Proof (AHL 1.15) is the topic where IB Mathematics Analysis & Approaches HL students lose the most marks per minute of work. The mathematics is rarely difficult — almost every proof in the syllabus reduces to a parity argument, a perfect-square inequality, or the irrationality of $\sqrt{N}$. What kills the marks is the language: writing "let" instead of "assume", forgetting to justify multiplication of an inequality by a positive quantity, or skipping the final conclusion sentence. Each of these costs a guaranteed R1, and there is no algebra hack to recover it.
This cheatsheet condenses every framework, template, trick and trap from AHL 1.15 into one revision page — direct proof, proof by contradiction, and counterexample. Scroll to the bottom for the printable PDF, the full Notes, the worked tutorials and the marked-up solutions in the gated student library.
§1 — Direct Proof AHL 1.15
The 4-step framework
- Name your integers. Write $n = 2k$, $n = 2k + 1$, $n = mk$, etc., with explicit $k \in \mathbb{Z}$.
- Substitute and expand. Do the algebra carefully — keep both sides tidy.
- Identify the structure. Factor out $m$, identify even/odd, spot the divisibility.
- State the conclusion. Explicitly say why the result follows. Never leave it implicit.
Parity and divisibility
TrickExpress, don't just state. Always write out the algebraic form. If $n$ is even, write $n = 2k$ where $k \in \mathbb{Z}$. If $n$ is odd, write $n = 2k + 1$. Then substitute and compute.
TrickProduct of consecutives is always even: $n(n + 1)$ — one of $n, n+1$ is always even, so the product is even.
Trap"$n$ is even, so $n^2$ is even" is NOT a proof. You must show it: $n = 2k \Rightarrow n^2 = 4k^2 = 2(2k^2)$, divisible by 2. ✓ Just asserting "even × even = even" earns 0 marks.
TrapConsecutive integers — choose your representation wisely. For the sum of 3 consecutives, $n - 1, n, n + 1$ gives sum $= 3n$ immediately. Using $n, n + 1, n + 2$ gives $3n + 3 = 3(n + 1)$ — still works but messier.
NoteDivisible by $m$ means $= m \times (\text{integer})$. To prove $E$ is divisible by 12, show $E = 12k$ for some $k \in \mathbb{Z}$, or factor as $E = 12 \times (\ldots)$ where the bracket is an integer.
TrickDifference of squares shortcut: $(a + b)^2 - (a - b)^2 = 4ab$ always. Use this to spot divisibility quickly.
NoteThe "hence" signal. When a question says "hence prove", the previous part has set up exactly what you need. Use the result you just proved — don't rederive it. This is tested every year.
Key representations table
| Type | Write as | Key property |
|---|---|---|
| Even integer | $n = 2k$, $k \in \mathbb{Z}$ | $n^2 = 4k^2 \equiv 0 \pmod 4$ |
| Odd integer | $n = 2k + 1$, $k \in \mathbb{Z}$ | $n^2 = 4k^2 + 4k + 1 \equiv 1 \pmod 4$ |
| Multiple of $m$ | $n = mk$, $k \in \mathbb{Z}$ | — |
| Rational number | $r = p/q$, $\gcd(p, q) = 1$, $q \neq 0$ | In lowest terms |
Algebraic identity proofs
TrickFor LHS = RHS proofs, always expand the more complicated side first (or expand both sides towards a common middle). Never "work from both ends simultaneously" — the IB expects a single directed chain.
TrapDon't assume what you're proving. In an identity proof, you cannot start with "$\text{LHS} = \text{RHS}$" and then manipulate — that assumes the result. Always start with one side and arrive at the other.
TrickSum of squares of consecutive odds: $(2n - 1)^2 + (2n + 1)^2 = 8n^2 + 2 = 2(4n^2 + 1)$ is always even. Know this pattern.
Conclusion language
NoteHow to end a direct proof — always finish with an explicit conclusion: "Since $4n^2 + 1 \in \mathbb{Z}$, the expression $2(4n^2 + 1)$ is divisible by 2, hence even." Or: "Since $2n \in \mathbb{Z}$, the expression $12(2n)$ is a multiple of 12." The final R1 mark requires this statement — algebra alone does not earn it.
