Proof by induction, contradiction and counterexample Notes | Mathematics AA HL

Meaning of mathematical proof

A mathematical proof is a finite chain of valid logical deductions that establishes a statement from accepted definitions, axioms, or previously proved results. A proof explains why a result must be true, not merely why it appears true in several examples.

A theorem usually has hypotheses and a conclusion. In an implication $P\implies Q$ , P is the hypothesis and Q is the conclusion. The proof may assume P and must justify Q.

A universal statement claims that a property holds for every object in a stated domain. An existential statement claims that at least one object satisfies a property. The quantifiers and domain are part of the statement and cannot be ignored.

Numerical checks can suggest a conjecture or reveal an error, but checking any finite list does not prove a claim about infinitely many cases.

Key terms

Proof: A logically valid argument establishing that a mathematical statement must be true.
Theorem: A statement established by proof.
Hypothesis: An assumption supplied by a theorem or implication.
Conclusion: The result that must follow from the hypotheses.
Conjecture: A plausible statement not yet proved.

Worked examples

Evidence is not proof

A student checks $2^n>n$ for $n = 1$ , 2, 3, 4, 5. Has the statement been proved for every positive integer?

1Only five cases were checked.
2The claim concerns infinitely many positive integers.

Answer: No. The checks are evidence; a general argument such as induction is required.

Identify hypothesis and conclusion

In the statement ‘if n is divisible by 4, then n is even’, identify the two parts.

1The phrase after ‘if’ is the hypothesis.
2The phrase after ‘then’ is the conclusion.

Answer: Hypothesis: $4\mid n$ . Conclusion: $2\mid n$ .

A valid existential proof

Prove that there exists an even prime.

1An existential statement needs one valid witness.
2Choose 2 and verify that it is even and prime.

Answer: The witness 2 proves the statement.

Rules and formulas

Implication

P\implies Q

Universal statement

\forall x\in D,\ P(x)

Existential statement

\exists x\in D\text{ such that }P(x)

Choosing the appropriate proof method

Choose a proof method by reading the logical structure rather than by copying a familiar template. Induction is designed for a chain of integer-indexed cases. Contradiction is useful when the negation creates strong algebraic or logical consequences. A counterexample disproves a universal claim.

Direct proof is often best when definitions immediately transform the hypothesis into the conclusion. Do not use induction simply because the statement contains n; many integer statements have shorter direct proofs.

To prove $P\implies Q$ by contradiction, assume P and $\neg Q$ . To disprove $\forall x\in D, P(x)$ , find one $c\in D$ with $\neg P(c)$ .

Before writing, identify the domain, quantifier, hypotheses, exact conclusion, and any restrictions such as $n\ge n_0$ .

Key terms

Direct proof: A proof that derives the conclusion directly from the hypotheses.
Induction: A proof method connecting a verified starting case to every later integer case.
Contradiction: A proof that assumes the negation and derives an impossibility.
Counterexample: One valid domain value that makes a universal claim false.

Worked examples

Choose induction

Which method naturally proves $1+2+\cdots+n=\frac{n(n+1)}2$ for every positive integer n?

1The statement is indexed by consecutive positive integers.
2The formula for k can help build the formula for k + 1.

Answer: Mathematical induction is natural.

Choose contradiction

Which method naturally proves that $\sqrt2$ is irrational?

1Assuming rationality gives a lowest-terms fraction.
2Parity then forces numerator and denominator both even, contradicting lowest terms.

Answer: Proof by contradiction is natural.

Choose a counterexample

How should the claim ‘every prime is odd’ be handled?

1The claim is universal.
2Search for one prime that is not odd.

Answer: The counterexample 2 disproves the claim.

Rules and formulas

Contradiction form for an implication

P\land\neg Q\implies\bot

Disproving a universal claim

\neg(\forall x\in D, P(x))\iff\exists x\in D\text{ with }\neg P(x)

Mathematical induction: the complete logical chain

Suppose ( $P(n)$ ) is a statement for every integer $n\ge n_0$ . Mathematical induction has two logical obligations: prove $P(n_0)$ , then prove $P(k)\implies P(k+1)$ for an arbitrary integer $k\ge n_0$ .

The first obligation is the base case. The second contains the inductive hypothesis and inductive step. Together they establish ( $P(n)$ ) for all integers from $n_0$ onward.

A domino chain illustrating a verified starting case and a repeatable next-case implication. A complete induction proof needs both the verified first case and a valid implication from each case to the next.

Open graph full size

The domino analogy separates the logic: placing or knocking the first domino corresponds to the base case; proving each falling domino knocks the next corresponds to the implication $P(k)\implies P(k+1)$ . Either part alone is insufficient.

