McIntyre MathsMathematics Extension 2
Extension 2 › Proof › Lesson 2

Lesson 2 · Quantifiers, converse and contrapositive

Mathematics Extension 2 · Proof · Outcome ME2-12-01 · Term 4 · Week 1
Learning intentionTo use quantifiers, identify sufficient and necessary conditions, and form the converse and contrapositive of an implication.
Success criteria · I can…
  • write and negate statements using ∀\forall and ∃\exists
  • identify sufficient and necessary conditions in P⇒QP \Rightarrow Q
  • write the converse, decide if it is true, and write an iff statement when both directions hold
  • write the contrapositive and explain why P⇒Q≃¬Q⇒¬PP \Rightarrow Q \simeq \neg Q \Rightarrow \neg P

Lesson flow

LaunchWarm-up · Negations from Lesson 1Slide 2RecapLesson 1 recap · Working with symbolsSlide 3HSC practiceHSC practice · Negating an implicationSlide 4DefinitionsQuantifiersSlide 5DefinitionsNegating quantifiersSlide 6ExamplesExample 3 · Quantified statementsSlide 7HSC practiceHSC practice · Negating “there exists”Slide 8HSC practiceHSC practice · The order of quantifiersSlide 9DefinitionsSufficient and necessarySlide 10DefinitionsConverse and equivalenceSlide 11ExamplesExample 6 · ConverseSlide 12HSC practiceHSC practice · The converseSlide 13DefinitionsThe contrapositiveSlide 14ExamplesExample 8 · ContrapositiveSlide 15HSC practiceHSC practice · The contrapositiveSlide 16HSC practiceHSC practice · Which cards?Slide 17Exam practiceExam practice · Putting it togetherSlide 18ThinkSpot the flaw · ContrapositiveSlide 19Exit ticketExit ticketSlide 20

Warm-up · Negations from Lesson 1

Slide 2

Write the negation of each statement.

(a) x>3x > 3 and y≤1y \le 1
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Answer
x≤3x \le 3 or y>1y > 1.
Why
Negating “and” gives “or”, and each part is negated: ¬(P∧Q)≃¬P∨¬Q\neg(P\wedge Q)\simeq\neg P\vee\neg Q.
(b) If it rains, then the game is cancelled.
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Answer
It rains and the game is not cancelled.
Why
¬(P⇒Q)≃P∧¬Q\neg(P\Rightarrow Q)\simeq P\wedge\neg Q. An implication only fails when PP is true and QQ is false.
(c) nn is prime or nn is even.
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Answer
nn is not prime and nn is odd.
Why
Negating “or” gives “and”: both parts must fail. (For an integer, not even means odd.)
(d) True or false: “If 2+2=52 + 2 = 5, then 7 is even.”
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Answer
True.
Why
The hypothesis is false, so the implication is vacuously true. Indeed P⇒Q≃¬P∨QP\Rightarrow Q\simeq\neg P\vee Q, and ¬P\neg P is true.

Lesson 1 recap · Working with symbols

Slide 3
(a) Simplify ¬(P∨¬Q)\neg(P \vee \neg Q).
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Answer
¬P∧Q\neg P \wedge Q
Why
¬(P∨¬Q)≃¬P∧¬(¬Q)≃¬P∧Q\neg(P \vee \neg Q) \simeq \neg P \wedge \neg(\neg Q) \simeq \neg P \wedge Q, because a double negation cancels.
(b) Show that ¬(¬P∨Q)≃P∧¬Q\neg(\neg P \vee Q) \simeq P \wedge \neg Q.
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Answer
¬(¬P∨Q)≃¬(¬P)∧¬Q≃P∧¬Q\neg(\neg P \vee Q) \simeq \neg(\neg P) \wedge \neg Q \simeq P \wedge \neg Q
Why
This is ¬(P⇒Q)\neg(P \Rightarrow Q), as expected, since ¬P∨Q≃P⇒Q\neg P \vee Q \simeq P \Rightarrow Q. It confirms the rule for negating “if…then”.
(c) “If this is an owl, then it is a bird.” Write it using “or”, then write its negation using “and”.
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Answer
“This is not an owl or it is a bird.” Negation: “This is an owl and it is not a bird.”
Why
Use P⇒Q≃¬P∨QP \Rightarrow Q \simeq \neg P \vee Q, then ¬(P⇒Q)≃P∧¬Q\neg(P \Rightarrow Q) \simeq P \wedge \neg Q.