§2 — Proof by Contradiction AHL 1.15
The 5-step framework
- State the assumption. Write: "Assume for contradiction that [negation of claim]."
- Justify any manipulations. If multiplying an inequality, state the sign first.
- Derive the algebra. Rearrange, expand, substitute — towards a contradiction.
- Name the contradiction. Explicitly identify what is impossible (even = odd, square $< 0$, etc.).
- Conclude. "This is a contradiction. Therefore [original claim] is proved."
The assumption word — most important rule
Trap"Let" earns M0 — use "Assume" or "Suppose". Wrong: "Let $\sqrt{3} = p/q$…" — this defines, it doesn't contradict. Correct: "Assume for contradiction that $\sqrt{3}$ is rational, so $\sqrt{3} = p/q$ where $p, q \in \mathbb{Z}$, $q \neq 0$, $\gcd(p, q) = 1$." The IB mark scheme explicitly awards M0 for "let", "consider", or "given that".
Negation guide
| Original claim $P$ | What to assume ($\neg P$) |
|---|---|
| $\sqrt{N}$ is irrational | Assume $\sqrt{N} = p/q$ with $\gcd(p, q) = 1$ |
| $f(x) = 0$ has no integer roots | Assume $\exists\, \alpha \in \mathbb{Z}$ such that $f(\alpha) = 0$ |
| $p^2 - 8q - 11 \neq 0$ for $p, q \in \mathbb{Z}$ | Assume $p^2 - 8q - 11 = 0$ for some $p, q \in \mathbb{Z}$ |
| $f(x) \geq c$ on interval | Assume $f(x) < c$ for some $x$ in interval |
| $a$ and $b$ cannot both be odd | Assume $a$ and $b$ are both odd |
| There are infinitely many primes | Assume there are only $n$ primes: $p_1, \ldots, p_n$ |
| $a + b$ irrational ($a$ rational, $b$ irrational) | Assume $a + b$ is rational |
The three contradiction types (by IB frequency)
Trick#1 — Parity mismatch (most common). Derive: even integer = odd integer. Pattern: equation with even-coefficient LHS and odd RHS. e.g. $2\alpha^3 + 6\alpha = -1$ — LHS even, RHS odd. Done.
Trick#2 — Perfect square $< 0$ (inequality type). Derive: $(2x - 1)^2 < 0$ or $(2x - y)^2 < 0$. Pattern: assume $f(x) < c$, multiply through, rearrange to get a completed square that must be negative. Squares are always $\geq 0$. Contradiction.
Trick#3 — Both $p$ and $q$ share a factor (irrationality). Derive: $\sqrt{N} = p/q$ in lowest terms, then show both $p$ and $q$ are divisible by $N$ (or 2, etc.). But $\gcd(p, q) = 1$ — contradiction.
§3 — Irrationality Proof Template AHL 1.15
Proving $\sqrt{N}$ is irrational (where $N$ is prime):
- Assume $\sqrt{N}$ is rational: $\sqrt{N} = \dfrac{p}{q}$, $p, q \in \mathbb{Z}$, $q \neq 0$, $\gcd(p, q) = 1$.
- $\Rightarrow N q^2 = p^2 \Rightarrow N \mid p^2 \Rightarrow N \mid p$ (since $N$ is prime) $\Rightarrow p = Nk$.
- $\Rightarrow N q^2 = N^2 k^2 \Rightarrow q^2 = N k^2 \Rightarrow N \mid q^2 \Rightarrow N \mid q$.
- Contradiction: $N \mid p$ and $N \mid q$, but $\gcd(p, q) = 1$. $\therefore \sqrt{N}$ is irrational. $\blacksquare$
§4 — Inequality Proof Template AHL 1.15
Proving $f(x) \geq c$ by contradiction:
- Assume $f(x) < c$ for some valid $x$.
- Justify sign: state that the denominator/quantity you will multiply by is positive (earns R1).
- Rearrange: multiply both sides, collect terms, complete the square $\Rightarrow (\ldots)^2 < 0$.
