IB Maths AA HL Topic 1 — Number & Algebra Paper 1 & 2 HL only ~9 min read

Proof by Contradiction

Sometimes a result is hard to prove directly but easy to prove “from the wrong side”. Proof by contradiction means: assume the opposite of what you want to prove, then chase the algebra until something impossible falls out. Since the assumption led to nonsense, the assumption itself must have been wrong — and so the original claim is true. This is the classic trick used to prove things like “√2 is irrational” and “there are infinitely many primes”. The technique is short and powerful, but it relies on writing the negation correctly — that’s where most marks are won or lost.

📘 What you need to know

To prove a statement by contradiction: assume its negation, derive a contradiction, conclude the statement must be true.
The structure is always three steps: assume the opposite → derive contradiction → conclude.
Writing the negation correctly is the most important skill. Words like “all”, “exists”, “and”, “or” all flip in specific ways.
The classic exam scenarios: prove a number is irrational (assume it’s rational a/b in lowest terms, find a contradiction), prove infinitely many primes exist, prove there is no biggest […].
A useful result you’ll lean on: if n² is a multiple of a prime p, then n is a multiple of p. This is what unlocks irrationality proofs.
Questions don’t always say “use proof by contradiction” — you need to recognise when it’s the right tool. Two-option claims and “no biggest…” claims are usually contradiction.

How proof by contradiction works

Suppose you want to prove some statement P is true. The contradiction approach starts by assuming P is false. You then do algebra with that assumption until two of your statements clash — they can’t both hold at the same time. That clash is the “contradiction”. The only way to escape it is to abandon the original assumption, so P must be true after all.

The shape of every proof by contradiction

🤔 Why does this work logically?

A statement is either true or false — there’s no third option. If you assume the statement is false and that leads to an impossibility, then “false” wasn’t a valid choice. The only remaining option is “true”. This is sometimes called the law of the excluded middle.

Writing the negation correctly

Before anything else, you need to know what the opposite of the claim is. Get this wrong and the whole proof falls apart. Here are the negations you’ll meet most often.

Original statement

Negation

x is rational

x is irrational

n is even

n is odd

Every integer has property X

There exists an integer that does NOT have property X

There exists a number with property X

Every number does NOT have property X

n is even and n is prime

n is not even, OR n is not prime

n is a multiple of 4 or 6

n is not a multiple of 4 AND not a multiple of 6

If A then B

A is true AND B is false

There are infinitely many primes

There are only finitely many primes

🧠 Two patterns to remember

“All” flips to “exists” (and vice versa). “And” flips to “or” (and vice versa) — this is De Morgan’s law in disguise. Just memorise these two pairs and most negations fall into place.

The three-step template

Assume the negation

Write down the opposite of the claim and call it the assumption.

Derive a contradiction

Use algebra and known results to reach two facts that can’t both be true.

Conclude

State the contradiction shows the assumption is wrong, so the original claim must be true.

Standard concluding sentences (memorise) “This contradicts [the earlier statement]. So the assumption is false.
Therefore [the original claim] is true.”

Three classic setups you should recognise

Setup A — Proving a number is irrational

Standard recipe: assume rational, write as a/b with a, b integers and no common factors (already simplified to lowest terms). Square or rearrange. Show that a and b must both be divisible by some prime — contradicting “no common factors”.

Setup B — Proving “there is no biggest…”

Assume there is a biggest one — call it M. Construct something larger than M using M itself. Contradiction: you said M was the biggest.

Setup C — Proving “if A then B” by contradiction

Negation is “A and not B”. Assume both at once and use algebra to derive a contradiction with a known fact (like “n is even” vs “n is odd”).

If a question asks you to prove a number like √3 is irrational, contradiction is almost always the right tool — there’s no easy direct way to show “this isn’t a fraction”. When you see “show that there is no maximum” or “show there are infinitely many”, reach for contradiction.

Worked examples

WE 1

Prove √3 is irrational

Use proof by contradiction to show that √3 is irrational.

STEP 1 — Assume the negation assume √3 is rational, so √3 = a/b for some a, b ∈ ℤ, b ≠ 0, where a/b is in lowest terms (no common factors) STEP 2 — Square both sides and rearrange 3 = a²/b² → a² = 3b² so a² is a multiple of 3, which means a is a multiple of 3 let a = 3k for some k ∈ ℤ substitute: (3k)² = 3b² → 9k² = 3b² → b² = 3k² so b² is a multiple of 3, hence b is a multiple of 3 but then a and b share factor 3 — contradicts “no common factors” STEP 3 — Conclude The assumption is false, therefore √3 is irrational. ✓ the same template works for √2, √5, √7 — any prime square root

WE 2

If n² is odd, then n is odd

Prove by contradiction: for any integer n, if n² is odd, then n is odd.