Induction does not assume the statement for every n. It assumes one arbitrary case $P(k)$ temporarily in order to prove the next case.

Key terms

Induction proposition: The statement $P(n)$ to be proved for all integers in a specified range.
Base case: Verification of the first required integer $n_{0}$ .
Inductive hypothesis: The temporary assumption that $P(k)$ holds for an arbitrary eligible k.
Inductive step: The proof that $P(k)$ implies $P(k + 1)$ .

Worked examples

Identify $P(n)$

For the claim $1+3+\cdots+(2n-1)=n^2$ , state $P(n)$ .

1Include both the indexed sum and the proposed result.
2State the integer domain.

Answer: $P(n):\sum\limits_{r=1}^{n}(2r-1)=n^2$ for $n\ge1$ .

Identify the starting index

For a claim stated for every integer $n\ge4$ , which base case is required?

1Use the first integer in the stated range.

Answer: Verify $P(4)$ , not $P(1)$ .

Explain the two failures

Why does checking $P(1)$ without proving $P(k)$ ⇒ $P(k + 1)$ not establish every case?

1The base starts the chain.
2No rule has been proved to reach later cases.

Answer: Both the base case and the inductive implication are necessary.

Rules and formulas

Induction principle

P(n_0)\land\big(\forall k\ge n_0,\ P(k)\implies P(k+1)\big)\implies\forall n\ge n_0,\ P(n)

The base case

The base case substitutes the first required integer $n_0$ into the original statement and verifies both sides or the required property.

Write the base case explicitly. For an identity, calculate the left-hand side and right-hand side separately. For divisibility, show the required integer multiple.

If the statement starts at $n = 0$ , checking $n = 1$ leaves a gap. If it starts at $n = 3$ , checking $n = 1$ does not start the required chain at its stated first case.

Some statements require more than one base case when the inductive step jumps by more than one index or depends on several previous terms.

Key terms

Starting index: The smallest integer $n_{0}$ covered by the theorem.
Verification: A calculation or argument confirming the proposition at $n_{0}$ .

Worked examples

Base case for a sum

Verify the base case for $\sum\limits_{r=1}^{n}r=\frac{n(n+1)}2$ , $n\ge1$ .

1At $n = 1$ , the left side is 1.
2The right side is $\frac{1(2)}2=1$ .

Answer: $P(1)$ is true.

Base case for divisibility

Verify the base case for $7\mid(8^n-1)$ , $n\ge1$ .

1At $n = 1$ , $8^1-1=7$ .
2Seven is divisible by 7.

Answer: $P(1)$ is true.

A non-standard start

Verify the base case for $2^n>n^2$ , $n\ge5$ .

1At $n = 5$ , $2^5=32$ .
2Also, $5^2=25$ , so $32>25$ .

Answer: $P(5)$ is true.

Rules and formulas

Base-case instruction

n=n_0:\quad\text{verify }P(n_0)

The inductive hypothesis

In the inductive hypothesis, let k be an arbitrary integer with $k\ge n_0$ and assume $P(k)$ is true. Write the complete statement $P(k)$ , not only the phrase ‘assume true for k’.

The hypothesis is temporary and local to the inductive step. It is not the result being proved, because the conclusion concerns all n after the step and base case are combined.

Do not assume $P(k + 1)$ . That is exactly what the inductive step must establish.

The hypothesis must be used. If the calculation of $P(k + 1)$ never depends on $P(k)$ , the central link has probably not been proved.

Key terms

Arbitrary k: A general integer in the required range, not a chosen example.
Temporary assumption: A statement permitted within the implication proof and discharged afterward.

Worked examples

Write a sum hypothesis

State the inductive hypothesis for $\sum\limits_{r=1}^{n}r=\frac{n(n+1)}2$ .

1Replace n by k on both sides.

Answer: Assume $\sum\limits_{r=1}^{k}r=\frac{k(k+1)}2$ for an arbitrary $k\ge1$ .

Write a divisibility hypothesis

State the hypothesis for $7\mid(8^n-1)$ .

1Write divisibility as an integer multiple if it will be used algebraically.

Answer: Assume $8^k-1=7m$ for some integer m.

Write a recurrence hypothesis

For $u_1=2$ , $u_{n+1}=3u_n+2$ , state the hypothesis for the claim $u_n=3^n-1$ .

1Replace n by k in the claimed explicit formula.

Answer: Assume $u_k=3^k-1$ for an arbitrary $k\ge1$ .

