For all sets $A$, $B$ and $C$, show that $(A - B) \cap (C - B) = A - (B \cup C)$.

For all sets A, B and C, show that (A - B) ∩ (C - – Class 11 Mathematics

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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 $A \setminus B$ ) 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 \mid x \in A \text{ and } x \notin 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\}, \qquad 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 $B$ .

$1 \in A$ , $1 \notin B$ → keep
$2 \in A$ , $2 \notin B$ → keep
$3 \in A$ , $3 \in B$ → remove
$4 \in A$ , $4 \in B$ → remove
$5 \in A$ , $5 \in B$ → remove

A - B = \{1, 2\}

Now compute the other direction:

B - A = \{6, 7\}

Notice $A - B \neq B - A$ — set difference is not commutative.

Key properties

Property	Statement
Not commutative	$A - B \neq B - A$ in general
Difference with itself	$A - A = \emptyset$
Difference with empty set	$A - \emptyset = A$ , and $\emptyset - A = \emptyset$
Difference with universal set	$U - A = A^c$ (the complement of $A$ )
Disjoint sets	If $A \cap B = \emptyset$ , 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 $A^c$ (complement of $A$ ). The difference is what you're comparing against:

Complement $A^c$ 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: $A^c = U - A$ .

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 \mid 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 $\mid$ means "such that".

Example:

A = \{ n \mid n \in \mathbb{N}, n \text{ 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 \iff (\forall x)(x \in A \iff x \in 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 $\in$ (belongs to).

x \in S \quad \text{or} \quad x \notin 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:

\mathbb{N} = \{ n \mid n \text{ 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 $\emptyset$ (or $\{\}$ ) is the set with no elements. …

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The statement is false. Writing each difference as $X - Y = X \cap Y^c$ , the left side simplifies to $(A \cap C) - B$ , while the right side is $A \cap B^c \cap C^c$ -- these are different sets in general, as a counterexample confirms.

The expression $(A-B)\cap(C-B)$ asks for elements that are in $A$ but not $B$ , AND in $C$ but not $B$ -- in other words, elements common to $A$ and $C$ that also avoid $B$ . The expression $A-(B\cup C)$ asks for elements of $A$ that avoid BOTH $B$ and $C$ . These are different requirements, so the two sides need not be equal.

Step 1: Simplify the left side.

For any sets $X, Y$ : $X - Y = X \cap Y^c$ (elements in $X$ that are not in $Y$ ). So:

(A-B)\cap(C-B) = (A\cap B^c)\cap(C\cap B^c) = A\cap C\cap B^c = (A\cap C) - B

Step 2: Simplify the right side.

By De Morgan's law, $(B\cup C)^c = B^c\cap C^c$ , so:

A - (B\cup C) = A\cap(B\cup C)^c = A\cap B^c\cap C^c

Step 3: Compare.

Left side (simplified): $A\cap C\cap B^c$ -- requires the element to be IN $C$ .

Right side: $A\cap B^c\cap C^c$ -- requires the element to be NOT in $C$ .

These are opposite conditions on $C$ , so the two sides are not equal in general.

Step 4: Counterexample. …

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

Showing the 12 most recent of 17 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$A = \{2, 3, 4\}$
(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 $A$ :

$2 \in A$ , $2 \notin B$ → keep …
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CBSE 2025Set ANNUAL1 markMCQ
$U = \{1, 2, \ldots, 15\}$
(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 $B = \{2, 4, 6, 8, 10, 12, 14\}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$U = \{1, 2, \ldots, 15\}$
(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 $C = \{2, 3, 5, 7, 11, 13\}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$U = \{1, 2, \ldots, 15\}$
(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 $A$ : $2, 3, 5 \in A$ (removed); $7, 11, 13 \notin A$ (kept). …

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CBSE 2025Set ANNUAL1 markMCQ
$U = \{1, 2, \ldots, 15\}$
(a) $\{4, 6, 8, 10, 12, 14\}$
(b) $\{1, 3, 5, 15\}$
(c) $\{4, 6, 15\}$
(d) $\phi$
›Reveal solutionSolution

$B - A = \{4, 6, 8, 10, 12, 14\}$ .

Given $B = \{2, 4, 6, 8, 10, 12, 14\}$ and $A = \{1, 2, 3, 5, 15\}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$U = \{1, 2, \ldots, 15\}$
(a) $\{3, 5, 7, 13\}$
(b) $\{3, 5, 7, 2, 13\}$
(c) $\{3, 5, 7, 11, 13\}$
(d) $\phi$
›Reveal solutionSolution

$C - B = \{3, 5, 7, 11, 13\}$ .

Given $C = \{2, 3, 5, 7, 11, 13\}$ and $B = \{2, 4, 6, 8, 10, 12, 14\}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$U = \{1, 2, \ldots, 15\}$
(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 $C = \{2, 3, 5, 7, 11, 13\}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If A = {1, 3, 4, 5, 6}, B = {2, 4, 6, 7, 8}, then A - B is:$
(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, $A - B = \{x : x \in A \text{ and } x \notin B\}$ .

Check each element of $A$ :

$1 \notin B$ → keep

$3 \notin B$ → keep …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}; A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}, find (B - C):$
(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 $U=\{1,2,\dots,9\}$ , $A=\{1,2,3,4\}$ , $B=\{2,4,6,8\}$ , $C=\{3,4,5,6\}$ .

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 $C$ ).

…

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CBSE 2023Set ANNUAL1 mark
$Q. If R is the set of real numbers and Q is the set of rational numbers, then what is R - Q?$
›Reveal solutionSolution

$R-Q$ is the set of all irrational numbers.

The real numbers $\mathbb R$ are partitioned into rationals $\mathbb Q$ and irrationals. Removing all rational numbers from $\mathbb R$ leaves exactly the numbers that cannot be expresse …

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CBSE 2023Set ANNUAL1 markMCQ
$A$
(a) $(A \cap B) \cup (A \cup B)$
(b) $(A \cup B) - (A \cap B)$
(c) $A - (A \cap B)$
(d) $(A \cup B) - B$
›Reveal solutionSolution

$(A-B)\cup(B-A)=(A\cup B)-(A\cap B)$ ; option (b).

$(A-B)\cup(B-A)$ collects elements in exactly one of $A,B$ — the symmetric difference. Equivalently it is everything in $A\cup B$ that is not common to both …

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CBSE 2022Set TERM11 markMCQ
$A=\{1,2,3,4,5,6\}$
(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 $A$ : $2\in A$ (drop), $4\in A$ (drop), …

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Q.For all sets AAA, BBB and CCC, show that (A−B)∩(C−B)=A−(B∪C)(A - B) \cap (C - B) = A - (B \cup C)(A−B)∩(C−B)=A−(B∪C).

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Q.For all sets $A$ , $B$ and $C$ , show that $(A - B) \cap (C - B) = A - (B \cup C)$ .

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