- Contradiction: a squared real expression is always $\geq 0$. $\therefore f(x) \geq c$. $\blacksquare$
Example: Prove $\dfrac{1}{x(1 - x)} \geq 4$ for $0 < x < 1$.
Assume $\dfrac{1}{x(1 - x)} < 4$. Since $x(1 - x) > 0$: $1 < 4x(1 - x) \Rightarrow 4x^2 - 4x + 1 < 0 \Rightarrow (2x - 1)^2 < 0$. Contradiction.
§5 — Parity Argument Template AHL 1.15
Integer/parity contradiction pattern:
- Assume $f(p, q) = 0$ for some $p, q \in \mathbb{Z}$.
- Step 1: Determine parity of $p$ from the equation (check: if $p$ even $\Rightarrow$ LHS even, RHS odd $\Rightarrow$ impossible, so $p$ must be odd).
- Step 2: Write $p = 2k + 1$, substitute, expand.
- Step 3: Show one side is even and the other odd.
- Contradiction: An even number cannot equal an odd number. $\blacksquare$
Traps to avoid in contradiction proofs
TrapWrong inequality direction. If proving $f \geq 4$, assume $f \mathbf{< 4}$ (strict), not $f \leq 4$. If proving $f > 4$, assume $f \mathbf{\leq 4}$. Wrong direction $\Rightarrow$ M0 on the assumption mark.
TrapNo contradiction statement. Just arriving at "$(2x - 1)^2 < 0$" is not enough. You must write: "This is a contradiction since $(2x - 1)^2 \geq 0$ for all $x \in \mathbb{R}$." Without this, the final R1 is lost.
TrapNot justifying multiplication of an inequality. When $0 < x < 1$, before multiplying by $x(1 - x)$, you must state: "Since $x > 0$ and $1 - x > 0$, $x(1 - x) > 0$, so the inequality direction is preserved." The examiner awards a specific R1 for this.
NoteEuclid's proof — know this cold. Assume finitely many primes $p_1, \ldots, p_n$. Let $N = p_1 p_2 \cdots p_n + 1$. Then $N$ is divisible by no $p_k$ (remainder 1 each time), and $N > 1$, so $N$ has a prime factor not in the list. Contradiction.
§6 — Counterexample AHL 1.15
The 3-step framework
- State the value. Give the specific value(s): "Let $n = 4$" or "Let $a = 1$, $b = 1$".
- Compute both sides. Show the full calculation — never just assert the result.
- Explain why it fails. State clearly which property of the original claim is violated.
NoteOne counterexample is sufficient — but it must be verified. You do not need multiple examples. But the IB requires both the specific value and the verification. Just writing "$n = 4$" without checking earns partial credit at best.
How to find a counterexample efficiently
TrickTry small values first. Test $n = 1, 2, 3, 4, 5$ for integer claims. For algebraic identities try $a = 1, b = 2$ or $a = b = 1$. For prime claims, look for the first composite output.
TrickThink about what could break it. For "always prime", look for an $n$ that makes the expression composite (e.g. $n = $ the constant itself). For "always divisible", look for $n$ where the LCM argument fails.
TrickThe LCM trap counterexample: "Divisible by $a$ and $b$ $\Rightarrow$ divisible by $ab$". Find a number divisible by both $a$ and $b$ but NOT $ab$ — answer: $\operatorname{lcm}(a, b)$ when $\gcd(a, b) > 1$. e.g. $a = 6$, $b = 10$: $\operatorname{lcm} = 30$. But $30 \div 60 = 0.5$. Done.
TrapDon't pick a value that happens to satisfy the claim. Before presenting your counterexample, verify it actually violates the statement. A "counterexample" that satisfies the claim scores 0.