Rules and formulas

Inductive hypothesis

P(k)\text{ is assumed for an arbitrary }k\ge n_0

Divisibility hypothesis

d\mid A_k\iff A_k=dm\text{ for some }m\in\mathbb Z

The inductive step and conclusion

The inductive step starts from the left-hand side or defining expression for $P(k + 1)$ , separates the new contribution, and uses the inductive hypothesis to reach the required right-hand side or property.

For a sum, write the sum to k + 1 as the sum to k plus the new term. For a recurrence, use the recurrence to express $u_{k+1}$ in terms of $u_k$ . For divisibility, rearrange the next expression so that the assumed divisible part appears.

End the step with the exact statement $P(k + 1)$ , then discharge the temporary assumption and state the induction conclusion for every integer in the domain.

Avoid circular reasoning: do not begin by assuming the formula for k + 1 or manipulate the desired equation as though it were already true.

Key terms

New term: The additional term appearing when a sum grows from k to k + 1.
Discharge: Completing an implication so the temporary hypothesis is no longer assumed globally.
Induction conclusion: The final statement that $P(n)$ holds for every integer n in the required range.

Worked examples

Complete a sum step

Assuming $\sum\limits_{r=1}^{k}r=\frac{k(k+1)}2$ , prove the k + 1 case.

1Write $\sum\limits_{r=1}^{k+1}r=\sum\limits_{r=1}^{k}r+(k+1)$ .
2Use the hypothesis to obtain $\frac{k(k+1)}2+(k+1)$ .
3Factor and simplify to $\frac{(k+1)(k+2)}2$ .

Answer: This is $P(k + 1)$ .

Complete a recurrence step

Assume $u_k=3^k-1$ and $u_{k+1}=3u_k+2$ . Prove the next formula.

1Substitute the hypothesis: $u_{k+1}=3(3^k-1)+2$ .
2Simplify to $3^{k+1}-1$ .

Answer: Therefore $P(k + 1)$ holds.

Write the conclusion

How should a complete induction proof finish?

1Refer to both the verified base and proved implication.
2Restate the domain.

Answer: Therefore, by mathematical induction, $P(n)$ holds for every integer $n\ge n_0$ .

Rules and formulas

Sum extension

\sum\limits_{r=1}^{k+1}a_r=\sum\limits_{r=1}^{k}a_r+a_{k+1}

Inductive implication

P(k)\implies P(k+1)

Induction proofs involving sums

For a finite-sum identity, define $P(n)$ precisely with correct sigma limits. In the inductive step, isolate the term produced when the upper limit changes from k to k + 1.

Substitute the inductive hypothesis for the entire sum to k. Then simplify until the result has exactly the claimed form with n replaced by k + 1.

Factorization is often the decisive final step. Write enough algebra to show how the target expression appears.

The same structure works for arithmetic, geometric, polynomial, and telescoping sums when the new term is identified correctly.

Key terms

Finite sum: A sum with a specified lower and upper index.
Sigma notation: Compact notation for indexed addition.
Target form: The claimed formula after replacing n by k + 1.

Worked examples

Sum of odd integers

Prove $\sum\limits_{r=1}^{n}(2r-1)=n^2$ for $n\ge1$ .

1Base: at $n = 1$ , both sides equal 1.
2Assume $\sum\limits_{r=1}^{k}(2r-1)=k^2$ .
3Then the next sum is $k^2+(2(k+1)-1)=k^2+2k+1=(k+1)^2$ .

Answer: The identity holds for every positive integer n by induction.

Geometric sum

Prove $\sum\limits_{r=0}^{n}3^r=\frac{3^{n+1}-1}{2}$ for $n\ge0$ .

1Base: $n = 0$ gives $1=\frac{3-1}{2}$ .
2Assume the formula for k.
3Add $3^{k+1}$ and simplify: $\frac{3^{k+1}-1}{2}+3^{k+1}=\frac{3^{k+2}-1}{2}$ .

Answer: The formula holds for every integer $n\ge0$ .

A polynomial summand

Prove $\sum\limits_{r=1}^{n}r(r+1)=\frac{n(n+1)(n+2)}3$ .

1Verify $n = 1$ : both sides equal 2.
2Assume the formula for k.
3Add $(k+1)(k+2)$ , factor $(k+1)(k+2)$ , and obtain $\frac{(k+1)(k+2)(k+3)}3$ .

Answer: The identity holds for all positive integers n.

Rules and formulas

Natural-number sum

\sum\limits_{r=1}^{n}r=\frac{n(n+1)}2

Odd-number sum

\sum\limits_{r=1}^{n}(2r-1)=n^2

Finite geometric sum

\sum\limits_{r=0}^{n}q^r=\frac{q^{n+1}-1}{q-1}

$q$: real or complex ratio with q not equal to 1

Induction proofs involving divisibility

A divisibility statement $d\mid A_n$ means $A_n=dm$ for some integer m. This representation makes the inductive hypothesis usable.

