›Reveal solutionSolution
A−B={1,3,5,15}, i.e., the elements of A that are not in B.
Given A={1,2,3,5,15} and B={2,4,6,8,10,12,14}.
…
- CBSE 2025Set ANNUAL1 markMCQQ.Given U={1,2,…,15}, A={1,2,3,5,15}, B={2,4,6,8,10,12,14}, C={2,3,5,7,11,13}. B−C=(a) {4,6,8,10,12,14}(b) {3,5,7,11,13}(c) {2}(d) {4,6,13}
›Reveal solutionSolution
B−C={4,6,8,10,12,14}.
Given B={2,4,6,8,10,12,14} and C={2,3,5,7,11,13}.
…
- CBSE 2025Set ANNUAL1 markMCQQ.Given U={1,2,…,15}, A={1,2,3,5,15}, B={2,4,6,8,10,12,14}, C={2,3,5,7,11,13}. C−A=(a) {1,2,3,5}(b) {1,2,7,11,13}(c) {3,7,11,13}(d) {7,11,13}
›Reveal solutionSolution
C−A={7,11,13}.
Given C={2,3,5,7,11,13} and A={1,2,3,5,15}.
C−A keeps elements of C not in A: 2,3,5∈A (removed); 7,11,13∈/A (kept). …
- CBSE 2025Set ANNUAL1 markMCQQ.Given U={1,2,…,15}, A={1,2,3,5,15}, B={2,4,6,8,10,12,14}, C={2,3,5,7,11,13}. B−A=(a) {4,6,8,10,12,14}(b) {1,3,5,15}(c) {4,6,15}(d) ϕ
›Reveal solutionSolution
B−A={4,6,8,10,12,14}.
Given B={2,4,6,8,10,12,14} and A={1,2,3,5,15}.
…
- CBSE 2025Set ANNUAL1 markMCQQ.Given U={1,2,…,15}, A={1,2,3,5,15}, B={2,4,6,8,10,12,14}, C={2,3,5,7,11,13}. C−B=(a) {3,5,7,13}(b) {3,5,7,2,13}(c) {3,5,7,11,13}(d) ϕ
›Reveal solutionSolution
C−B={3,5,7,11,13}.
Given C={2,3,5,7,11,13} and B={2,4,6,8,10,12,14}.
…
- CBSE 2025Set ANNUAL1 markMCQQ.Given U={1,2,…,15}, A={1,2,3,5,15}, B={2,4,6,8,10,12,14}, C={2,3,5,7,11,13}. A−C=(a) {2,3,5}(b) {1,2,3,5}(c) {1,5,15}(d) {1,15}
›Reveal solutionSolution
A−C={1,15}.
Given A={1,2,3,5,15} and C={2,3,5,7,11,13}.
…
- CBSE 2025Set ANNUAL1 markMCQQ.If A = {1, 3, 4, 5, 6}, B = {2, 4, 6, 7, 8}, then A − B is:(a) {-1, -1, -2, -2, -2}(b) {1, 3, 5}(c) {2, 7, 8}(d) None of these
›Reveal solutionSolution
A−B contains exactly the elements of A that do not belong to B.
Given A={1,3,4,5,6} and B={2,4,6,7,8}.
By definition, A−B={x:x∈A and x∈/B}.
Check each element of A:
- 1∈/B → keep
- 3∈/B → keep …
- CBSE 2024Set ANNUAL1 markMCQQ.Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}; A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}, find (B - C):(a) {1, 3, 4, 5, 6, 7, 9}(b) {1, 4, 7, 8, 9}(c) {3, 4, 6, 8}(d) {2, 4, 5, 6, 7, 8}
›Reveal solutionSolution
B−C={2,8}, and its complement in U is {1,3,4,5,6,7,9}, matching option (a).
Given U={1,2,…,9}, A={1,2,3,4}, B={2,4,6,8}, C={3,4,5,6}.
Step 1: Compute B−C (elements of B not in C).
B−C={2,8} (4 and 6 are removed since they also lie in C).
…
- CBSE 2023Set ANNUAL1 markQ.If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?
›Reveal solutionSolution
R−Q is the set of all irrational numbers.
The real numbers R are partitioned into rationals Q and irrationals. Removing all rational numbers from R leaves exactly the numbers that cannot be expresse …
- CBSE 2023Set ANNUAL1 markMCQQ.If A, B and C are non-empty subsets of a set then (A−B)∪(B−A) equals(a) (A∩B)∪(A∪B)(b) (A∪B)−(A∩B)(c) A−(A∩B)(d) (A∪B)−B
›Reveal solutionSolution
(A−B)∪(B−A)=(A∪B)−(A∩B); option (b).
(A−B)∪(B−A) collects elements in exactly one of A,B — the symmetric difference. Equivalently it is everything in A∪B that is not common to both …
- CBSE 2022Set TERM11 markMCQQ.If A={1,2,3,4,5,6} and B={2,4,6,8} then B−A will be(a) {8}(b) {2,4,6}(c) {2,4,6,8}(d) none of these
›Reveal solutionSolution
B−A keeps only the elements of B that are absent from A.
A={1,2,3,4,5,6}, B={2,4,6,8}. Check each element of B against A: 2∈A (drop), 4∈A (drop), …
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