Sets · Teacher reference · answers included

← Senior 1 · short mathematical inquiries

Teacher reference: contains answers, conditions and proofs.

01 Sets · g10-01-01 · 5–10 min · TEACHER · ANSWERS

An element or a subset?

Does changing 3 into {3} change the question?

Predict: Predict whether a number, its singleton and the empty set belong to A or are subsets of A.

Compare	Computed observation
A member Number n = 2; Object = n; Set A = Even numbers	The object 2 is an element; subset status is not applicable.
Its singleton Number n = 2; Object = {n}; Set A = Even numbers	The object {2} is not an element; subset status is true.
The empty set Number n = 3; Object = Empty set; Set A = Even numbers	The object ∅ is not an element; subset status is true.

Conclusion: A contains numbers. A number can be an element; a singleton is a subset exactly when its number belongs to A. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: The universe is {1,…,8}; A is the chosen numeric set. A subset question requires a set as its left operand.

Justification: {n} ⊆ A means every element of {n}, namely n, is in A. The empty set has no element violating inclusion.

Ask: Ask pupils to name the type of the object before choosing a symbol.

Watch for: Membership and inclusion are different relations; the empty set is not the number zero.

Extension: Construct a set whose elements are themselves sets, then reconsider membership.

Exit: For A={2,4}, compare 2∈A, {2}⊆A and {2}∈A.

True, true, false; A contains numbers, not singleton sets.

Before: Numbers and braces · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

01 Sets · g10-01-02 · 5–10 min · TEACHER · ANSWERS

Containment, equality and the empty set

When is a subset proper?

Predict: Compare A={1,…,a} and B={1,…,b}; predict the relation when a=b or a=0.

Compare	Computed observation
Proper inclusion Upper bound a = 3; Upper bound b = 5	A⊆B: true; A⊊B: true.
Equal sets Upper bound a = 4; Upper bound b = 4	A⊆B: true; A⊊B: false.
Both empty Upper bound a = 0; Upper bound b = 0	A⊆B: true; A⊊B: false.

Conclusion: A⊆B allows equality. A proper subset additionally requires an element of B outside A. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: a and b are integers from 0 to 8; an upper bound of zero denotes the empty set.

Justification: For these initial segments, a≤b gives inclusion; a>b leaves a in A outside B. Equality holds iff a=b.

Ask: Request an explicit witness whenever inclusion fails.

Watch for: An empty set is a subset of itself but not a proper subset of itself.

Extension: Describe all pairs (a,b) giving A⊊B without listing every set.

Exit: Is ∅ a proper subset of {1}? Is ∅ a proper subset of ∅?

Yes; no. The second pair is equal.

Before: Set notation · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

01 Sets · g10-01-03 · 5–10 min · TEACHER · ANSWERS

Operations select different elements

Can union, intersection and difference use the same rule?

Predict: Choose A and B, predict which chips remain, then change only the operation.

Compare	Computed observation
Intersection A ends at a = 5; B starts at b = 3; Operation = Intersection	Result = {3, 4, 5}
Difference A ends at a = 5; B starts at b = 3; Operation = A minus B	Result = {1, 2}
Empty A, complement A ends at a = 0; B starts at b = 9; Operation = Complement of A	Result = {1, 2, 3, 4, 5, 6, 7, 8}

Conclusion: The decision for each element is respectively OR, AND, A-and-not-B, or not-A in U. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: U={1,…,8}; A={1,…,a}; B={b,…,8}. b=9 means B is empty.

Justification: For any x∈U, membership of the result is determined by its two truth values x∈A and x∈B. The displayed four regions partition U.

Ask: Ask for one element distinguishing two operations.

Watch for: A−B need not equal B−A; a complement depends on its universe.

Extension: Use element-by-element decisions to explain a De Morgan identity.

Exit: If A={1,2}, B={2,3}, find A−B and A∩B.

A−B={1}; A∩B={2}.

Before: Membership · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

01 Sets · g10-01-04 · 5–10 min · TEACHER · ANSWERS

Can these class counts coexist?

How many pupils choose neither option?

Predict: Predict four disjoint category counts before changing the overlap.

Compare	Computed observation
A feasible survey Class total N = 40; Choose A = 24; Choose B = 21; Choose both = 10	Feasible counts sum to 40.
No one chooses neither Class total N = 30; Choose A = 20; Choose B = 18; Choose both = 8	Feasible counts sum to 30.
Impossible margins Class total N = 30; Choose A = 20; Choose B = 18; Choose both = 5	Impossible counts: a category is negative.

Conclusion: The overlap is counted twice in a+b, so subtract it once. Invalid negative categories signal inconsistent data. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: Counts are integers; 0≤overlap≤min(a,b), and a+b−overlap≤N.

Justification: The categories are a−o, b−o, o and N−a−b+o. Nonnegativity gives max(0,a+b−N)≤o≤min(a,b).

Ask: Keep an impossible case on screen and ask which assumption it violates.

Watch for: An overlap cannot be chosen independently of the class total and margins.

Extension: Derive the smallest and largest possible overlap from N,a,b.

Exit: N=30,a=20,b=18: can the overlap be 5?

No; the union would be 33, exceeding 30. The overlap must be at least 8.

Before: Set operations and integer counts · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.