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 ā 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), ā” (“identical for all values”)
- Number sets: ā ā ā¤ ā ā ā ā
- 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.
ā Reals
ā Rationals
ā¤ Integers
ā Naturals
ā ā ā¤ ā ā ā ā (each set is contained in the next)
āNaturals = {0, 1, 2, 3, ā¦}
ā¤Integers = {0, Ā±1, Ā±2, ā¦}. Use ā¤āŗ for positives only.
āRationals = numbers ab where a, b ā ā¤, b ā 0
āReals = every number on the number line. āāŗ = positives only.
Quick read: “for all x ā ā” means it must work for any real ā integers, fractions, decimals, negatives, zero. The exam usually writes = instead of ā”, but they mean the same thing in proofs.
Direct Proof ā 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
ā
2Expand & simplify
ā
3Reach the other side
“= RHS as required”
The structure of every identity proof in IB AA SL
Pick the messier side first ā easier to simplify down than build up.
ā
Example 1 ā Prove an identity
Prove that (2x ā 2)(x ā 3) + 2(x ā 1) = 2(x ā 2)(x ā 1) for all x ā ā.
Answer:
Step 1: work with LHS (it’s messier).
LHS = (2x ā 2)(x ā 3) + 2(x ā 1)
Step 2: expand the brackets (FOIL).
= 2xĀ² ā 6x ā 2x + 6 + 2x ā 2
Step 3: simplify carefully (mind the signs).
= 2xĀ² ā 6x + 4
Step 4: factor out 2.
= 2(xĀ² ā 3x + 2)
Step 5: factor the quadratic.
= 2(x ā 2)(x ā 1) = RHS ā
LHS = RHS as required ā
The ā symbol (or “QED”) closes the proof.
Proof by Deduction ā 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 ā ā¤
Two consecutive integers
n, n+1
or nā1, n
Two different integers
n, m
use different letters!
Even integer
2n
any multiple of 2
Odd integer
2n + 1
or 2n ā 1
Multiple of k
kn
e.g. multiple of 5 = 5n
Square / cube number
nĀ², nĀ³
Rational number
ab
a, b ā ā¤, b ā 0
Even ā¹ show it equals 2 Ć (integer)
Odd ā¹ show it equals 2 Ć (integer) + 1
Important: the part inside the brackets must be an integer. So 2(n + 13) is not even ā 13 isn’t an integer!
ā
Example 2 ā 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 ā 1 and 2n + 1
(next odd after 2n ā 1)
Step 2: add them.
(2n ā 1) + (2n + 1) = 4n
Step 3: write as 2 Ć integer.
= 2(2n)
2 Ć integer ā always even ā
ā
Example 3 ā 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 Ć anything = even.
n(n + 1) is always even ā
Divisibility Proofs
To prove an expression is a multiple of k, factor out k and check what’s left is an integer.
Expression = k Ć (integer) ā¹ multiple of k
ā
Example 4 ā Divisibility by 8
Prove that (2n + 1)Ā² ā (2n ā 1)Ā² is divisible by 8 for all n ā ā¤.
Answer:
Step 1: expand both squares.
(2n + 1)Ā² = 4nĀ² + 4n + 1
(2n ā 1)Ā² = 4nĀ² ā 4n + 1
Step 2: subtract.
(4nĀ² + 4n + 1) ā (4nĀ² ā 4n + 1)
= 8n
Step 3: write as 8 Ć integer.
= 8(n) where n ā ā¤
divisible by 8 ā
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 ā just produce one number that breaks it.
To prove TRUE
Use algebra ā
Must hold for all values ā testing examples is not enough.
To prove FALSE
One counter-example ā
A single value where the statement fails is enough.
ā
Example 5 ā Disprove with a counter-example
Disprove the statement: “nĀ² + n + 41 is prime for all n ā ā.”
Answer:
Strategy: try values until one fails.
n = 0: 41 ā prime
n = 1: 43 ā prime
n = 2: 47 ā prime
ā¦ it keeps working ā try bigger.
n = 40: 1600 + 40 + 41 = 1681
1681 = 41 Ć 41 ā not prime!
n = 40 is a counter-example ā
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”, “ā“ even”, or ā ā examiners reward the conclusion
- If you’re stuck on an identity, check if the RHS factors easily ā that often shows you the target
ā
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 “ā“ even” or ā. Marks are awarded for the conclusion.
- Not factoring out 2 to show “even”. Stopping at 4n + 6 is not enough ā 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 ā ā¤ā¦”, “let two consecutive odd numbers be 2nā1 and 2n+1ā¦”) 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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