What are the different types of mathematical proofs covered in Mathematical Proofs 2?

In Mathematical Proofs 2, students typically explore various types of proofs such as direct proofs, indirect proofs (including proof by contradiction), and mathematical induction. Each type has its unique approach and application, which are essential for solving different kinds of mathematical problems in the IB DP Mathematics curriculum.

How does proof by contradiction work in Mathematical Proofs 2?

Proof by contradiction involves assuming the opposite of what you want to prove, and then showing that this assumption leads to a contradiction. This method is powerful in IB DP Mathematics as it helps students understand the logical structure of arguments and is often used when direct proof is challenging.

Why is mathematical induction important in Mathematical Proofs 2?

Mathematical induction is crucial because it allows students to prove statements about integers, particularly those involving sequences or series. In the IB DP curriculum, understanding induction helps in developing a deeper comprehension of how properties hold for an infinite set of numbers by proving a base case and an inductive step.

What role do axioms play in constructing mathematical proofs?

Axioms are fundamental truths accepted without proof, serving as the starting point for further reasoning and arguments. In Mathematical Proofs 2, axioms are used to build logical arguments and derive theorems, helping students understand the foundational principles of mathematics in the IB DP program.

How can I improve my skills in writing mathematical proofs for the IB DP curriculum?

To improve your skills in writing mathematical proofs, practice is key. Start by understanding the problem, choose the appropriate proof technique, and write each step clearly and logically. Reviewing examples, seeking feedback from teachers, and collaborating with peers can also enhance your proficiency in constructing proofs in the IB DP Mathematics course.

Why is "Skipping the explicit 'By PMI' conclusion." a common mistake in Mathematical Proofs 2?

Why it happens: Considering it obvious. How to avoid it: Examiners require the conclusion sentence to award the final mark.

Why is "Using $P(k+1)$ implicitly during the inductive step." a common mistake in Mathematical Proofs 2?

Why it happens: Assuming what to prove. How to avoid it: Inductive step builds P(k+1) from P(k) — never assume P(k+1).

Why is "Starting induction at $n = 0$ when the claim is for $n \ge 1$." a common mistake in Mathematical Proofs 2?

Why it happens: Misreading the domain. How to avoid it: Match the base case to the stated domain — typically n = 1.

ProofsIB Maths AA HL

Mathematical Proofs 2

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Mathematical Proofs 2 — IB Maths AA HL: Proof by Mathematical Induction

HL-only topic. Proof by induction is a powerful technique for statements about natural numbers. AA HL exams reward strict adherence to the four-step structure: base case, inductive hypothesis, inductive step, conclusion.

What you’ll learn

Mapped to the IB DP Maths AA HL subject guide (2021 onwards (applies to 2026 exams)).

AO1 — State the principle of mathematical induction.
AO2 — Apply induction to sums, divisibility and inequalities.
AO3 — Lay out a four-step induction proof correctly to earn full marks.

The four-step structure

Exam-credit framework.

Principle of Mathematical Induction (PMI). Let $P(n)$ be a statement about $n \in \mathbb{N}$ . If:

Base: $P(1)$ is true (or the smallest relevant $n$ ).
Inductive step: For all $k \ge 1$ , $P(k) \Rightarrow P(k + 1)$ .

Then $P(n)$ is true for all $n \in \mathbb{N}$ ( $\ge$ base).

Four-step layout (must follow for full marks).

Base case. Verify $P(1)$ .
Inductive hypothesis. Assume $P(k)$ true for some $k \ge 1$ .
Inductive step. Prove $P(k + 1)$ .
Conclusion. State the result, citing PMI.

Worked example (sum). Prove $1 + 2 + 3 + \cdots + n = n(n+1)/2$ .

1. Base. $n = 1$ : LHS $= 1$ , RHS $= 1 \cdot 2/2 = 1$ . ✓.

2. Hypothesis. Assume $1 + 2 + \cdots + k = k(k+1)/2$ .

3. Step. Show $1 + 2 + \cdots + k + (k+1) = (k+1)(k+2)/2$ .

LHS = $k(k+1)/2 + (k+1) = (k+1)[k/2 + 1] = (k+1)(k+2)/2 =$ RHS. ✓.

