The Language of Proof
Mathematics is a language of precise statements — and proof is how those statements are verified. Before you can prove anything, you need to read implications correctly, distinguish a statement from its converse, and recognise when a quantifier ("for all", "there exists") changes the meaning entirely. This lesson builds the vocabulary every Extension 2 proof depends on.
Write the statement "If $x > 3$, then $x^2 > 9$" in symbolic form using $\Rightarrow$. Before checking — is the converse true? Find a counterexample if not. Sketch your reasoning below.
Every mathematical statement rewards two habits: identify the hypothesis and conclusion (what is being assumed, what is being concluded), then check the direction (one-way implication or two-way biconditional?). Mixing these up is the single biggest cause of error in proof questions.
The hypothesis-conclusion-direction reading: (1) box the hypothesis $P$, (2) box the conclusion $Q$, (3) decide whether the connective is $\Rightarrow$ (one-way) or $\Leftrightarrow$ (two-way).
Implication: $P \Rightarrow Q$ · Converse: $Q \Rightarrow P$ · Contrapositive: $\neg Q \Rightarrow \neg P$
Key facts
- $P \Rightarrow Q$ means "if $P$ then $Q$"; $P$ is the hypothesis, $Q$ the conclusion
- Converse of $P \Rightarrow Q$ is $Q \Rightarrow P$ (not equivalent)
- Contrapositive of $P \Rightarrow Q$ is $\neg Q \Rightarrow \neg P$ (logically equivalent)
- $\forall$ = for all; $\exists$ = there exists
Concepts
- Why a single counterexample disproves a universal statement
- Why proving the contrapositive proves the original statement
- Why $P \Leftrightarrow Q$ requires proving both $P \Rightarrow Q$ and $Q \Rightarrow P$
Skills
- Write the converse, contrapositive and negation of any implication
- Construct counterexamples to disprove false statements
- Translate quantified statements between English and symbolic form
Given any statement of the form "If $P$, then $Q$" (symbolically $P \Rightarrow Q$), three related statements arise. Knowing which are equivalent to the original is critical for proof strategy.
- Converse $Q \Rightarrow P$ — swap hypothesis and conclusion. Not equivalent in general.
- Contrapositive $\neg Q \Rightarrow \neg P$ — negate both and swap. Always equivalent.
- Inverse $\neg P \Rightarrow \neg Q$ — negate both without swapping. Equivalent to the converse, not the original.
Worked through the hook: Original: "If $n$ is even, then $n^2$ is even." ($P$: $n$ even; $Q$: $n^2$ even.)
- Converse: "If $n^2$ is even, then $n$ is even." (Also true — but requires its own proof.)
- Contrapositive: "If $n^2$ is odd, then $n$ is odd." (Automatically true, by equivalence with the original.)
- Because both original and converse hold, this is a biconditional: $n$ is even $\Leftrightarrow$ $n^2$ is even.
Three transformations: converse (swap), contrapositive (negate + swap), inverse (negate) · Original $\equiv$ contrapositive (always) · Original $\not\equiv$ converse in general; if both hold, the statement is a biconditional · Disproving a "for all" requires only one counterexample
Pause — copy the three transformations (converse, contrapositive, inverse), the equivalence $P \Rightarrow Q \equiv \neg Q \Rightarrow \neg P$, and the biconditional condition into your book.
Quick check: The statement "If $x$ is a multiple of 4, then $x$ is even" is true. Which of the following is also automatically true?
We just saw that $P \Rightarrow Q$ is always logically equivalent to its contrapositive $\neg Q \Rightarrow \neg P$, while the converse $Q \Rightarrow P$ is independent. That raises a question: how do we handle statements that quantify over a whole set — "for all" or "there exists"? This card answers it → the symbols $\forall$ and $\exists$, their negation rules, and why quantifier order matters.
A quantifier specifies how many elements of a set satisfy a property. Two suffice for HSC Extension 2:
- $\forall x \in S,\; P(x)$ — "for every $x$ in $S$, $P(x)$ is true".
- $\exists x \in S,\; P(x)$ — "there exists at least one $x$ in $S$ for which $P(x)$ is true".
Negation rules (essential for contradiction proofs):
- $\neg(\forall x,\; P(x)) \;\equiv\; \exists x,\; \neg P(x)$ — to disprove a "for all", find one $x$ where $P$ fails.
- $\neg(\exists x,\; P(x)) \;\equiv\; \forall x,\; \neg P(x)$ — to disprove a "there exists", show $P$ fails for every $x$.
$\forall$ = "for all"; $\exists$ = "there exists" · One counterexample disproves $\forall x, P(x)$ · One example proves $\exists x, P(x)$ · Negation flips the quantifier: $\neg \forall \to \exists \neg$ and $\neg \exists \to \forall \neg$ · Order matters: $\forall x \exists y \neq \exists y \forall x$ in general
Pause — copy the negation rules $\neg\forall \to \exists\neg$ and $\neg\exists \to \forall\neg$, and the quantifier-order warning $\forall x\,\exists y \neq \exists y\,\forall x$, into your book.