In the next case, rewrite $A_{k+1}$ using $A_k$ plus or times an additional expression. The hypothesis supplies one multiple of d; show the remaining part is also a multiple of d.

Conclude by explicitly identifying an integer multiplier. It is not enough to say that an expression ‘looks divisible’.

Modular arithmetic can guide the rearrangement, but the induction proof must still contain its base, hypothesis, step, and conclusion when induction is requested.

Key terms

Divisibility: $d$ divides $A$ when $A=dm$ for some integer $m$ .
Integer multiplier: The integer showing that an expression is an exact multiple of the divisor.
Congruence: A modular statement describing equal remainders.

Worked examples

Power divisibility

Prove $7\mid(8^n-1)$ for $n\ge1$ .

1Base: $8^1-1=7$ .
2Assume $8^k-1=7m$ .
3Then $8^{k+1}-1=8(8^k-1)+7=7(8m+1)$ .
4Since $8m+1$ is an integer, the next case is divisible by 7.

Answer: The result holds for all positive integers n.

Another power pattern

Prove $3\mid(4^n-1)$ for $n\ge1$ .

1Base: $4-1=3$ .
2Assume $4^k-1=3m$ .
3Then $4^{k+1}-1=4(4^k-1)+3=3(4m+1)$ .

Answer: The result holds for every positive integer n.

Choose a useful rearrangement

Why is $8^{k+1}-1=8(8^k-1)+7$ useful?

1It exposes the expression in the inductive hypothesis.
2The extra term 7 is also divisible by 7.

Answer: Both parts are known multiples of 7.

Rules and formulas

Divisibility definition

d\mid A\iff A=dm\text{ for some }m\in\mathbb Z

Useful power rearrangement

a^{k+1}-1=a(a^k-1)+(a-1)

Induction proofs involving sequences and recurrences

To prove an explicit formula for a recursively defined sequence, verify the formula at the initial index, assume it at index k, and substitute it into the recurrence for $u_{k+1}$ .

The recurrence is the bridge from k to k + 1. Without using it, the inductive step does not connect the sequence terms.

If a recurrence depends on two previous terms, strong induction or two initial cases may be required. The scope of the inductive hypothesis must match the recurrence.

After simplifying $u_{k+1}$ , compare it with the claimed explicit formula after replacing n by k + 1.

Key terms

Recursive sequence: A sequence whose later terms are defined using earlier terms.
Explicit formula: A formula giving $u_{n}$ directly as a function of n.
Strong induction: An induction form that assumes all relevant earlier cases up to k.

Worked examples

Linear recurrence

Given $u_1=2$ , $u_{n+1}=3u_n+2$ , prove $u_n=3^n-1$ .

1Base: $u_1=2=3^1-1$ .
2Assume $u_k=3^k-1$ .
3Then $u_{k+1}=3(3^k-1)+2=3^{k+1}-1$ .

Answer: The formula holds for every $n\ge1$ .

Arithmetic recurrence

Given $v_1=3$ , $v_{n+1}=v_n+2$ , prove $v_n=2n+1$ .

1Base: $v_1=3=2(1)+1$ .
2Assume $v_k=2k+1$ .
3Then $v_{k+1}=2k+1+2=2(k+1)+1$ .

Answer: The formula holds for every $n\ge1$ .

Recognize a two-term dependency

A recurrence uses $u_{n+2}=u_{n+1}+u_n$ . Why may one base case be insufficient?

1The next term depends on two previous terms.
2The induction needs both starting values and access to both earlier formulas.

Answer: Verify two initial cases and use an appropriately strengthened hypothesis.

Rules and formulas

One-step recurrence proof

u_{k+1}=f(u_k)\xrightarrow{P(k)}f(g(k))=g(k+1)

Strong induction hypothesis

P(n_0),P(n_0+1),\ldots,P(k)

Induction proofs involving inequalities

Induction can prove inequalities when the k + 1 expression can be bounded using the hypothesis. Every multiplication or division must respect sign conditions.

State the valid starting index carefully. Many exponential-versus-polynomial inequalities fail for small n and become true only after a threshold.

The inductive step often requires an additional simple inequality valid in the range, such as $k+1\ge1$ . State it rather than hiding it.

A strict inequality must remain strict; do not silently replace it by equality during algebra.

Key terms

Inductive inequality: An inequality whose next case is derived using the assumed current case.
Threshold: The first integer from which an inequality remains true.