4. Conclusion. By PMI, $1 + 2 + \cdots + n = n(n+1)/2$ for all $n \in \mathbb{N}$ .

Four steps — every one earns marks.
Always write hypothesis explicitly with 'some $k$ '.
Inductive step: connect $P(k)$ to $P(k+1)$ .

Divisibility and inequality induction

Beyond simple sums.

Divisibility induction. Prove $n^3 - n$ is divisible by 6 for all $n \ge 1$ .

Base. $n = 1$ : $1 - 1 = 0 = 6(0)$ ✓.

Hypothesis. Assume $k^3 - k = 6m$ for some $m \in \mathbb{Z}$ .

Step. $(k+1)^3 - (k+1) = k^3 + 3k^2 + 3k + 1 - k - 1 = (k^3 - k) + 3k^2 + 3k = 6m + 3k(k + 1)$ .

$k(k + 1)$ is product of consecutive integers, so even — say $= 2j$ . So $3k(k+1) = 6j$ , and total $= 6(m + j)$ , divisible by 6. ✓.

Conclusion. By PMI, $n^3 - n$ is divisible by 6 for all $n \in \mathbb{N}$ .

Inequality induction. Prove $2^n > n$ for all $n \ge 1$ .

Base. $n = 1$ : $2 > 1$ ✓.

Hypothesis. Assume $2^k > k$ for some $k \ge 1$ .

Step. $2^{k+1} = 2 \cdot 2^k > 2k$ (using hypothesis). We need $2k \ge k + 1$ , i.e. $k \ge 1$ ✓.

So $2^{k+1} > 2k \ge k + 1$ . ✓.

Conclusion. By PMI, $2^n > n$ for all $n \in \mathbb{N}$ .

Compound angle and De Moivre by induction. For $\theta \in \mathbb{R}, n \in \mathbb{N}$ :

$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta.$

(De Moivre's theorem — covered in Complex Numbers; the proof is by induction.)

Divisibility: target $= 6m + (\text{remaining})$ form.
Inequality: use hypothesis to bound LHS.
Use induction confidently for De Moivre too.

How it’s examined

Paper 1 or 2: classic 7-8 mark induction proofs (sums, divisibility, inequalities, De Moivre). The four-step structure earns method marks even if the algebra slips.

Sources: IB Diploma Programme Mathematics: Analysis and Approaches subject guide (IBO, official). Last reviewed 2026-05-31.

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Step-by-step worked examples — Mathematical Proofs 2

Step-by-step solutions to past-paper-style questions on mathematical proofs 2, written exactly the way a tutor would explain them at the board.

1Sum of squares
Building confidence• sum, induction
Question
Prove by induction: $1^2 + 2^2 + \cdots + n^2 = n(n+1)(2n+1)/6$ .
Step-by-step solution
1. Step 1
  Base: $n = 1$ : LHS = 1, RHS = $1 \cdot 2 \cdot 3/6 = 1$ . ✓.
2. Step 2
  Hypothesis: assume $\sum_{i=1}^k i^2 = k(k+1)(2k+1)/6$ .
3. Step 3
  Step: $\sum_{i=1}^{k+1} i^2 = k(k+1)(2k+1)/6 + (k+1)^2 = (k+1)[k(2k+1) + 6(k+1)]/6 = (k+1)(2k^2 + 7k + 6)/6$ .
4. Step 4
  Factor: $2k^2 + 7k + 6 = (k+2)(2k+3)$ . So sum $= (k+1)(k+2)(2k+3)/6$ , matching $P(k+1)$ . ✓.
5. Step 5
  By PMI, true for all $n \in \mathbb{N}$ .
Answer
Proved by induction. QED.
2Divisibility — $7^n - 1$ by 6
Building confidence• divisibility, induction
Question
Prove $7^n - 1$ is divisible by 6 for all $n \ge 1$ .
Step-by-step solution
1. Step 1
  Base: $n = 1$ : $7 - 1 = 6 = 6 \cdot 1$ . ✓.
2. Step 2
  Hypothesis: $7^k - 1 = 6m$ for some $m \in \mathbb{Z}$ .
3. Step 3
  Step: $7^{k+1} - 1 = 7 \cdot 7^k - 1 = 7(6m + 1) - 1 = 42m + 6 = 6(7m + 1)$ . Divisible by 6. ✓.
4. Step 4
  By PMI, $7^n - 1$ divisible by 6 for all $n \ge 1$ .
Answer
Proved by induction. QED.
3Inequality $n! > 2^n$
Stretch• inequality, induction
Question
Prove $n! > 2^n$ for all $n \ge 4$ .
Step-by-step solution
1. Step 1
  Base: $n = 4$ : $24 > 16$ ✓.
2. Step 2
  Hypothesis: $k! > 2^k$ for some $k \ge 4$ .
3. Step 3
  Step: $(k+1)! = (k+1) \cdot k! > (k+1) \cdot 2^k$ (by hypothesis).
4. Step 4
  Since $k \ge 4$ , $k + 1 \ge 5 > 2$ , so $(k+1) \cdot 2^k > 2 \cdot 2^k = 2^{k+1}$ . ✓.
5. Step 5
  By PMI, $n! > 2^n$ for all $n \ge 4$ .
Answer
Proved by induction. QED.
4De Moivre's theorem via induction
Stretch• De Moivre, induction
Question
Prove $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ for all $n \in \mathbb{N}$ .
Step-by-step solution
1. Step 1
  Base: $n = 1$ : trivial.
2. Step 2
  Hypothesis: $(\cos\theta + i\sin\theta)^k = \cos k\theta + i\sin k\theta$ .
3. Step 3
  Step: $(\cos\theta + i\sin\theta)^{k+1} = (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta)$ .
4. Step 4
  Expand: $\cos k\theta\cos\theta - \sin k\theta\sin\theta + i(\sin k\theta\cos\theta + \cos k\theta\sin\theta) = \cos(k+1)\theta + i\sin(k+1)\theta$ . ✓.
5. Step 5
  By PMI, holds for all $n \in \mathbb{N}$ .
Answer
Proved by induction (uses compound-angle identities).

