Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 7 · Chapter companion part
Use union, intersection and complements, read every region of two- and three-set Venn diagrams and justify the set laws element by element.
Class 8 English Version: part B of the chapter 7 companion. Eight sections, 30 written tasks and a notebook challenge. Selected companion coverage, not complete textbook coverage; use all three parts. Counts are checked, while rosters, equivalent set-builder rules, diagrams and general proofs need guardian or teacher review.
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A∪B contains elements in A or B or both. For A={1,2,3}, B={3,4,5}, their union is {1,2,3,4,5}. The shared element 3 is listed once. A Venn union shades both circles, including their overlap. In this definition or is inclusive; it does not mean exactly one of the two sets.
A∩B contains elements that belong to both A and B. For A={1,2,3}, B={3,4,5}, it is {3}. A Venn intersection shades only the overlap. An element in A but outside B is not in the intersection. The set A∖B contains elements in A and not B; reversing this difference generally changes the answer.
The sets {1,3,5} and {2,4,6} are disjoint because no element is shared, so their intersection is ∅. This says nothing by itself about whether their union fills the chosen universe. A set can be disjoint from ∅; the empty set has no shared member with any set. A nonempty set is not disjoint from itself.
For A⊆U, A∪∅=A, A∩∅=∅, A∪U=U and A∩U=A. Also A∪A=A and A∩A=A: repeating a condition does not add a second copy of each element. The operations describe membership, not arithmetic addition and multiplication of the written sets.
Union and intersection are commutative: A∪B=B∪A and A∩B=B∩A. They are associative: (A∪B)∪C=A∪(B∪C) and similarly for intersection. Mixed operations still need brackets. A∩(B∪C) is not in general (A∩B)∪C; distributing correctly gives (A∩B)∪(A∩C).
An element belongs to A∩(B∪C) exactly when it belongs to A and at least one of B,C. This is exactly the condition belongs to both A,B or to both A,C, so A∩(B∪C)=(A∩B)∪(A∩C). The dual law is A∪(B∩C)=(A∪B)∩(A∪C). A correct concrete example verifies that example; the membership reasoning establishes the general identity.
With one fixed U, (A∪B)ᶜ=Aᶜ∩Bᶜ: outside either circle’s union means outside both circles. Also (A∩B)ᶜ=Aᶜ∪Bᶜ: not in both means missing from at least one set. These are De Morgan’s laws. The complement applies to the entire bracket. A membership table uses 1 for in and 0 for out, and checks all four A/B cases; those symbols are logical states, not probabilities.
In the original three-set model, U={1,...,8}, A={1,2,4,5}, B={2,3,5,6}, C={4,5,6,7}. The central element 5 belongs to all three sets. Thus A∩B={2,5}, not just the pair-only element 2. The element 8 is outside every circle but still inside U. Region areas are illustrative; they are not proportional to numbers of elements.
For U={1,...,6}, A={1,2,3},B={3,4,5}, (A∪B)ᶜ={6}. The same element is obtained from Aᶜ∩Bᶜ. Both complements use the same universe.
A∖B generally differs from B∖A. State which first set supplies the remaining elements.
The written membership condition and adjacent rosters explain each shaded region without requiring colour recognition.
Use A={1,2,3},B={3,4,5},U={1,...,6}.
| Operation | Membership rule | Result |
|---|---|---|
| A∪B | In at least one | {1,2,3,4,5} |
| A∩B | In both | {3} |
| A∖B | In A but not B | {1,2} |
| B∖A | In B but not A | {4,5} |
| (A∪B)ᶜ | In U but neither | {6} |
Shared 3 remains in the inclusive union. Exactly one of A,B would instead give {1,2,4,5}.
Here 1 means belongs and 0 means does not belong; these are logical states, not scores or probabilities.
| In A | In B | Not (A or B) | Not A and not B | Not (A and B) | Not A or not B |
|---|---|---|---|---|---|
| 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 0 | 0 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 | 0 | 0 |
The third/fourth columns agree in all four cases, as do the fifth/sixth. This checks every possible membership state for an arbitrary element of U.
For A={1,2,4},B={2,3},C={4,5}, keep A in both alternatives.
A∩(B∪C) = {2,4} (A∩B)∪(A∩C) = {2}∪{4} = {2,4}
The finite example illustrates the identity. The membership argument establishes it for arbitrary sets.
Keep one universe and the same six members through every stage. Shading indicates membership, not population area.