Did you get this? True or false: the negation of "$\forall n \in \mathbb{N},\; n^2 \geq n$" is "$\exists n \in \mathbb{N},\; n^2 < n$".
Worked examples · 3 in a row, reveal as you go
Write the contrapositive of: "If $n$ is a prime greater than 2, then $n$ is odd." Then explain why proving the contrapositive proves the original.
Consider "If $x > 2$, then $x^2 > 4$." Write the converse and decide whether it is true. If false, give a counterexample.
Compare these two statements about $\mathbb{R}$: (i) $\forall x \,\exists y\; (y > x)$, and (ii) $\exists y \,\forall x\; (y > x)$. Decide which are true.
Fill the gap: The contrapositive of "$P \Rightarrow Q$" is " $\Rightarrow$ ", and it is always logically equivalent to the original statement.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the negation of "If $n$ is even, then $n^2$ is even" is "If $n$ is odd, then $n^2$ is odd".
Activities · practice with the ideas
Write the converse and contrapositive of: "If a quadrilateral is a square, then it is a rectangle." State whether the converse is true.
Disprove: "$\forall n \in \mathbb{N},\; n^2 + n + 41$ is prime." Find a counterexample.
Translate to symbolic form using quantifiers: "Every positive integer has a successor." Then write the negation.
A biconditional: "$n$ is divisible by 6 $\Leftrightarrow$ $n$ is divisible by both 2 and 3." Outline the two implications you would need to prove.
Write the negation of: "$\exists x \in \mathbb{R},\; \forall y \in \mathbb{R},\; xy = y$." Decide whether the original is true.
Odd one out: Three of these are logically equivalent to "$P \Rightarrow Q$". Which one is NOT?
Earlier you analysed "If $n$ is even, then $n^2$ is even" — writing its converse and contrapositive and judging truth.
The contrapositive ("If $n^2$ is odd, then $n$ is odd") is automatically true by equivalence. The converse ("If $n^2$ is even, then $n$ is even") also happens to be true — but it needed its own argument. Because both directions hold, the connection is a biconditional: $n$ is even $\Leftrightarrow$ $n^2$ is even. Recognising when a statement is a biconditional (and therefore needs two proofs) is the single most useful skill in Module 11.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Write the contrapositive of: "If $n^2$ is even, then $n$ is even." Then state whether the original statement is true. (2 marks)
Q2. Show that the statement "$\forall n \in \mathbb{N},\; n^2 - n + 11$ is prime" is false by finding a counterexample. (3 marks)
Q3. Decide whether each of the following is true. If true, give a brief reason; if false, give a counterexample. (a) $\forall x \in \mathbb{R},\; x^2 \geq 0$. (b) $\exists x \in \mathbb{R},\; x^2 < 0$. (c) $\forall x \in \mathbb{R},\; x > 0 \Rightarrow x^2 > x$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. Converse: "If a quadrilateral is a rectangle, then it is a square." False (a $2 \times 3$ rectangle is not a square). Contrapositive: "If a quadrilateral is not a rectangle, then it is not a square." True (equivalent to original).
2. The formula $n^2 + n + 41$ fails first at $n = 40$: $40^2 + 40 + 41 = 1600 + 40 + 41 = 1681 = 41^2$, not prime.
3. Symbolic: $\forall n \in \mathbb{Z}^+,\; \exists m \in \mathbb{Z}^+,\; m = n + 1$. Negation: $\exists n \in \mathbb{Z}^+,\; \forall m \in \mathbb{Z}^+,\; m \neq n + 1$ — i.e., some positive integer has no successor (false).
4. ($\Rightarrow$) Assume $6 \mid n$. Then $n = 6k$ so $2 \mid n$ and $3 \mid n$. ($\Leftarrow$) Assume $2 \mid n$ and $3 \mid n$. Since $\gcd(2,3) = 1$, $6 = 2 \times 3 \mid n$.
5. Negation: $\forall x \in \mathbb{R},\; \exists y \in \mathbb{R},\; xy \neq y$. Original is true: take $x = 1$, then $1 \cdot y = y$ for every $y$.
Q1 (2 marks): Contrapositive: "If $n$ is odd, then $n^2$ is odd" [1]. Original is true; proof via contrapositive: $n = 2k+1 \Rightarrow n^2 = 4k^2 + 4k + 1 = 2(2k^2+2k)+1$, odd [1].
Q2 (3 marks): Statement is "for all $n$"; one counterexample disproves [1]. Test: $n = 11$: $121 - 11 + 11 = 121 = 11^2$, not prime [1]. Therefore the statement is false [1].
Q3 (3 marks): (a) True: $x^2 \geq 0$ for all real $x$ [1]. (b) False: $x^2 \geq 0$ for every real $x$, so no $x$ satisfies $x^2 < 0$ [1]. (c) False: take $x = \tfrac{1}{2}$; then $x^2 = \tfrac{1}{4} < \tfrac{1}{2} = x$, so $x^2 > x$ fails [1].
Five timed questions on implication, converse, contrapositive and quantifiers. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering quick logic questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.