Key Definitions and Keywords — Mathematical Proofs 2

Definitions to memorise and the exact keywords mark schemes credit for mathematical proofs 2 answers — sharpened from recent examiner reports for the 2026 IB DP Maths AA HL sitting.

Principle of Mathematical Induction (PMI)
Examiner keyword
$P(1)$ true and $P(k) \Rightarrow P(k+1)$ for all $k \ge 1$ implies $P(n)$ true for all $n \in \mathbb{N}$ .
Base case
Examiner keyword
Smallest value of $n$ at which the statement is verified directly.
Inductive step
Examiner keyword
Proof that $P(k) \Rightarrow P(k+1)$ .

Common Mistakes and Misconceptions — Mathematical Proofs 2

The traps other students keep falling into on mathematical proofs 2 questions — taken from recent IB DP Maths AA HL examiner reports and mark schemes — and how to avoid them.

✕Skipping the explicit 'By PMI' conclusion.
Why it happens
Considering it obvious.
How to avoid it
Examiners require the conclusion sentence to award the final mark.
✕Using $P(k+1)$ implicitly during the inductive step.
Why it happens
Assuming what to prove.
How to avoid it
Inductive step builds $P(k+1)$ from $P(k)$ — never assume $P(k+1)$ .
✕Starting induction at $n = 0$ when the claim is for $n \ge 1$ .
Why it happens
Misreading the domain.
How to avoid it
Match the base case to the stated domain — typically $n = 1$ .

Past paper style quiz

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Mathematical Proofs 2 — frequently asked questions

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Mathematical Proofs 2

Short Study Notes in the form of Slides

Short Study Notes — Mathematical Proofs 2

Detailed Notes

What you’ll learn

How it’s examined

1Sum of squares

2Divisibility — 7n−17^n - 17n−1 by 6

3Inequality n!>2nn! > 2^nn!>2n

4De Moivre's theorem via induction

Principle of Mathematical Induction (PMI)

Base case

Inductive step

✕Skipping the explicit 'By PMI' conclusion.

✕Using P(k+1)P(k+1)P(k+1) implicitly during the inductive step.

✕Starting induction at n=0n = 0n=0 when the claim is for n≥1n \ge 1n≥1.

Past paper style quiz

Mathematical Proofs 2 — frequently asked questions

2Divisibility — $7^n - 1$ by 6

3Inequality $n! > 2^n$

✕Using $P(k+1)$ implicitly during the inductive step.

✕Starting induction at $n = 0$ when the claim is for $n \ge 1$ .