HSC practice · Negating an implication

Slide 4
Adapted from 2020 HSC Q8Consider: “If nn is even and a multiple of 3, then nn is a multiple of 6.” Which is its negation?
A nn is not even or not a multiple of 3, and nn is a multiple of 6.B nn is even and a multiple of 3, and nn is not a multiple of 6.C If nn is not a multiple of 6, then nn is not even or not a multiple of 3.D If nn is not even and not a multiple of 3, then nn is not a multiple of 6.
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Answer
B
Why
¬(P⇒Q)≃P∧¬Q\neg(P \Rightarrow Q) \simeq P \wedge \neg Q: keep the hypothesis, deny the conclusion. C is the contrapositive, which is equivalent to the original statement, not its negation. D is the inverse. A negates the wrong part. (Wording adapted from the paper.)

Quantifiers

Slide 5
Definition
Quantifiers ∀\forall and ∃\exists

∀\forall means for all (every). ∃\exists means there exists (at least one). ∈\in means “is an element of”.

Number sets: N\mathbb{N} natural, Z\mathbb{Z} integers, Q\mathbb{Q} rational, R\mathbb{R} real, C\mathbb{C} complex.

∀x∈E, x∈F\forall x \in E,\ x \in F says “if x∈Ex \in E then x∈Fx \in F”. Use clear English in answers when symbols get in the way.

Example 1

Write in English, then simplify: ∀m∈Z, ∃k∈Z\forall m \in \mathbb{Z},\ \exists k \in \mathbb{Z} such that (2m+1)2=4k+1(2m + 1)^2 = 4k + 1.
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Answer
For all integers mm, there exists an integer kk such that (2m+1)2=4k+1(2m+1)^2 = 4k+1. In English: the square of any odd number is one more than a multiple of 4.
Why
Every odd number has the form 2m+12m+1. Preview of Lesson 3: (2m+1)2=4m2+4m+1=4m(m+1)+1(2m+1)^2 = 4m^2 + 4m + 1 = 4m(m+1) + 1, so we can take k=m(m+1)k = m(m+1).

Negating quantifiers

Slide 6
Key idea
Swap the quantifier, negate the rest

The negation of “all AA are BB” is “there exists an AA that is not BB”.

The negation of “there exists an AA that is BB” is “all AA are not BB” (no AA is BB).

To disprove a “for all” statement: one counterexample. To prove a “there exists” statement: one example.

Example 2

Write in symbols: “Every real number has a square that is not negative.” Then write its negation.
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Answer
∀x∈R, x2≥0\forall x\in\mathbb{R},\ x^2\ge0. Negation: ∃x∈R\exists x\in\mathbb{R} such that x2<0x^2<0.
Why
This is the same idea as the “all cats are fluffy” example in Lesson 1. The negation is false, because the original statement is true for every real number.

Example 3 · Quantified statements

Slide 7

Write each in words. True or false? Then write its negation in symbols.

