Mathematical logic · Teacher reference · answers included

← Senior 1 · short mathematical inquiries

Teacher reference: contains answers, conditions and proofs.

02 Mathematical logic · g10-02-01 · 5–10 min · TEACHER · ANSWERS

A counterexample to an implication

When does p⇒q fail?

Predict: Test p: x>c and q: x>0; predict the one truth-value combination that refutes p⇒q.

Compare	Computed observation
Both true Test value x = 2; Threshold c = 1	This input is not a counterexample to p⇒q.
A refuting witness Test value x = -1; Threshold c = -2	This input refutes p⇒q: true premise, false conclusion.
False premise Test value x = -2; Threshold c = 1	This input is not a counterexample to p⇒q.

Conclusion: A counterexample needs p true and q false. A case with p false does not refute the implication. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: x and c are real numbers. A displayed test at one x is distinct from the claim for every x.

Justification: By the definition of implication, only true⇒false fails. For example c<0 permits a point c<x≤0 as a refutation.

Ask: Ask whether each example refutes, supports locally, or leaves a universal claim unproved.

Watch for: Finding q true does not establish p; the converse is a separate statement.

Extension: Choose c for which the forward implication always holds but the converse fails.

Exit: Does x=−1 refute x>−2 ⇒ x>0?

Yes: −1>−2 but −1 is not positive.

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

02 Mathematical logic · g10-02-02 · 5–10 min · TEACHER · ANSWERS

Necessary or sufficient?

Which interval is contained in which?

Predict: For p:x>a and q:x>b, predict both implication directions when a<b, a=b and a>b.

Compare	Computed observation
Sufficient only Threshold in p = 3; Threshold in q = 1	Sufficient, not necessary
Equivalent Threshold in p = 2; Threshold in q = 2	Necessary and sufficient
Necessary only Threshold in p = -1; Threshold in q = 2	Necessary, not sufficient

Conclusion: Sufficiency is containment of the p-solution set in the q-solution set; necessity reverses this containment. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: Both inequalities are strict and share the real-number universe.

Justification: If a≥b, x>a implies x>b. If a<b, the midpoint (a+b)/2 satisfies p and not q. Thus the comparison completely classifies the directions.

Ask: Require a general interval argument and a witness for the failing direction.

Watch for: A single common solution proves neither implication universally.

Extension: Replace one strict inequality by a non-strict one and examine equal thresholds.

Exit: x>3 is what condition for x>1?

Sufficient but not necessary; x=2 refutes the converse.

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

02 Mathematical logic · g10-02-03 · 5–10 min · TEACHER · ANSWERS

All or at least one?

Can a statement be existentially true but universally false?

Predict: For x²≥k or x²>k, compare a finite integer domain with all real numbers.

Compare	Computed observation
Existence without universality Finite bound n = 2; Square threshold k = 1; Comparison = x² ≥ k; Domain = Integers −n to n	All: false; exists: true. Domain: finite integers.
Boundary at zero Finite bound n = 2; Square threshold k = 0; Comparison = x² > k; Domain = All real numbers	All: false; exists: true. Domain: R.
Finite domain exhausted Finite bound n = 2; Square threshold k = 5; Comparison = x² ≥ k; Domain = Integers −n to n	All: false; exists: false. Domain: finite integers.

Conclusion: All requires every allowed input; existence requires one. Enlarging the domain can change these conclusions differently. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: The finite domain is {−n,…,n}, n≥1. The alternative domain is R.

Justification: The minimum square is 0. On the finite domain the maximum is n²; on R squares are unbounded. These bounds decide all and existence exactly.

Ask: Ask pupils to distinguish a witness from a complete proof over R.

Watch for: Checking the plotted real inputs is not checking every real number.

Extension: Find k for which the finite domain has no witness but R has one.

Exit: Over R, compare ∀x:x²≥1 and ∃x:x²≥1.

The universal claim is false at 0; the existential claim is true at 1.

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

02 Mathematical logic · g10-02-04 · 5–10 min · TEACHER · ANSWERS

Negating a quantified statement

Does negating “all are greater” mean “all are smaller”?

Predict: Predict the negation of ∀x:x²>k and ∃x:x²>k on [−n,n].

Compare	Computed observation
Negate all Interval bound n = 2; Threshold k = 1; Original quantifier = For every x	Negation: ∃x: x² ≤ k; True
Equality is essential Interval bound n = 2; Threshold k = 0; Original quantifier = For every x	Negation: ∃x: x² ≤ k; True
Negate existence at the maximum Interval bound n = 2; Threshold k = 4; Original quantifier = For at least one x	Negation: ∀x: x² ≤ k; True

Conclusion: Negation switches all and exists and replaces > by ≤. Equality belongs to the negation. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: The domain is the entire real interval [−n,n], with n>0.

Justification: On this interval 0≤x²≤n², with both bounds attained. ∀x:x²>k holds iff k<0; ∃x:x²>k holds iff k<n². Their negations are the complementary conditions.

Ask: Use k=0 to expose the missing equality in an incorrect negation.

Watch for: Not all is not the same as none.

Extension: Negate an inequality with ≥ and justify the new strict endpoint.

Exit: Negate “every real x has x²>0”.

There exists a real x with x²≤0; x=0 is a witness.

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