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hscscience Ext 2 · Y12
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Module 11 · L01 of 8 ~40 min ⚡ +90 XP available

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.

Today's hook, Consider the statement "If $n$ is even, then $n^2$ is even." Before reading on, write down (a) the converse, (b) the contrapositive, and (c) decide whether each of those is true. If the converse is true, you have a biconditional. Compare your answers after card 05.
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You are here

Recall, your gut answer first

Start with the warm-up and record your first reasoning.

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Recall, your gut answer first
+5 XP warm-up

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.

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The two moves for reading a statement

Orient yourself to the lesson goals, language, and core approach.

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The two moves for reading a statement
+5 XP to read

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: $\neg Q \Rightarrow \neg P$

Sequence layout for the read-classify-direction flow with a final check note
$P \Rightarrow Q \;\equiv\; \neg Q \Rightarrow \neg P$
Converse is NOT equivalent
$P \Rightarrow Q$ does not imply $Q \Rightarrow P$. A statement can be true while its converse is false. Always test the converse separately.
Contrapositive IS equivalent
$P \Rightarrow Q$ is logically equivalent to $\neg Q \Rightarrow \neg P$. Proving the contrapositive is a valid way to prove the original.
Quantifiers change scope
"For all $x$" ($\forall$) and "there exists $x$" ($\exists$) are not interchangeable. Disprove a "for all" with one counterexample; disprove a "there exists" by showing it fails for every value.
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What you'll master
Know

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
Understand

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 $Q \Rightarrow P$
Can do

Skills

  • Write the converse, contrapositive and negation of any implication
  • Construct counterexamples to disprove false statements
  • Translate quantified statements between English and symbolic form
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Key terms
Implication ($P \Rightarrow Q$)"If $P$ then $Q$". $P$ is the hypothesis (sufficient condition); $Q$ is the conclusion (necessary condition). False only when $P$ is true and $Q$ is false.
ConverseThe converse of $P \Rightarrow Q$ is $Q \Rightarrow P$. The truth of one does not guarantee the truth of the other.
ContrapositiveThe contrapositive of $P \Rightarrow Q$ is $\neg Q \Rightarrow \neg P$. It is always logically equivalent to the original statement.
Biconditional ($P \Leftrightarrow Q$)"$P$ if and only if $Q$" (iff). True exactly when $P$ and $Q$ have the same truth value. Requires proving both directions.
Universal quantifier ($\forall$)"For all". The statement $\forall x \in S, P(x)$ means $P(x)$ holds for every $x$ in $S$. One counterexample disproves it.
Existential quantifier ($\exists$)"There exists". $\exists x \in S, P(x)$ means at least one $x$ in $S$ makes $P(x)$ true. One example proves it.
MEX-P1NESA outcome (Nature of Proof): uses the language and notation of mathematical proof, including implication, converse, contrapositive, biconditional and quantifiers.
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Implication, converse and contrapositive

Work through the concept and complete its check.

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Implication, converse and contrapositive
core concept

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.

  1. Converse $Q \Rightarrow P$, swap hypothesis and conclusion. Not equivalent in general.
  2. Contrapositive $\neg Q \Rightarrow \neg P$, negate both and swap. Always equivalent.
  3. 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.
Connecting to proof. When the original statement is hard to prove directly, the contrapositive often gives an easier route. For example, proving "$n^2$ is even $\Rightarrow n$ is even" directly is awkward; the contrapositive "$n$ is odd $\Rightarrow n^2$ is odd" is a one-line algebra exercise.

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?

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Quantifiers: $\forall$ and $\exists$

Build the next proof move and complete its check.

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Quantifiers: $\forall$ and $\exists$
core concept

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 $x$.
$$\neg(\forall x \in S,\; P(x)) \;\equiv\; \exists x \in S,\; \neg P(x)$$
Common mistake. The order of mixed quantifiers matters. "$\forall x \,\exists y$" (for every $x$ there is some $y$) is generally weaker than "$\exists y \,\forall x$" (one $y$ works for every $x$). Switching them changes the statement.

$\forall$ = "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 "$\exists n \in \mathbb{N},\; n^2 < n$".

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Worked examples · 3 in a row, reveal as you go

Follow the worked examples, then complete the check.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · WRITE THE CONTRAPOSITIVE

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.

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Identify $P$ and $Q$: $P$ = "$n$ is a prime greater than 2"; $Q$ = "$n$ is odd". Original is $P \Rightarrow Q$.
Always pin down the hypothesis and conclusion before manipulating. Misidentifying them produces the wrong contrapositive.
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PROBLEM 2 · COUNTEREXAMPLE TO THE CONVERSE

Consider "If $x > 2$, then $x^2 > 4$." Write the converse and decide whether it is true. If false, give a counterexample.

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Original: $(x > 2) \Rightarrow (x^2 > 4)$. Converse: $(x^2 > 4) \Rightarrow (x > 2)$.
Swap hypothesis and conclusion, do not negate. Converses are obtained by reversing the arrow, nothing more.
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PROBLEM 3 · QUANTIFIER SCOPE

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.

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(i) reads: "For every real $x$, there exists a real $y$ with $y > x$." Test: given any $x$, choose $y = x + 1$. Then $y > x$ ✓.
In a $\forall x \,\exists y$ statement the choice of $y$ may depend on $x$. Here $y = x+1$ is a valid $x$-dependent witness.
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Fill the gap: The contrapositive of "$P \Rightarrow Q$" is " $\Rightarrow$ ", and it is always logically equivalent to the original statement.

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Misconceptions to fix · the 3 traps that cost marks

Correct the common traps and test your understanding.

Misconceptions to fix · the 3 traps that cost marks

Trap 01
Confusing converse with contrapositive
The converse $Q \Rightarrow P$ swaps; the contrapositive $\neg Q \Rightarrow \neg P$ swaps AND negates. Only the contrapositive is equivalent to the original. Treating the converse as equivalent leads to invalid proofs.
Trap 02
Negating an implication wrongly
The negation of $P \Rightarrow Q$ is NOT $\neg P \Rightarrow \neg Q$. It is $P \wedge \neg Q$, that is, $P$ is true and $Q$ is false. Use this in contradiction proofs: assume $P$ but not $Q$, derive an impossibility.
Trap 03
Swapping quantifier order
"For every $x$ there exists $y$" is weaker than "There exists $y$ such that for every $x$". The first lets $y$ depend on $x$; the second demands one $y$ that works universally. Reordering changes the meaning.

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".

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Revisit your thinking

Complete the practice activities and revisit your opening thinking.

Activities · practice with the ideas

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Write the converse and contrapositive of: "If a quadrilateral is a square, then it is a rectangle." State whether the converse is true.

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Disprove: "$\forall n \in \mathbb{N},\; n^2 + n + 41$ is prime." Find a counterexample.

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Translate to symbolic form using quantifiers: "Every positive integer has a successor." Then write the negation.

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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.

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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?

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Revisit your thinking

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.

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Show what you have learned

Multiple choice, then short answer under exam conditions.

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer
ApplyBand 32 marks

Q1. Write the contrapositive of: "If $n^2$ is even, then $n$ is even." Then state whether the original statement is true. (2 marks)

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ApplyBand 43 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)

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AnalyseBand 53 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)

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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].