(a) ∀n∈N, n2≥n\forall n\in\mathbb{N},\ n^2\ge n
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Answer
Every natural number is at most its square. True. Negation: ∃n∈N\exists n\in\mathbb{N} such that n2<nn^2<n.
Why
For n≥1n\ge1, n2−n=n(n−1)≥0n^2-n=n(n-1)\ge0.
(b) ∃x∈R\exists x\in\mathbb{R} such that x2=2x^2=2
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Answer
Some real number squares to 2. True. Negation: ∀x∈R, x2≠2\forall x\in\mathbb{R},\ x^2\ne2.
Why
x=2x=\sqrt2 is an example, so the “there exists” statement is true.
(c) ∀x∈R, x2>0\forall x\in\mathbb{R},\ x^2>0
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Answer
Every real number has a positive square. False. Negation: ∃x∈R\exists x\in\mathbb{R} such that x2≤0x^2\le0.
Why
x=0x=0 is a counterexample, since 02=00^2=0 is not greater than 0.
(d) ∃n∈Z\exists n\in\mathbb{Z} such that n2=5n^2=5
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Answer
Some integer squares to 5. False. Negation: ∀n∈Z, n2≠5\forall n\in\mathbb{Z},\ n^2\ne5.
Why
The only real solutions of n2=5n^2=5 are ±5\pm\sqrt5, which are not integers.

HSC practice · Negating “there exists”

Slide 8
Adapted from 2022 HSC Q3Points AA, BB, PP are given. Which is the negation of “There exists a real λ\lambda such that AP→=λAB→\vec{AP} = \lambda\vec{AB}”?
A There exists a real λ\lambda such that AP→≠λAB→\vec{AP} \ne \lambda\vec{AB}.B For all real λ\lambda, AP→≠λAB→\vec{AP} \ne \lambda\vec{AB}.C For all real λ\lambda, AP→=λAB→\vec{AP} = \lambda\vec{AB}.D There is no real λ\lambda such that AP→≠λAB→\vec{AP} \ne \lambda\vec{AB}.
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Answer
B
Why
Negating “there exists” gives “for all”, and == becomes ≠\ne. Geometrically, PP is not on the line ABAB. D says the same as C (“for all λ\lambda, AP→=λAB→\vec{AP}=\lambda\vec{AB}”), which is not the negation. A keeps the quantifier unchanged. (Wording adapted from the paper.)

HSC practice · The order of quantifiers

Slide 9
Adapted from 2023 HSC Q4Which statement says that ln⁡xx3\dfrac{\ln x}{x^3} can be made smaller than any positive number?
A ∃x>1\exists x > 1 such that ∀r>0\forall r > 0, ln⁡xx3<r\dfrac{\ln x}{x^3} < rB ∀x>1\forall x > 1, ∃r>0\exists r > 0 such that ln⁡xx3<r\dfrac{\ln x}{x^3} < rC ∀r>0\forall r > 0, ∃x>1\exists x > 1 such that ln⁡xx3<r\dfrac{\ln x}{x^3} < rD ∃r>0\exists r > 0 such that ∀x>1\forall x > 1, ln⁡xx3<r\dfrac{\ln x}{x^3} < r
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Answer
C
Why
Whatever positive rr is chosen first, a suitable xx can be found. A is false: it would need ln⁡xx3≤0\frac{\ln x}{x^3} \le 0 for some x>1x>1, but ln⁡xx3>0\frac{\ln x}{x^3}>0 there. B and D say something much weaker (that the values are bounded above). The order of quantifiers matters: “∀r ∃x\forall r\ \exists x” is not the same as “∃x ∀r\exists x\ \forall r”. (Wording adapted from the paper.)

Sufficient and necessary

Slide 10
Definition
In a proven implication P⇒QP \Rightarrow Q

PP is a sufficient condition for QQ: knowing PP is enough to guarantee QQ.

QQ is a necessary condition for PP: QQ must hold whenever PP does, but QQ alone does not guarantee PP.

Example 4

(a) On a die: “If the number is prime, then it is a Fibonacci number.” Is it true? Name the sufficient and necessary conditions.
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Answer
True. Sufficient: the number is prime. Necessary: the number is a Fibonacci number.
Why
The primes on a die are 2, 3, 5 and the Fibonacci numbers on a die are 1, 2, 3, 5, so every prime is Fibonacci. QQ does not guarantee PP: 1 is Fibonacci but not prime.
(b) Explain why m>5m > 5 is sufficient for m>3m > 3, and why m>3m > 3 is not sufficient for m>5m > 5.
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Answer
Any number greater than 5 is greater than 3, so m>5m>5 is sufficient for m>3m>3. But m=4m=4 satisfies m>3m>3 and not m>5m>5.
Why
So m>3m>3 is not sufficient for m>5m>5 (equivalently, m>5m>5 is not necessary for m>3m>3).