Worked examples

Basic exponential inequality

Prove $2^n\ge n+1$ for $n\ge0$ .

1Base: $2^0=1=0+1$ .
2Assume $2^k\ge k+1$ .
3Then $2^{k+1}=2\cdot2^k\ge2(k+1)$ .
4Since $2(k+1)\ge k+2$ for $k\ge0$ , $P(k + 1)$ holds.

Answer: The inequality holds for every integer $n\ge0$ .

Check a threshold

Why can the claim $2^n>n^2$ not start at $n = 2$ ?

1At $n = 2$ , both sides equal 4.
2The strict inequality fails at the proposed base.

Answer: A valid theorem needs a later starting index, such as $n = 5$ .

Preserve direction

If the hypothesis is $a_k>0$ , when can both sides be multiplied by c without reversing the inequality?

1The sign of c determines the direction.

Answer: The direction is preserved when $c>0$ ; it reverses when $c<0$ .

Rules and formulas

Example inequality chain

2^{k+1}=2\cdot2^k\ge2(k+1)\ge k+2

Proof by contradiction: logical structure

To prove a statement S by contradiction, assume its negation $\neg S$ . Use valid deductions until an impossibility appears, then conclude that the assumption was false and S must be true.

A contradiction may be a direct impossibility such as $0=1$ , a statement and its negation, a violation of a definition, or a conflict with an established hypothesis such as ‘the fraction is in lowest terms’.

For an implication $P\implies Q$ , assume P and $\neg Q$ . Do not discard the original hypothesis P.

The contradiction must depend on the negated assumption. An unrelated false statement does not prove the desired conclusion.

Key terms

Negation: The logical statement that S is false.
Contradiction: An impossible conclusion, often written ⊥.
Lowest terms: A fraction whose numerator and denominator have greatest common divisor 1.

Worked examples

Negate an implication correctly

To prove ‘if $n^2$ is even, then n is even’ by contradiction, what should be assumed?

1Keep the hypothesis that $n^2$ is even.
2Negate the conclusion that n is even.

Answer: Assume $n^2$ is even and n is odd.

Recognize a contradiction

A proof assumes $\gcd(a,b)=1$ and later proves both $a$ and $b$ are even. What is contradicted?

1If both are even, they share factor 2.
2That conflicts with greatest common divisor 1.

Answer: The lowest-terms assumption is contradicted.

Finish the logic

After deriving a contradiction from $\neg S$ , what may be concluded?

1The negated assumption cannot be true.
2Therefore S is true.

Answer: Conclude S, explicitly citing contradiction.

Rules and formulas

Contradiction schema

\neg S\implies\bot\quad\therefore\quad S

Implication by contradiction

P\land\neg Q\implies\bot\quad\therefore\quad P\implies Q

Worked proofs by contradiction

Classic contradiction proofs are valuable because their structure is reusable: assume the opposite, translate it into precise algebra, derive incompatible consequences, and state the conclusion.

In irrationality proofs, write a rational number as a lowest-terms fraction. Without that condition, proving numerator and denominator share a factor is not a contradiction.

In parity proofs, use definitions: an even integer is $2m$ , and an odd integer is $2m+1$ for some integer m.

In existence arguments such as infinitely many primes, construct a new object that cannot fit the supposedly complete list.

Key terms

Irrational: A real number that cannot be expressed as a ratio of integers with non-zero denominator.
Parity: The property of an integer being even or odd.
Constructive contradiction: A contradiction produced by building an object excluded by the assumption.

Worked examples

Irrationality of square root 2

Prove $\sqrt2$ is irrational.

1Assume $\sqrt2=\frac ab$ in lowest terms, with integers a, b and $b\ne0$ .
2Then $a^2=2b^2$ , so $a^2$ and hence a are even; write $a=2m$ .
3Substitute to get $b^2=2m^2$ , so b is even.
4Then a and b share factor 2, contradicting lowest terms.

Answer: Therefore $\sqrt2$ is irrational.

Even square implies even integer

Prove that if $n^2$ is even, then n is even.

1Assume $n^2$ is even but n is odd.
2Write $n=2m+1$ .
3Then $n^2=4m^2+4m+1=2(2m^2+2m)+1$ , which is odd.
4This contradicts the hypothesis that $n^2$ is even.

Answer: Therefore n is even.

Infinitely many primes

Prove that there are infinitely many primes.

1Assume there are finitely many primes $p_1,p_2,\ldots,p_m$ .
2Form $N=p_1p_2\cdots p_m+1$ .
3No listed prime divides N, because division leaves remainder 1.
4But every integer $N > 1$ has a prime divisor, producing a prime not on the complete list.

Answer: The finite-list assumption is contradictory, so infinitely many primes exist.

