IB Maths AA HL · Unit 1: Number and Algebra
IB Maths AA HL Proof Questions
Exam-style IB Maths AA HL proof questions with worked solutions. Start with the 3 fully worked examples below — each is solved step by step the way an IB examiner expects — then try the practice questions and check your working against the mark scheme.
- 18 questions
- Paper 1: 15
- Paper 2: 3
- 5 easy
- 5 medium
- 5 hard
- 3 starter
- 3 worked examples
Practise Proof questions →
AA HL formula booklet
What's examined in AA HL proof
The question bank covers these proof question types (number of questions in brackets):
- Mathematical Induction (10)
- Proof by Contradiction (5)
- Proof by Counterexample (3)
Key formulas
- Proof by induction (skeleton)
- \(P(1)\text{ true};\ P(k) \Rightarrow P(k+1);\ \therefore P(n)\text{ true for all }n\in\mathbb{Z}^+\)
In the same notation as the IB formula booklet. All AA HL formulas →
Proof worked examples
Worked example 1: Disproving statements by counterexample · easy
Disprove the following statement using a valid counterexample: "For all positive integers $n$, if $n$ is prime, then $2^n - 1$ is also prime."
1. Understand the logic that a single case where the hypothesis ($n$ is prime) is true but the conclusion ($2^n - 1$ is prime) is false will perfectly disprove the entire statement.
2. Test small prime numbers. For $n = 2$, $2^2 - 1 = 3$ (prime). For $n = 3$, $2^3 - 1 = 7$ (prime).
3. Test $n = 5$, giving $2^5 - 1 = 31$ (prime).
4. Test the next prime, $n = 7$, giving $2^7 - 1 = 127$ (prime).
5. Test the next prime, $n = 11$, giving $2^{11} - 1 = 2047$.
6. Show that $2047$ is composite by testing divisibility by small primes, revealing $2047 = 23 \times 89$. Since $n=11$ yields a composite number, the statement is disproven.
Examiner tip: When hunting for a counterexample, students should be systematic. It is vital they explicitly state the factors of the resulting composite number to formally validate the disproof to the examiner.
Worked example 2: Proof by contradiction basics · medium
Prove by contradiction that $\log_2 5$ is an irrational number.
1. Assume the opposite of the statement to be true: Assume that $\log_2 5$ is a rational number.
2. Express the rational number as a fraction in its simplest form: $\log_2 5 = \frac{p}{q}$, where $p$ and $q$ are integers with no common factors, and $q \neq 0$.
3. Rewrite the logarithmic equation into its equivalent exponential form: $2^{\frac{p}{q}} = 5$.
4. Raise both sides to the power of $q$ to eliminate the fraction from the exponent: $2^p = 5^q$.
5. Analyze the parity (even/odd nature) of both sides. Since $p$ and $q$ must be non-zero positive integers, $2^p$ is always an even integer, and $5^q$ is always an odd integer.
6. State the contradiction: An even number cannot equal an odd number. Therefore, our initial assumption must be false, proving $\log_2 5$ is irrational.
Examiner tip: A formal proof by contradiction must always begin with an explicit opening statement assuming the opposite, and must conclude by explicitly identifying exactly where the logical contradiction occurs.
Worked example 3: Proof by mathematical induction · hard
Use mathematical induction to prove that $5^n - 1$ is a multiple of $4$ for all $n \in \mathbb{Z}^+$.
1. Establish the base case for $n=1$: $5^1 - 1 = 4$, which is a multiple of $4$. The statement therefore holds for $n=1$.
2. State the inductive assumption: Assume the statement is true for some integer $k$, such that $5^k - 1 = 4M$ for some integer $M$.
3. Set up the inductive step to prove true for $n = k+1$: Consider the expression $5^{k+1} - 1$.
4. Manipulate the expression algebraically to substitute the assumption: $5^{k+1} - 1 = 5(5^k) - 1 = 5(4M + 1) - 1$.
5. Expand and simplify the resulting expression: $20M + 5 - 1 = 20M + 4 = 4(5M + 1)$.
6. Conclude: Since $5M+1$ is an integer, $5^{k+1}-1$ is a multiple of 4. As it holds for $n=1$ and $n=k \implies n=k+1$, it is true for all $n \in \mathbb{Z}^+$.
Examiner tip: The final concluding sentence in mathematical induction (linking the base case and inductive step to the set of all positive integers) is strictly required by IB mark schemes; omitting it will frequently cost the final reasoning (R1) mark.
Try these IB Maths AA HL proof questions
Three questions from the bank, easiest first. Mark schemes and AI marking of your written working are in the practice area.
Question 1 · easy · 2 marks · Paper 2
A student claims that "If \(n\) is a prime number, then \(2^n - 1\) is also a prime number."
Provide a specific counterexample to prove that this statement is false.
Question 2 · medium · 7 marks · Paper 1
Use mathematical induction to prove that for all \(n \in \mathbb{Z}^+\):
\[\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}\]
Question 3 · hard · 8 marks · Paper 1
[Calculus & Induction Synthesis]
Let \(f(x) = x e^x\). Use mathematical induction to prove that the \(n\)-th derivative of \(f(x)\) is given by:
\[f^{(n)}(x) = (x + n)e^x \quad \text{for all } n \in \mathbb{Z}^+\]
All 18 proof questions with mark schemes →
FAQ
How many IB Maths AA HL proof questions are there?
There are 18 exam-style proof questions in the AA HL question bank (Paper 1: 15 · Paper 2: 3), graded 5 easy, 5 medium, 5 hard, 3 starter. Every question has a full IB-style mark scheme (M, A and R marks).
Is proof on Paper 1 or Paper 2?
Both. In the bank, Paper 1: 15 · Paper 2: 3. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.
Where can I get the mark schemes?
Open the AA HL Unit 1 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.
Proof in other IB Maths courses
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