Converse and equivalence

Slide 11
Definition
Converse and “if and only if”

The converse of P⇒QP\Rightarrow Q is Q⇒PQ\Rightarrow P. The converse of a true implication may or may not be true.

Equivalence P⇔QP\Leftrightarrow Q (or P↔QP\leftrightarrow Q): both P⇒QP\Rightarrow Q and Q⇒PQ\Rightarrow P are true. Read “PP if and only if QQ”, written “PP iff QQ”.

Example 5

(a) Write the converse of “If x=3x=3, then x2=9x^2=9.” Is it true?
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Answer
Converse: if x2=9x^2=9, then x=3x=3. It is false.
Why
x=−3x=-3 gives x2=9x^2=9 but x≠3x\ne3. One counterexample shows the converse is false.
(b) “A triangle is isosceles iff its base angles are equal.” Write the two implications.
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Answer
If a triangle is isosceles, then its base angles are equal. If the base angles are equal, then the triangle is isosceles.
Why
Both are true, so the equivalence ⇔\Leftrightarrow holds. To prove an “iff” statement you must prove both directions.

Example 6 · Converse

Slide 12

Write the converse. If it is true, write an iff statement. If not, give a counterexample.

(a) If a triangle has two equal sides, then it has two equal angles.
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Answer
Converse: if a triangle has two equal angles, then it has two equal sides. True, so a triangle has two equal sides if and only if it has two equal angles.
Why
Both directions are true (the isosceles triangle theorem and its converse).
(b) If Fred murdered Alf, then Alf is dead.
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Answer
Converse: if Alf is dead, then Fred murdered him. False.
Why
Alf may have died in some other way, so being dead does not guarantee that Fred murdered him.
(c) ABCDABCD is a square ⇒\Rightarrow all angles of ABCDABCD are 90∘90^\circ.
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Answer
Converse: if all angles of ABCDABCD are 90∘90^\circ, then ABCDABCD is a square. False.
Why
A rectangle that is not a square is a counterexample.
(d) For x∈Rx\in\mathbb{R}: x>2⇒x2>4x>2\Rightarrow x^2>4.
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Answer
Converse: if x2>4x^2>4, then x>2x>2. False.
Why
Take x=−3x=-3: x2=9>4x^2=9>4 but x≯2x\not>2.

HSC practice · The converse

Slide 13
Adapted from 2024 HSC Q3Which is the converse of “If a polygon is a square, then it is a rectangle”?
A If a polygon is not a square, then it is not a rectangle.B If a polygon is not a rectangle, then it is not a square.C If a polygon is a rectangle, then it is a square.D A polygon is a square and it is not a rectangle.
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Answer
C
Why
The converse swaps the two parts. A is the inverse, B is the contrapositive and D is the negation. (Wording adapted from the paper.)
Adapted from 2023 HSC Q2PP: an animal is a herbivore. QQ: the animal eats meat. The statement “If an animal is a herbivore, then it does not eat meat” is P⇒¬QP \Rightarrow \neg Q. Which is its converse?
A ¬P⇒Q\neg P \Rightarrow QB Q⇒¬PQ \Rightarrow \neg PC P⇒QP \Rightarrow QD ¬Q⇒P\neg Q \Rightarrow P
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Answer
D
Why
The converse of P⇒¬QP \Rightarrow \neg Q swaps the two parts: ¬Q⇒P\neg Q \Rightarrow P (“if an animal does not eat meat, then it is a herbivore”). B is the contrapositive. A is the inverse. C is neither. (Wording adapted from the paper.)

