IB Maths
Paper 1 & 2
15 min read
Proof & Reasoning
A proof is a chain of logical steps showing a result is true for every value — not just a few you tried. In IB AA SL, you’ll mostly use algebra to prove identities, prove things about whole numbers, and disprove statements with a single counter-example. Let’s go.
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What you need to know
- Testing 2 or 3 examples is not a proof
- Symbols: LHS (left side), RHS (right side), ≡ (“identical for all values”)
- Number sets: ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ
- Even = 2n, Odd = 2n + 1, Multiple of k = kn
- To disprove a statement, just one counter-example is enough
Number sets & notation
Each number set sits inside the next bigger one. The naturals are inside the integers, integers inside the rationals, and everything sits inside the reals.
ℝ Reals
ℚ Rationals
ℤ Integers
ℕ Naturals
ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ (each set is contained in the next)
ℕNaturals = {0, 1, 2, 3, …}
ℤIntegers = {0, ±1, ±2, …}. Use ℤ⁺ for positives only.
ℚRationals = numbers ab where a, b ∈ ℤ, b ≠ 0
ℝReals = every number on the number line. ℝ⁺ = positives only.
Quick read: “for all x ∈ ℝ” means it must work for any real — integers, fractions, decimals, negatives, zero. The exam usually writes = instead of ≡, but they mean the same thing in proofs.
Direct Proof — Identities (LHS = RHS)
You’re given LHS = RHS and asked to prove it. The rule is simple: pick one side, expand and simplify, and show it becomes the other side. Never work both sides at once.
1Start with LHS or RHS
→
2Expand & simplify
→
3Reach the other side
“= RHS as required”
The structure of every identity proof in IB AA SL
Pick the messier side first — easier to simplify down than build up.
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Example 1 — Prove an identity
Prove that (2x − 2)(x − 3) + 2(x − 1) = 2(x − 2)(x − 1) for all x ∈ ℝ.
Answer:
Step 1: work with LHS (it’s messier).
LHS = (2x − 2)(x − 3) + 2(x − 1)
Step 2: expand the brackets (FOIL).
= 2x² − 6x − 2x + 6 + 2x − 2
Step 3: simplify carefully (mind the signs).
= 2x² − 6x + 4
Step 4: factor out 2.
= 2(x² − 3x + 2)
Step 5: factor the quadratic.
= 2(x − 2)(x − 1) = RHS ✓
LHS = RHS as required ∎
The ∎ symbol (or “QED”) closes the proof.
Proof by Deduction — Integers
To prove a statement about whole numbers, first write the integers algebraically, then operate on them. This little table is your starter kit.
What you want
Write it as
Note
Any integer
n
where n ∈ ℤ
Two consecutive integers
n, n+1
or n−1, n
Two different integers
n, m
use different letters!
Even integer
2n
any multiple of 2
Odd integer
2n + 1
or 2n − 1
Multiple of k
kn
e.g. multiple of 5 = 5n
Square / cube number
n², n³
Rational number
ab
a, b ∈ ℤ, b ≠ 0
Even ⟹ show it equals 2 × (integer)
Odd ⟹ show it equals 2 × (integer) + 1
Important: the part inside the brackets must be an integer. So 2(n + 13) is not even — 13 isn’t an integer!
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Example 2 — Sum of two consecutive odd numbers
Prove that the sum of any two consecutive odd numbers is always even.
Answer:
Step 1: let two consecutive odd numbers be
2n − 1 and 2n + 1
(next odd after 2n − 1)
Step 2: add them.
(2n − 1) + (2n + 1) = 4n
Step 3: write as 2 × integer.
= 2(2n)
2 × integer → always even ∎
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Example 3 — Product of two consecutive integers
Prove that the product of any two consecutive integers is always even.
Answer:
Step 1: let consecutive integers be n and n + 1.
Step 2: their product is
n(n + 1)
Step 3: one of n, n+1 must be even.
(consecutive integers always alternate even/odd)
Step 4: even × anything = even.
n(n + 1) is always even ∎
Divisibility Proofs
To prove an expression is a multiple of k, factor out k and check what’s left is an integer.
Expression = k × (integer) ⟹ multiple of k
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Example 4 — Divisibility by 8
Prove that (2n + 1)² − (2n − 1)² is divisible by 8 for all n ∈ ℤ.
Answer:
Step 1: expand both squares.
(2n + 1)² = 4n² + 4n + 1
(2n − 1)² = 4n² − 4n + 1
Step 2: subtract.
(4n² + 4n + 1) − (4n² − 4n + 1)
= 8n
Step 3: write as 8 × integer.
= 8(n) where n ∈ ℤ
divisible by 8 ∎
Disproof by Counter-Example
To disprove a statement, you only need one case where it fails. You don’t need to show it always fails — just produce one number that breaks it.
To prove TRUE
Use algebra ✓
Must hold for all values — testing examples is not enough.
To prove FALSE
One counter-example ✗
A single value where the statement fails is enough.
✎
Example 5 — Disprove with a counter-example
Disprove the statement: “n² + n + 41 is prime for all n ∈ ℕ.”
Answer:
Strategy: try values until one fails.
n = 0: 41 ✓ prime
n = 1: 43 ✓ prime
n = 2: 47 ✓ prime
… it keeps working — try bigger.
n = 40: 1600 + 40 + 41 = 1681
1681 = 41 × 41 ✗ not prime!
n = 40 is a counter-example ∎
One failure is all you need. Statement disproved.
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Tips
- For counter-examples, try small values first (0, 1, 2), then negatives, then fractions if allowed
- Always close with “as required”, “∴ even”, or ∎ — examiners reward the conclusion
- If you’re stuck on an identity, check if the RHS factors easily — that often shows you the target
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Common mistakes
- Working LHS and RHS at the same time. Pick one side and transform it. Never write “LHS = RHS” then manipulate both.
- Same letter for two different integers. “Any two integers” needs n and m, not n and n.
- Testing examples and calling it proof. Checking n = 1, 2, 3 is not a proof. Use algebra.
- Forgetting the final line. Always close with “= RHS as required” or “∴ even” or ∎. Marks are awarded for the conclusion.
- Not factoring out 2 to show “even”. Stopping at 4n + 6 is not enough — write it as 2(2n + 3).
- Confusing “any” with “specific”. 2n + 1 means any odd number; 7 is just one odd number.
Final word: get the algebraic setup right (“let n ∈ ℤ…”, “let two consecutive odd numbers be 2n−1 and 2n+1…”) and the rest of the proof flows on its own. Most marks come from a clean setup and a clear final line.
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