§7 — Famous IB Counterexamples to Know AHL 1.15
| False claim | Counterexample | Why it fails |
|---|---|---|
| $n^2 + n + 41$ always prime | $n = 40$: $40^2 + 40 + 41 = 41^2$ | $41 \times 41$, not prime |
| $n^2 + 41n + 41$ always prime (IB) | $n = 41$: $41^2 + 41^2 + 41 = 41 \times 83$ | $41$ is a factor |
| $n^2 + n + 5$ always prime | $n = 4$: $16 + 4 + 5 = 25 = 5^2$ | Not prime |
| $2^n - 1$ always prime | $n = 4$: $2^4 - 1 = 15 = 3 \times 5$ | Not prime |
| $(m + n)^2 = m^2 + n^2$ | $m = 1, n = 1$: $4 \neq 2$ | Neglects $2mn$ term |
| Div by 6 and 10 $\Rightarrow$ div by 60 | $n = 30$ | $30 \div 60 = 0.5$ |
| $x^2 + y^2 = 10$ has no positive int sols | $x = 1, y = 3$: $1 + 9 = 10$ | Solution exists |
Prove or disprove questions
NoteStrategy for "prove or disprove" questions: (1) test a few small values to guess whether it is true or false; (2) if it seems false, find your counterexample and verify; (3) if it seems true, attempt a direct proof or contradiction proof; (4) if wrong either way (counterexample fails, proof breaks), reconsider. Never spend more than a minute guessing before committing to a direction.
TrapA counterexample is not enough for a universal proof. If the question asks you to "prove or disprove", and you find a counterexample, you must state that this disproves the claim. Simply presenting the calculation without saying "therefore the claim is false" loses the reasoning mark.
§8 — Decision Plan & Trigger Table AHL 1.15
Decision flow:
- Read the question carefully.
- Does it say prove (true) or disprove / not always true?
- If prove: try direct first. If the negation creates a tangible constraint (parity, irrationality, inequality), use contradiction.
- If disprove: find a counterexample, verify, explain.
If you see… then reach for…
| You see in the question… | Technique to use |
|---|---|
| Equation with all-even coefficients + odd constant | Parity contradiction: LHS even, RHS odd |
| Prove $f(x) \geq c$ or $f(x) > c$ | Contradiction: assume $f < c$, rearrange to $(\ldots)^2 < 0$ |
| Prove $\sqrt{N}$ irrational ($N$ prime) | Contradiction + irrationality template |
| "$a$ and $b$ cannot both be…" | Contradiction: assume they ARE both that way |
| $a^2 + b^2$ divisible by 4 | Write $a = 2m + 1$, $b = 2n + 1$; sum $\equiv 2 \pmod 4$, contradiction |
| Prove $\sqrt{XY} \leq \tfrac{X + Y}{2}$ type | Contradiction: assume $>$, square both sides, get $(X - Y)^2 < 0$ |
| "Show the statement is not always true" | Counterexample: find $n$, verify it fails, explain |
| "$f(n)$ is not always prime" | Counterexample: find $n$ that makes $f(n)$ composite, show factorisation |
| "Prove or disprove" | Test small values first; pick the correct approach based on what you find |
§9 — Mark Scheme Codes & Final Checklist AHL 1.15
| Code | Name | What it means in proof questions |
|---|---|---|
| M1 | Method mark | Usually awarded for the correct assumption statement (contradiction) or the correct setup (direct) |
| A1 | Accuracy mark | Correct algebra, expansion, substitution |
| (A1) | Implied accuracy | Correct intermediate step; often awarded even if not explicitly shown |
| R1 | Reasoning mark | Logical conclusion: naming the contradiction, stating why result follows, identifying parity |
| AG | Answer given | Answer is printed in the question; you must show full working to earn all preceding marks |
5-second proof checklist before submitting
- Used "assume" or "suppose" (not "let" or "consider") in contradiction proofs.
- Named all integers with types: "where $k \in \mathbb{Z}$".
- Justified the sign when multiplying an inequality.
- Named the contradiction explicitly ("even = odd", "square $< 0$", etc.).
- Written a conclusion sentence restating the original claim.
- For counterexamples: stated the value AND computed AND explained why.
- Not assumed what you're trying to prove (no circular argument).
Worked Example — IB-Style Proof by Contradiction
Question (HL Paper 1 style — 7 marks)
Prove, by contradiction, that there are no integers $p$ and $q$ such that $p^2 - 8q = 11$.
