What is the language of implication?

A statement of the form "if P then Q" is the implication P Q; P is the hypothesis and Q the conclusion. Two related statements matter:

What is proof by contrapositive?

Since Q P is equivalent to P Q, you may prove the contrapositive instead. This is useful when assuming Q gives more to work with than assuming P, as in "if n^2 is even then n is even", where assuming n odd is concrete.

What are "Proving" a universal claim by checking cases?

Verifying a statement for n = 1, 2, 3 is not a proof; it could fail at the next value. Give a general argument.

What is a counterexample that does not satisfy the hypothesis?

A valid counterexample must meet every condition of the statement and yet fail the conclusion.

SEAB Original-style practice: Prove by contradiction that sqrt(2) is irrational.

Suppose, for contradiction, that sqrt(2) is rational. Then sqrt(2) = pq for integers p, q with q ≠ 0 and the fraction in lowest terms (so p and q share no common factor). Square both sides: 2 = p^2q^2, so p^2 = 2q^2. Hence p^2 is even, so p is even (the square of an odd number is odd). Write p = 2k. Then (2k)^2 = 2q^2 4k^2 = 2q^2 q^2 = 2k^2, so q^2 is even and hence q is even. But then p and q are both even, contradicting that the fraction was in lowest terms. The assumption is false, so sqrt(2) is irrational. Markers reward the opening assumption, p^2 = 2q^2, the deduction that both p and q are even, and identifying the contradiction with the lowest-terms assumption.

SingaporeFurther MathsSyllabus dot point

What are the standard methods of mathematical proof, and how do we structure a rigorous argument?

Construct rigorous mathematical arguments using direct proof, proof by contradiction, proof by contrapositive, and disproof by counterexample

A focused answer to the H2 Further Mathematics outcome on methods of proof. Direct proof, proof by contradiction, proof by contrapositive, disproof by counterexample, and the logical language of implication, converse and equivalence.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to construct and present rigorous mathematical arguments using the standard methods of proof, and to know the logical vocabulary that frames them. You should be able to give a direct proof, a proof by contradiction, a proof by contrapositive, and to disprove a false statement with a single counterexample, while using "implies", "converse" and "if and only if" correctly.

The answer

The language of implication

A statement of the form "if $P$ then $Q$ " is the implication $P \Rightarrow Q$ ; $P$ is the hypothesis and $Q$ the conclusion. Two related statements matter:

the converse $Q \Rightarrow P$ , which is a different statement and need not be true;
the contrapositive $\lnot Q \Rightarrow \lnot P$ , which is logically equivalent to the original.

If both $P \Rightarrow Q$ and $Q \Rightarrow P$ hold, we write $P \iff Q$ (" $P$ if and only if $Q$ "), and proving an equivalence means proving both directions.

Direct proof

A direct proof of $P \Rightarrow Q$ assumes $P$ and reasons forward through valid steps to reach $Q$ . Most algebraic and identity proofs are direct: start from the definitions or given facts and manipulate to the conclusion.

Proof by contrapositive

Since $\lnot Q \Rightarrow \lnot P$ is equivalent to $P \Rightarrow Q$ , you may prove the contrapositive instead. This is useful when assuming $\lnot Q$ gives more to work with than assuming $P$ , as in "if $n^2$ is even then $n$ is even", where assuming $n$ odd is concrete.

Proof by contradiction

To prove a statement $S$ by contradiction, assume $S$ is false and derive a logical impossibility (a contradiction with a known fact or with the assumption itself). The contradiction shows the assumption was untenable, so $S$ must be true. The classic example is the irrationality of $\sqrt{2}$ .

Disproof by counterexample

A universal statement "for all $x$ , $P(x)$ " is false if even one $x$ fails it. To disprove such a claim, exhibit a single explicit counterexample; you do not need to prove the statement fails in general, just to display one case. To prove a universal statement, however, a single example is never enough.

Worked example

Prove that the sum of a rational number and an irrational number is irrational.

Step 1: Set up a proof by contradiction

Let $a$ be rational and $b$ irrational, and suppose, for contradiction, that $a + b$ is rational. Call it $c$ , so $a + b = c$ with $c$ rational.

Step 2: Isolate the irrational quantity

Rearrange: $b = c - a$ .

Step 3: Use closure of the rationals

Both $c$ and $a$ are rational, and the rationals are closed under subtraction, so $c - a$ is rational. Hence $b = c - a$ is rational.

Step 4: Identify the contradiction

This contradicts the assumption that $b$ is irrational. Therefore our supposition was false.

Step 5: Conclude

The sum $a + b$ cannot be rational, so the sum of a rational and an irrational number is irrational.

Examples in context

Example 1. Justifying an algebraic identity. Proving a trigonometric or algebraic identity is a direct proof: start from one side and transform it, step by valid step, into the other, which is why "show that" questions reward a clean forward chain.

Example 2. Refuting a plausible conjecture. The claim " $n^2 + n + 41$ is prime for every positive integer $n$ " looks compelling for small $n$ , but $n = 41$ gives a multiple of $41$ . One counterexample settles it, illustrating why testing cases never constitutes a proof.

Try this

Q1. State the contrapositive of "if it is raining then the ground is wet". [1 mark]

Cue. If the ground is not wet then it is not raining.

Q2. Explain why one counterexample is enough to disprove "every prime is odd". [2 marks]

Cue. The statement claims all primes are odd; the single prime $2$ is even, so the universal claim fails.

Q3. Outline how you would begin a proof by contradiction that there is no largest integer. [2 marks]

Cue. Assume there is a largest integer $N$ ; then $N + 1$ is an integer larger than $N$ , contradicting that $N$ is largest.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksProve by contradiction that

\sqrt{2}

is irrational.

Show worked answer →

Suppose, for contradiction, that $\sqrt{2}$ is rational. Then $\sqrt{2} = \dfrac{p}{q}$ for integers $p, q$ with $q \neq 0$ and the fraction in lowest terms (so $p$ and $q$ share no common factor).

Square both sides: $2 = \dfrac{p^2}{q^2}$ , so $p^2 = 2q^2$ . Hence $p^2$ is even, so $p$ is even (the square of an odd number is odd). Write $p = 2k$ .

Then $(2k)^2 = 2q^2 \Rightarrow 4k^2 = 2q^2 \Rightarrow q^2 = 2k^2$ , so $q^2$ is even and hence $q$ is even.

But then $p$ and $q$ are both even, contradicting that the fraction was in lowest terms. The assumption is false, so $\sqrt{2}$ is irrational.

Markers reward the opening assumption, $p^2 = 2q^2$ , the deduction that both $p$ and $q$ are even, and identifying the contradiction with the lowest-terms assumption.

Original4 marksDetermine whether the following statement is true: for all integers

n

, if

n^2

is even then

n

is even. Prove it or give a counterexample.

Show worked answer →

The statement is true. Prove it by contrapositive: the contrapositive of "if $n^2$ is even then $n$ is even" is "if $n$ is odd then $n^2$ is odd".

Suppose $n$ is odd, so $n = 2k + 1$ for some integer $k$ . Then

n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1,

which is odd. So

n

odd implies

n^2

odd, and the contrapositive holds; therefore the original statement holds.

Markers reward stating the correct contrapositive, the algebra $n = 2k+1 \Rightarrow n^2 = 2(\cdots) + 1$ , and the conclusion that proving the contrapositive proves the original.