The contrapositive

Slide 14
Definition
Contrapositive ¬Q⇒¬P\neg Q\Rightarrow\neg P

The contrapositive of “if PP then QQ” is “if not QQ then not PP”. An implication and its contrapositive are logically equivalent:

P⇒Q  ≃  ¬Q⇒¬PP\Rightarrow Q\;\simeq\;\neg Q\Rightarrow\neg P

This is the basis of proof by contrapositive (Lesson 4).

StatementConverseInverse (not syllabus)Contrapositive
P⇒QP\Rightarrow QQ⇒PQ\Rightarrow P¬P⇒¬Q\neg P\Rightarrow\neg Q¬Q⇒¬P\neg Q\Rightarrow\neg P

Diagonal pairs are logically equivalent: the statement and its contrapositive, and the converse and the inverse. The inverse is not a syllabus term; it is shown so that you can recognise it as a distractor.

Example 7

PP: It is raining. QQ: I carry an umbrella. Write P⇒QP\Rightarrow Q, its converse and its contrapositive in words.
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Answer
Statement: if it is raining, then I carry an umbrella. Converse: if I carry an umbrella, then it is raining. Contrapositive: if I do not carry an umbrella, then it is not raining.
Why
The statement and its contrapositive are equivalent. The converse is not equivalent to the statement: I may carry an umbrella on a sunny day.

Example 8 · Contrapositive

Slide 15

Write the contrapositive of each statement. Is the statement true?

(a) If n2n^2 is even, then nn is even.
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Answer
If nn is odd, then n2n^2 is odd. True.
Why
An odd number is 2k+12k+1, and (2k+1)2=2(2k2+2k)+1(2k+1)^2=2(2k^2+2k)+1 is odd. Because the contrapositive is true, so is the original.
(b) If x>5x>5, then x2>25x^2>25.
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Answer
If x2≤25x^2\le 25, then x≤5x\le 5. True.
Why
If x2≤25x^2\le25 then −5≤x≤5-5\le x\le5, so x≤5x\le5.
(c) If a quadrilateral is a rectangle, then its diagonals are equal.
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Answer
If the diagonals are not equal, then it is not a rectangle. True.
Why
The diagonals of a rectangle are equal, so the contrapositive follows.
(d) If ab=0ab=0, then a=0a=0 or b=0b=0.
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Answer
If a≠0a\ne0 and b≠0b\ne0, then ab≠0ab\ne0. True.
Why
Negating “or” gives “and”. A product of two non-zero real numbers is non-zero.

HSC practice · The contrapositive

Slide 16
Adapted from 2021 HSC Q4Which is the contrapositive of “For all integers nn, if nn is a multiple of 6, then nn is a multiple of 2”?
A There is an integer nn that is a multiple of 6 but not a multiple of 2.B For all integers nn, if nn is a multiple of 2, then nn is a multiple of 6.C For all integers nn, if nn is not a multiple of 2, then nn is not a multiple of 6.D For all integers nn, if nn is not a multiple of 6, then nn is not a multiple of 2.
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Answer
C
Why
The contrapositive of P⇒QP\Rightarrow Q is ¬Q⇒¬P\neg Q\Rightarrow\neg P, and the “for all integers nn” stays in front. A is the negation, B is the converse and D is the inverse. (Wording adapted from the paper.)

HSC practice · Which cards?

Slide 17
Adapted from 2021 HSC Q9Each card has a colour on one side and a word on the other. The rule is: “If a card has RED on one side, then it has WIN on the other.” The cards show: RED, BLUE, WIN, TRY AGAIN. Which cards must be turned over to check the rule? Explain using the contrapositive.
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Answer
RED and TRY AGAIN.
Why
RED must be turned over to check P⇒QP \Rightarrow Q directly: its back must say WIN. TRY AGAIN (not WIN) must be turned over to check the contrapositive ¬Q⇒¬P\neg Q \Rightarrow \neg P (not WIN ⇒\Rightarrow not RED): its back must not be RED. BLUE and WIN cannot break the rule, because WIN with BLUE behind it is allowed. A common error is turning WIN, which tests the converse. (Card labels adapted from the paper.)