STEP 1 — Negation of “if A then B” is “A and not B” assume n² is odd AND n is even STEP 2 — Derive contradiction n is even, so let n = 2k for some k ∈ ℤ then n² = (2k)² = 4k² = 2(2k²) since 2k² ∈ ℤ, n² is even but we assumed n² is odd — contradiction ✗ STEP 3 — Conclude The assumption is false, so if n² is odd then n is odd. ✓ “if A then B” by contradiction means assuming A and ¬B together

WE 3

Sum of a rational and an irrational

Prove by contradiction: the sum of a rational number and an irrational number is irrational.

STEP 1 — Assume the negation let r be rational and x be irrational assume r + x is rational, so r + x = a/b for some a, b ∈ ℤ, b ≠ 0 STEP 2 — Rearrange and use known facts x = (a/b) − r since r is rational, write r = c/d for c, d ∈ ℤ, d ≠ 0 x = a/b − c/d = (ad − bc)/(bd) numerator and denominator are integers, denominator ≠ 0 so x is rational — but we said x is irrational ✗ STEP 3 — Conclude The assumption is false. So rational + irrational = irrational ✓ the contradiction is that x is supposed to be irrational, yet it ended up as a fraction of integers

WE 4

No smallest positive rational

Prove by contradiction: there is no smallest positive rational number.

STEP 1 — Assume the negation assume there IS a smallest positive rational, call it q so q ∈ ℚ and q > 0, and any positive rational is ≥ q STEP 2 — Build something smaller consider q/2 since q is positive, q/2 is also positive since q is rational, q/2 = q × ½ is rational (rational × rational) but q/2 < q, contradicting “q is the smallest” ✗ STEP 3 — Conclude The assumption is false. So no smallest positive rational exists ✓ “no smallest/biggest” proofs typically construct a smaller/bigger candidate from the assumed one

WE 5

There are infinitely many primes (Euclid’s proof)

Prove by contradiction that there are infinitely many prime numbers.

STEP 1 — Assume the negation assume there are only finitely many primes list them all: p₁, p₂, …, p_n STEP 2 — Build a number bigger than every prime in the list let N = p₁ × p₂ × … × p_n + 1 N is bigger than every p_i, so N is not in the list divide N by any p_i: remainder is 1, not 0 so N is NOT divisible by any prime in the list therefore either N itself is prime, or N has a prime factor not in the list either way: a prime exists outside the list — contradiction ✗ STEP 3 — Conclude The assumption is false, so there are infinitely many primes ✓ this is Euclid’s classic proof from ~300 BC — one of the oldest in mathematics

💡 Top tips

Get the negation right first. If the negation is wrong, the whole proof is invalid no matter how good the algebra is.
For irrationality proofs, always say “in lowest terms” when writing a/b. The contradiction comes from showing they share a factor.
“All” ↔ “exists” and “and” ↔ “or” are the two flips you’ll use most. Memorise both.
For “if A then B” proofs by contradiction, the negation is “A and not B”, NOT “if A then not B”.
If a question doesn’t say which proof method to use, look at the claim. Words like “irrational”, “no biggest/smallest”, or “infinitely many” almost always mean contradiction.
Use the same wording every time in the conclusion: “this contradicts … so the assumption is false … therefore [original claim] is true.”
The result “if n² is a multiple of p (prime), then n is a multiple of p“ is the engine of every irrationality proof. Know it.

⚠ Common mistakes

Wrong negation. The negation of “all primes are odd” is “some prime is not odd”, not “no primes are odd”. “All” doesn’t flip to “none”.
Forgetting “no common factors” in irrationality proofs. Without that, you can’t reach the contradiction at the end.
Negating “if A then B” as “if A then not B”. The correct negation is “A and not B” — both at once.
Using a circular argument. The result you’re trying to prove can’t be used to derive the contradiction.
Missing the conclusion sentence. Just reaching a contradiction isn’t enough — you must explicitly state that the assumption is false and the original claim is true.
Writing the negation of “and” as “and not”. The negation of “P and Q” is “not P, or not Q” — at least one fails, not both.
Stopping too early. If you’ve shown a is a multiple of 3 in an irrationality proof, you still need to show b is too — that’s where the contradiction lands.

Proof by contradiction is the technique behind some of mathematics’ most famous results — including Euclid’s proof of infinite primes and the irrationality of √2 from over two thousand years ago. The pattern is short and powerful: assume the opposite, follow the logic, watch it crumble. Get the negation right and the rest is mostly mechanical algebra.

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