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Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original membership model, not copied source artwork. Circle areas and distances do not encode member counts. Read the labels and element rosters.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original membership model, not copied source artwork. Circle areas and distances do not encode member counts. Read the labels and element rosters.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original membership model, not copied source artwork. Circle areas and distances do not encode member counts. Read the labels and element rosters.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original membership model, not copied source artwork. Circle areas and distances do not encode member counts. Read the labels and element rosters.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original membership model, not copied source artwork. Circle areas and distances do not encode member counts. Read the labels and element rosters.
Show your working on paper. Check your answer, then choose Next.
5
List {1,2,3,4,5}; shared 3 appears once.
1
The only common element is 3.
3
Three belongs to both sets.
0
No member occurs in both, so the intersection is empty.
0
The empty set contributes no shared member.
3
Union with empty leaves A unchanged.
3
Repeating A does not duplicate its members.
2
Both 2 and central 5 are shared.
5
Only 5 belongs to all three sets.
8
The seven values 1 through 7 occur in at least one circle; 8 is outside all.
6
The union is {1,2,3,4,5}; only 6 remains in U.
5
Remove the shared element 3 from U.
2
The result is {2,4}, matching (A∩B)∪(A∩C).
4
Both union groupings give {1,2,3,4}.
2
A-only elements are 1 and 2.
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The union admits a member in A, B or both. It is {1,2,3,4,5}, including the shared 3 once. Exclusive or would wrongly omit that shared member.
Explain the overlap, not just the combined list.
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They are {3}, {1,2} and {4,5} respectively. Intersection requires both; each difference uses only the named first set while excluding the second.
Order matters for difference, unlike intersection.
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For example {1,3} and {2,4} share no member, but their union leaves 5,6,7,8 outside both. Disjoint does not mean complementary.
Verify both the empty intersection and the missing universal elements.
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Nothing belongs to ∅, so no element can satisfy both conditions for intersection. For union, membership in A or in ∅ reduces to membership in A.
State the two logical conditions in ordinary words.
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A∪B and B∪A both give {1,2,3,4}; A∩B and B∩A both give {3}. The result depends on membership, not the order in which the sets are named.
The example illustrates the laws; explain the symmetric membership conditions too.
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(A∪B)∪C={1,2,3}∪{3,4}={1,2,3,4}. A∪(B∪C)={1,2}∪{2,3,4} gives the same result.
Compute intermediate sets and keep duplicates only once.
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B∪C={2,3,4,5}, so the left side is {2,4}. On the right A∩B={2} and A∩C={4}; their union is {2,4}.
Then explain why the belongs-to-A condition applies to both alternatives.
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If an element is in A, it belongs to both right-hand unions. If it is outside A, the right side requires it to be in both B and C, exactly the remaining condition on the left.
Cover membership in A and outside A; one numerical example alone is not a proof.
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The union is {1,2,3,4,5}, with complement {6}. Separately Aᶜ={4,5,6},Bᶜ={1,2,6}; their intersection is {6}.
Use the same U for every complement and shade outside both circles.
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A∩B={3}, so its complement is {1,2,4,5,6}. Aᶜ={4,5,6} and Bᶜ={1,2,6}; their union gives exactly those five elements.
Do not keep the same intersection sign after complementing the whole bracket.
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An element may be in A only, in B only or in neither; all three fail the in-both condition. Outside both is just the neither case. This distinction explains why the complement of an intersection is a union of complements.
Account for all four membership combinations, including in both.
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They are {2,5}, {4,5} and {5,6}. Central element 5 belongs to every pairwise intersection as well as the triple intersection.
Do not confuse a pairwise intersection with a pair-only region.
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A-only 1; B-only 3; C-only 7; AB-only 2; AC-only 4; BC-only 6; all three 5; outside all 8. Each element appears in exactly one of these eight regions.
Check every roster against the circle membership and the universe.
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Take A={1},B=∅,C={2}. The left side is empty but the right side is {2}. The correct distributed expression also intersects A with C.
Compute both sides and explain the missing condition.
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A finite example confirms the identity for those particular sets. A general proof must show the same membership condition for every element of an arbitrary universe, using all relevant cases. A Venn diagram can encode those cases if every region is accounted for.
Use a membership argument or all logical cases rather than claiming one example proves every instance.
Make a two-set operation chart and an eight-region three-set diagram.
Keep your work on paper or in a local file. This page does not upload it or automatically grade the project.
This original Kidemy practice follows the topic order of 2026 Class 8 Mathematics · English Version, chapter 7, printed pages 121–132. It does not replace the complete chapter or its practical teaching.
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