Proof techniques (HL only)

IB Mathematics: Analysis and Approaches HL· Unit 1: Number & Algebra, Topic 6· 25 min read

1. Direct Proof & Proof by Contrapositive★★☆☆☆⏱ 6 min

📘 Definition

Direct Proof

A proof that starts from accepted true axioms or assumptions, and uses sequential logical steps to reach the desired conclusion.

Example:

Proving the sum of two even integers is even

For statements of the form $P \Rightarrow Q$ , direct proof assumes $P$ is true, then shows $Q$ must follow. This works best when assumptions are easy to work with directly. When the negation of $P$ and $Q$ are simpler to use, we use proof by contrapositive.

📘 Definition

Proof by Contrapositive

Uses the logical equivalence $P \Rightarrow Q \equiv \neg Q \Rightarrow \neg P$ . Instead of proving the original statement, we prove the contrapositive is true.

Example:

Proving if $n^{2}$ is even, $n$ is even

📐 Worked Example

Prove that if $n^{2}$ is even, then $n$ is even.

1
Original statement is $P \Rightarrow Q$ , where $P : n^{2}$ even, $Q : n$ even. The contrapositive is $\neg Q \Rightarrow \neg P$ : if $n$ is odd, then $n^{2}$ is odd.
2
Assume $n$ is odd, so we can write it as:
3
$n = 2k + 1 \text{ for some integer } k$
4
Square both sides:
5
$n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$
6
Since $2 k^{2} + 2 k$ is an integer, $n^{2}$ is odd. The contrapositive is true, so the original statement is true.

2. Proof by Contradiction★★★☆☆⏱ 7 min

📘 Definition

Proof by Contradiction

To prove a statement $P$ is true, you first assume $P$ is false, then show this assumption leads to a clear logical contradiction. This means the original assumption is wrong, so $P$ must be true.

Example:

Proving $2$ is irrational

Contradiction is particularly useful for proving statements about irrationality, infinity, and non-existence, which are hard to prove directly. The method works by ruling out the only alternative to your claim.

📐 Worked Example

Prove that $2$ is irrational.

1
Assume for contradiction that $2$ is rational, so we can write it in lowest terms as:
2
$\sqrt{2} = \frac{p}{q}, \quad p,q \text{ coprime integers}, q > 0$
3
Square both sides and rearrange:
4
$p^2 = 2q^2$
5
This means $p^{2}$ is even, so $p$ must be even. Let $p = 2 k$ for some integer $k$ , substitute back:
6
$(2k)^2 = 2q^2 \implies 4k^2 = 2q^2 \implies q^2 = 2k^2$
7
By the same logic, $q^{2}$ is even so $q$ is even. This is a contradiction: we assumed $p$ and $q$ are coprime, but both are even so share a factor of 2.
8
Therefore, the original assumption that $2$ is rational is false, so $2$ is irrational.

3. Principle of Mathematical Induction★★★★☆HL only⏱ 8 min

📘 Definition

Mathematical Induction

A technique to prove statements that hold for all integers $n \geq n_{0}$ , consisting of three core logical steps: base case, inductive hypothesis, and inductive step.

Induction is one of the most frequently tested proof techniques in IB AA HL, used for summation formulas, divisibility statements, inequalities, and recursive sequence results.

📐 Worked Example

Prove by induction that for all positive integers $n$ , $\sum_{r = 1}^{n} r = \frac{n ( n + 1 )}{2}$ .

1
Step 1: Base case, test $n = 1$ :
2
$\text{LHS} = 1, \quad \text{RHS} = \frac{1(1+1)}{2} = 1$
3
The statement holds for $n = 1$ , so base case is proven.
4
Step 2: Inductive hypothesis: assume the statement holds for $n = k$ , for some positive integer $k$ :
5
$\sum_{r=1}^k r = \frac{k(k+1)}{2}$
6
Step 3: Inductive step: prove the statement holds for $n = k + 1$ :
7
$\sum_{r=1}^{k+1} r = \sum_{r=1}^k r + (k+1)$
8
Substitute the inductive hypothesis and simplify:
9
$= \frac{k(k+1)}{2} + (k+1) = (k+1)\left(\frac{k}{2} + 1\right) = \frac{(k+1)(k+2)}{2}$
10
This matches the formula for $n = k + 1$ , so if the statement holds for $k$ , it holds for $k + 1$ . By the principle of mathematical induction, the statement is true for all positive integers $n$ .

4. Identifying Invalid Proofs★★★☆☆⏱ 4 min

IB exams often ask you to spot errors in incomplete or invalid proofs. The most common invalid arguments rely on incorrect logic, even if the final conclusion is right.

✓ Quick check

Which of the following is a valid proof that the sum of two even numbers is even?

- I tested 2+4=6 (even), 8+12=20 (even), so the rule is true
- If $a$ and $b$ are even, $a = 2 k$ and $b = 2 m$ for integers $k, m$ , so $a + b = 2 (k + m)$ , which is even by definition
- All even numbers I checked work, so the statement must be true
Reveal answer
1 —
Correct! This is a general direct proof that works for all even numbers. Testing specific cases never proves a general statement.

5. Common Pitfalls

Wrong move:

Prove a general statement by checking specific examples

Why:

A statement is only proven if it holds for all cases, not just the ones you tested

Correct move:

Use a general argument that applies to all values satisfying the problem's assumptions

Wrong move:

Skip writing the inductive hypothesis or final conclusion for induction

Why:

IB examiners award explicit marks for these steps, even if the algebra is correct

Correct move:

Always write 'assume the statement holds for $n = k$ ' and end with 'by induction, the statement holds for all $n \geq n_{0}$ '

Wrong move:

Divide by a variable without checking it cannot equal zero

Why:

Division by zero is undefined, which creates a fatal flaw in your proof

Correct move:

Explicitly confirm the variable is non-zero before dividing, or handle the zero case separately

Wrong move:

Mix up contrapositive and contradiction

Why:

They are logically related but distinct methods, misapplication leads to invalid proofs

Correct move:

Remember: contrapositive proves $P \Rightarrow Q$ by proving $\neg Q \Rightarrow \neg P$ ; contradiction assumes $\neg P$ and derives a contradiction

Wrong move:

Assume what you are trying to prove (circular reasoning)

Why:

This means your proof does not actually demonstrate the result

Correct move:

Always start from known true statements and work towards the conclusion you need to prove

6. Quick Reference Cheatsheet

Proof Technique	When to Use	Core Steps
Direct Proof	Working forward from assumptions is straightforward	Assume $P$ true 2. Derive $Q$ logically
Contrapositive	Prove $P \Rightarrow Q$ , negation is easier to use	State contrapositive $\neg Q \Rightarrow \neg P$ 2. Prove directly
Contradiction	Prove irrationality, non-existence, or negative claims	Assume claim false 2. Derive contradiction 3. Conclude original claim true
Induction	Prove for all integers $n \geq n_{0}$	Prove base case 2. Assume true for $n = k$ 3. Prove for $n = k + 1$ 4. Conclude by induction

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2021 · 1
Proof by contradiction of irrationality
2022 · 1
Induction for divisibility
2023 · 2
Proof by contrapositive

What's Next

Proof techniques are the foundation of all advanced mathematics, and they appear throughout the IB AA HL course, from calculus to complex numbers. Mastering these techniques will help you construct logical arguments for extended response questions, which carry a large proportion of marks in HL papers. You will encounter induction again when working with sequences, series, and recursive functions, while contradiction and contrapositive are used in topics like number theory and calculus. Building a strong understanding of proof will also prepare you for university-level mathematics studies.