Exam practice · Putting it together

Slide 18

Consider the statement: “For all integers nn, if n2n^2 is divisible by 4, then nn is even.”

Exam practice(i) 1 markWrite the statement using quantifier and implication notation.
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Answer
∀n∈Z, (n2 is divisible by 4)⇒(n is even)\forall n\in\mathbb{Z},\ (n^2\text{ is divisible by }4)\Rightarrow(n\text{ is even})
Why
Quantifier first, then the implication.
(ii) 2 marksWrite the converse and prove that the converse is true.
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Answer
Converse: for all integers nn, if nn is even then n2n^2 is divisible by 4. Proof: let n=2kn=2k with k∈Zk\in\mathbb{Z}. Then n2=4k2n^2=4k^2, which is divisible by 4.
Why
To prove a “for all” implication, assume the hypothesis for an arbitrary integer and show the conclusion.
(iii) 1 markWrite the negation of the original statement.
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Answer
There exists an integer nn such that n2n^2 is divisible by 4 and nn is odd.
Why
The negation of ∀\forall is ∃\exists, and ¬(P⇒Q)≃P∧¬Q\neg(P\Rightarrow Q)\simeq P\wedge\neg Q. The original statement is true (it is proved by contrapositive in Lesson 4), so its negation is false.

Spot the flaw · Contrapositive

Slide 19

A student wrote this. Find the flaw, explain why it matters, then write a correct version.

Claim

For real xx, if x2>4x^2 > 4, then x>2x > 2.

  1. The contrapositive is: if x2≤4x^2 \le 4, then x≤2x \le 2.
  2. If x2≤4x^2 \le 4, then −2≤x≤2-2 \le x \le 2, so x≤2x \le 2.
  3. The contrapositive is true, so the claim is true. □\square
What is the flaw, and what is a correct version?
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Answer
Line 1 is not the contrapositive. “If x2≤4x^2\le4 then x≤2x\le2” is ¬P⇒¬Q\neg P\Rightarrow\neg Q (the inverse), which is equivalent to the converse, not to the claim. The claim is false: x=−3x=-3 gives x2=9>4x^2=9>4 but x<2x<2.
Why
The contrapositive of P⇒QP\Rightarrow Q is ¬Q⇒¬P\neg Q\Rightarrow\neg P, so here it is: if x≤2x\le2, then x2≤4x^2\le4. This is false (x=−3x=-3), as it must be, since it is equivalent to the claim. Correct version: disprove the claim with the counterexample x=−3x=-3. (The true statement is the converse: if x>2x>2, then x2>4x^2>4.) Line 2 is correct reasoning about the wrong statement, so always write PP and QQ out before forming the contrapositive.

Exit ticket

Slide 20

For n∈Zn\in\mathbb{Z}, let PP: nn is divisible by 6 and QQ: nn is even.

(a) Write P⇒QP\Rightarrow Q, its converse and its contrapositive in words. Which are true?
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Answer
If nn is divisible by 6 then nn is even: true. Converse: if nn is even then nn is divisible by 6: false. Contrapositive: if nn is odd then nn is not divisible by 6: true.
Why
The converse fails for n=2n=2. The contrapositive has the same truth value as the original statement.
(b) Write the negation of P⇒QP\Rightarrow Q. Is the negation true or false? Is PP sufficient for QQ? Is PP necessary for QQ?
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Answer
Negation: nn is divisible by 6 and nn is odd. It is false. PP is sufficient for QQ, but PP is not necessary for QQ.
Why
The negation of a true statement is false. P⇒QP\Rightarrow Q is true, so PP is sufficient for QQ. PP is not necessary for QQ because n=2n=2 is even without being divisible by 6.