Rules and formulas

Even integer

n=2m,\qquad m\in\mathbb Z

Odd integer

n=2m+1,\qquad m\in\mathbb Z

Euclid construction

N=p_1p_2\cdots p_m+1

Disproving universal statements with counterexamples

A counterexample is one object in the stated domain for which a universal claim fails. Since ‘for every’ allows no exceptions, one valid exception disproves the whole statement.

A counterexample must satisfy every hypothesis while violating the conclusion. An object outside the domain or one that fails the hypothesis proves nothing about the claim.

Counterexamples disprove universal claims; they do not prove a different universal statement. Likewise, several supporting examples do not prove the original claim.

To disprove an implication $P\implies Q$ , find a case where P is true and Q is false. Cases with false P are not counterexamples.

Key terms

Counterexample: A valid domain case that makes a universal statement false.
Universal claim: A statement asserting a property for every member of a domain.
Converse: The implication Q ⇒ P formed by reversing P ⇒ Q.

Worked examples

Prime counterexample

Disprove ‘every prime number is odd’.

1Choose 2.
2Verify that 2 is prime and not odd.

Answer: The counterexample 2 disproves the claim.

Polynomial counterexample

Disprove ‘ $n^2+n+41$ is prime for every non-negative integer n’.

1Choose $n = 41$ , which is in the domain.
2Calculate $41^2+41+41=41(43)$ .
3This value is composite.

Answer: $n = 41$ is a counterexample.

Counterexample to a product claim

Disprove ‘if ab is even, then a and b are both even’ for integers a and b.

1Choose $a = 2$ and $b = 3$ .
2Then $ab = 6$ is even, so the hypothesis is true.
3But $b = 3$ is not even, so the conclusion is false.

Answer: $(a,b)=(2,3)$ is a counterexample.

An invalid attempted counterexample

Can $n = 3$ disprove ‘if n is divisible by 4, then n is even’?

1At $n = 3$ , the hypothesis ‘divisible by 4’ is false.
2A counterexample to an implication needs a true hypothesis.

Answer: No. $n = 3$ is not a counterexample.

Rules and formulas

Counterexample to implication

P(c)\text{ true and }Q(c)\text{ false}

Universal negation

\neg(\forall x\in D,P(x))\iff\exists c\in D,\neg P(c)

Common mistakes and IB HL exam strategy

A complete induction proof needs a defined $P(n)$ , correct base case, arbitrary k, explicit hypothesis, valid step using that hypothesis, and final conclusion. Omitting any link weakens the proof.

A contradiction proof must assume the exact negation and derive an impossibility that depends on it. A counterexample must belong to the domain and satisfy the original hypotheses.

Do not use phrases such as ‘obviously’ to hide a necessary algebraic or logical step. Concision is valuable only when every implication remains justified.

Keep notation consistent: distinguish n, the arbitrary k, and the next index k + 1. State integer conditions and divisibility multipliers explicitly.

After finishing, audit the proof: Did every line follow from a definition, hypothesis, previous line, or cited theorem? Did the final sentence answer the exact quantified claim?

Key terms

Circular reasoning: Using the desired conclusion as an assumption in its own proof.
Logical gap: An unsupported transition between two claims.
Proof audit: Checking each inference, condition and conclusion after writing.

Worked examples

Spot a circular induction step

A student begins the step with the formula for $P(k + 1)$ and simplifies it to itself. What is wrong?

1The target was assumed rather than derived.
2The inductive hypothesis did no work.

Answer: The argument is circular and does not prove $P(k)$ ⇒ $P(k + 1)$ .

Spot a missing conclusion

A proof verifies $P(1)$ and $P(k)$ ⇒ $P(k + 1)$ , then stops. What should be added?

1Combine both established components using the induction principle.

Answer: State that $P(n)$ holds for all integers in the specified range by mathematical induction.

Spot an invalid disproof

A student gives five cases where a universal claim is true and says it is proved. What is wrong? Assume the five cases do not exhaust the stated domain.

1A non-exhaustive set of supporting cases leaves untested values.
2A proof covering every allowed case is still required. Exhaustive verification is valid on a finite domain.

Answer: The argument provides evidence, not proof.

Rules and formulas

Induction checklist

P(n_0);\quad P(k)\text{ assumed};\quad P(k+1)\text{ proved};\quad\text{conclude for all }n\ge n_0

Contradiction checklist

\text{assume negation}\to\text{deduce impossibility}\to\text{reject negation}

IB HL-style practice with complete solutions

Attempt each proof before opening the solution. The set includes sums, divisibility, sequences, inequalities, contradiction and counterexample arguments.