Solution
- Assume for contradiction that there exist $p, q \in \mathbb{Z}$ such that $p^2 - 8q = 11$. (M1)
- Rearrange: $p^2 = 8q + 11$. (A1)
- Determine the parity of $p$. The RHS is $8q + 11$. Since $8q$ is even and $11$ is odd, $8q + 11$ is odd. So $p^2$ is odd, which forces $p$ to be odd. (R1)
- Write $p = 2k + 1$ for some $k \in \mathbb{Z}$. Then
$p^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k + 1) + 1$. (M1)(A1) - Note that $k(k + 1)$ is the product of two consecutive integers, so $k(k + 1)$ is even. Write $k(k + 1) = 2m$ for some $m \in \mathbb{Z}$. Then $p^2 = 8m + 1$. (R1)
- Substitute back: $8m + 1 = 8q + 11 \Rightarrow 8m - 8q = 10 \Rightarrow 8(m - q) = 10 \Rightarrow m - q = \tfrac{10}{8} = \tfrac{5}{4}$.
But $m - q \in \mathbb{Z}$, while $\tfrac{5}{4} \notin \mathbb{Z}$. Contradiction. Therefore there are no integers $p, q$ such that $p^2 - 8q = 11$. $\blacksquare$ (R1)
Examiner's note: The most common errors in this question are (i) starting with "Let $p, q \in \mathbb{Z}$…" — losing the assumption mark — and (ii) failing to spot that $p^2 \equiv 1 \pmod 8$ for odd $p$, which is the key step. The Photon Academy approach is to memorise the four squares modulo 8: $0, 1, 4$ — only those three residues occur, never 11 mod 8 = 3.
Common Student Questions
Why does writing "Let" instead of "Assume" lose marks in a proof by contradiction?
Because "Let" defines something, while "Assume" or "Suppose" explicitly negates the claim you are trying to disprove. The IB mark scheme awards M0 for "let", "consider" or "given that" at the start of a contradiction proof — the examiner needs to see that you understand you are assuming the OPPOSITE of what you are about to prove. Always start with: "Assume for contradiction that…" followed by the negation of the claim.
When proving an inequality by contradiction, why must I justify the sign before multiplying?
Because multiplying both sides of an inequality by a negative quantity flips the direction. Before multiplying through by, say, $x(1 - x)$ on the interval $0 < x < 1$, you must state: "since $x > 0$ and $1 - x > 0$, the product $x(1 - x) > 0$, so the inequality direction is preserved." The IB awards a dedicated R1 for this justification — without it, even a perfect algebraic chain loses a mark.
How do I prove that $\sqrt{N}$ is irrational when $N$ is prime?
Assume $\sqrt{N} = p/q$ with $\gcd(p, q) = 1$. Then $N q^2 = p^2$, so $N \mid p^2$, so $N \mid p$ (since $N$ is prime). Write $p = Nk$. Substitute: $N q^2 = N^2 k^2$, giving $q^2 = N k^2$, so $N \mid q$. But then $N$ divides both $p$ and $q$, contradicting $\gcd(p, q) = 1$. The crucial step is "$N \mid p^2 \Rightarrow N \mid p$", which only works because $N$ is prime.
Is one counterexample enough to disprove a "for all" claim?
Yes — one counterexample is sufficient, but you must do three things: (1) state the specific value(s), e.g. "Let $n = 4$"; (2) compute both sides of the claim explicitly, never just assert the result; (3) explain clearly why the original claim fails, e.g. "$25 = 5 \times 5$, so it is not prime, contradicting the claim that $n^2 + n + 5$ is always prime". Skipping step (2) or (3) typically loses a mark.
What is the difference between "show that" and "prove"?
In IB mark schemes, "show that" and "prove" are essentially synonymous and both are AG (Answer Given) questions — meaning the answer is printed in the question and you must present full working to earn the marks. You cannot skip steps. The only practical difference is tone: "prove" usually appears with formal logical statements (irrationality, divisibility, induction), while "show that" often appears with algebraic identities. Both demand explicit justification and a clear conclusion sentence.
Get the printable PDF version
Same cheatsheet, formatted for A4 print — keep it next to your study desk. Free for signed-in users.