When reviewing, check the logical structure as carefully as the algebra. A correct final formula does not repair a missing hypothesis or incomplete root of the argument.

Key terms

Complete solution: A proof containing every required logical stage and justified calculation.
Proof method mark: Credit for selecting and correctly structuring an appropriate proof method.

Worked examples

Practice 1: natural-number sum

Prove by induction that $\sum\limits_{r=1}^{n}r=\frac{n(n+1)}2$ for $n\ge1$ .

1Base: $n = 1$ gives $1=\frac{1(2)}2$ .
2Assume $\sum\limits_{r=1}^{k}r=\frac{k(k+1)}2$ .
3Then $\sum\limits_{r=1}^{k+1}r=\frac{k(k+1)}2+(k+1)=\frac{(k+1)(k+2)}2$ .
4This is $P(k + 1)$ , so the result follows by induction.

Answer: The formula holds for every positive integer n.

Practice 2: odd-number sum

Prove $\sum\limits_{r=1}^{n}(2r-1)=n^2$ for $n\ge1$ .

1Base: at $n = 1$ , both sides equal 1.
2Assume the sum to k equals $k^2$ .
3Add the next odd number: $k^2+2(k+1)-1=(k+1)^2$ .
4Conclude by induction.

Answer: The identity holds for every positive integer n.

Practice 3: geometric sum

Prove $\sum\limits_{r=0}^{n}3^r=\frac{3^{n+1}-1}{2}$ for $n\ge0$ .

1Base: $n = 0$ gives $1=\frac{3-1}{2}$ .
2Assume the formula for k.
3Add $3^{k+1}$ : $\frac{3^{k+1}-1}{2}+3^{k+1}=\frac{3^{k+2}-1}{2}$ .
4Conclude by induction.

Answer: The formula holds for every integer $n\ge0$ .

Practice 4: divisibility

Prove $3\mid(4^n-1)$ for every positive integer n.

1Base: $4^1-1=3$ .
2Assume $4^k-1=3m$ for some integer m.
3Then $4^{k+1}-1=4(4^k-1)+3=3(4m+1)$ .
4Since $4m+1$ is an integer, the next case holds.

Answer: The divisibility statement follows by induction.

Practice 5: recurrence

Given $u_1=2$ and $u_{n+1}=3u_n+2$ , prove $u_n=3^n-1$ .

1Base: $u_1=2=3^1-1$ .
2Assume $u_k=3^k-1$ .
3Then $u_{k+1}=3(3^k-1)+2=3^{k+1}-1$ .
4Conclude by induction.

Answer: The explicit formula holds for every $n\ge1$ .

Practice 6: inequality

Prove $2^n\ge n+1$ for every integer $n\ge0$ .

1Base: $2^0=1=0+1$ .
2Assume $2^k\ge k+1$ .
3Then $2^{k+1}\ge2(k+1)$ .
4Since $2(k+1)-(k+2)=k\ge0$ , $2^{k+1}\ge k+2$ .

Answer: The inequality follows by induction.

Practice 7: irrationality

Prove by contradiction that $\sqrt2$ is irrational.

1Assume $\sqrt2=\frac{a}{b}$ in lowest terms.
2Then $a^2=2b^2$ , so a is even; write $a=2m$ .
3Substitution gives $b^2=2m^2$ , so b is even.
4Both are even, contradicting lowest terms.

Answer: Therefore $\sqrt2$ is irrational.

Practice 8: parity contradiction

Prove that if $n^2$ is even, then n is even.

1Assume $n^2$ is even and n is odd.
2Write $n=2m+1$ .
3Then $n^2=2(2m^2+2m)+1$ , which is odd.
4This contradicts the even-square hypothesis.

Answer: Therefore n is even.

Practice 9: infinitely many primes

Prove by contradiction that there are infinitely many primes.

1Assume the complete finite list is $p_1,\ldots,p_m$ .
2Form $N=p_1p_2\cdots p_m+1$ .
3No listed prime divides N, but $N > 1$ must have a prime divisor.
4That divisor is a prime outside the supposedly complete list.

Answer: Therefore there are infinitely many primes.

Practice 10: prime counterexample

Disprove ‘every prime number is odd’.

1Choose 2.
2Verify that 2 is prime.
3Verify that 2 is even, not odd.

Answer: The counterexample 2 disproves the claim.

Practice 11: polynomial counterexample

Disprove ‘ $n^2+n+41$ is prime for every non-negative integer n’.

1Choose $n = 41$ .
2Calculate $41^2+41+41=41(43)$ .
3This is composite.

Answer: $n = 41$ is a valid counterexample.

Practice 12: implication counterexample

Disprove ‘if ab is even, then integers a and b are both even’.

1Choose $a = 2$ and $b = 3$ .
2Then $ab = 6$ is even, so the hypothesis holds.
3But b is odd, so the conclusion fails.

Answer: $(a,b)=(2,3)$ is a valid counterexample.

Rules and formulas

Proof workflow

\text{identify quantifier}\to\text{choose method}\to\text{justify every step}\to\text{state conclusion}

Topic summary and cheat sheet

Rigorous proof depends on matching a method to the statement's logic. Induction proves an integer chain, contradiction rejects an impossible negation, and a counterexample destroys a universal claim with one genuine exception.

Use the checklist below to audit structure, domain, hypotheses, deductions and conclusion before submitting an IB HL proof.

Key terms

Proof fluency: Selecting a valid method and communicating every logical dependency clearly.

Worked examples

Final proof audit

What should be checked before submitting a proof?

1The exact domain and statement are identified.
2Every assumption is permitted and every deduction justified.
3The chosen method is complete.
4The final sentence states the requested conclusion.

Answer: A proof is complete only when its logic and algebra both withstand the audit.

Rules and formulas

Induction structure

P(n_0);\quad P(k)\implies P(k+1);\quad\therefore\ P(n)\text{ for all }n\ge n_0

Contradiction structure

\neg S\implies\bot\quad\therefore\quad S

Counterexample structure

c\in D,\quad P(c)\text{ is false}\quad\therefore\quad\forall x\in D,P(x)\text{ is false}

Topic summary

A proof establishes why a statement must be true using valid deductions.
Checking finitely many cases does not prove a universal statement.
Always identify the domain, hypotheses, conclusion and quantifiers.
Use induction for a chain of integer-indexed statements.
Define the induction proposition $P(n)$ precisely.
Verify the first required case $P(n_0)$ .
Assume $P(k)$ for an arbitrary eligible integer k.
Use the inductive hypothesis to prove $P(k + 1)$ .
Do not assume $P(k + 1)$ or reason circularly.
Finish by invoking mathematical induction over the stated domain.
For sums, separate the k + 1 sum into the k sum and its new term.
For divisibility, express the hypothesis as an integer multiple.
For recurrences, use the recurrence to connect $u_k$ and $u_{k+1}$ .
For inequalities, justify every sign and threshold condition.
A contradiction proof assumes the exact negation of the desired statement.
For an implication by contradiction, assume the hypothesis and negate the conclusion.
The contradiction must depend on the negated assumption.
Irrationality proofs using fractions require a lowest-terms assumption.
A counterexample disproves a universal statement with one valid exception.
A counterexample to $P\implies Q$ must make P true and Q false.
An example outside the domain or failing the hypothesis is not a counterexample.
One witness can prove an existential claim, but examples cannot prove a universal claim.
Use explicit logical signposts so every proof stage is visible.
Audit every inference and restate the exact conclusion before submitting.

Meaning of mathematical proof

Key terms

Worked examples

Evidence is not proof

Identify hypothesis and conclusion

A valid existential proof

Rules and formulas

Implication

Universal statement

Existential statement

Choosing the appropriate proof method

Key terms

Worked examples

Choose induction

Choose contradiction

Choose a counterexample

Rules and formulas

Contradiction form for an implication

Disproving a universal claim

Mathematical induction: the complete logical chain

Key terms

Worked examples

Identify P(n)P(n)P(n)

Identify the starting index

Explain the two failures

Rules and formulas

Induction principle

The base case

Key terms

Worked examples

Base case for a sum

Base case for divisibility

A non-standard start

Rules and formulas

Base-case instruction

The inductive hypothesis

Key terms

Worked examples

Write a sum hypothesis

Write a divisibility hypothesis

Write a recurrence hypothesis

Rules and formulas

Inductive hypothesis

Divisibility hypothesis

The inductive step and conclusion

Key terms

Worked examples

Complete a sum step

Complete a recurrence step

Write the conclusion

Rules and formulas

Sum extension

Inductive implication

Induction proofs involving sums

Key terms

Worked examples

Sum of odd integers

Geometric sum

A polynomial summand

Rules and formulas

Natural-number sum

Odd-number sum

Finite geometric sum

Induction proofs involving divisibility

Key terms

Worked examples

Power divisibility

Another power pattern

Choose a useful rearrangement

Rules and formulas

Divisibility definition

Useful power rearrangement

Induction proofs involving sequences and recurrences

Key terms

Worked examples

Linear recurrence

Arithmetic recurrence

Recognize a two-term dependency

Rules and formulas

One-step recurrence proof

Identify $P(n)$