Proofs in Mathematics (Advanced) (Appendix)

Every result you have used since Class IX — the quadratic formula, the Pythagoras theorem, the rule that a one-one function never sends two different inputs to the same output — rests on a proof somewhere underneath it. A proof is not a worked example and it is not "checking a few cases and being fairly sure." It is a finite chain of statements, each licensed by a definition, an axiom, or a theorem already established, that carries you from what is given to what you claim. Vladimir Arnold's line captures the discipline: "Proofs are to Mathematics what calligraphy is to poetry." The poem — the theorem — can be admired on its own, but the calligraphy — the proof — is what makes it trustworthy rather than merely plausible. This appendix chapter steps back from computing answers and asks a different question: how do we know a mathematical statement is true, and how do we write down that knowledge so anyone can check it link by link? NCERT organises the answer into two families. A direct proof starts from what is given and marches forward — by one straight chain of implications, by the self-propagating machinery of mathematical induction, or by splitting the hypothesis into exhaustive cases. An indirect proof works with an equivalent statement instead — assuming the negation and hunting a contradiction, proving the contrapositive, or, when a general claim is simply false, killing it outright with one counterexample. This is enrichment material — no board or competitive exam sets numerical "proof theory" questions the way it sets calculus or algebra problems — but it is exactly the material that tells you why every other chapter is allowed to say "hence proved." We will build three complete proofs step by step, using genuine NCERT examples: a matrix power formula proved by induction, the infinitude of primes proved by contradiction, and a triangle identity proved by splitting into cases.

Key takeaways

A proof is a finite chain of statements, each justified by a definition, axiom, or previously established theorem, that leads from the given hypothesis to the claimed conclusion using only valid rules of logic.
Direct proof starts from what is given (or assumed) and reasons forward. It has three flavours in this chapter: the straightforward chain of implications, mathematical induction, and proof by cases (exhaustion).
Mathematical induction rests on one axiom about the natural numbers: if $1 \in S$ and $k \in S$ implies $k+1 \in S$ , then $S = \mathbb{N}$ . Practically, you prove $P(1)$ (the base case) and then prove $P(k) \Rightarrow P(k+1)$ (the inductive step).
Proof by cases splits the hypothesis $p$ into exhaustive alternatives $p = r \lor s \lor t$ , proves $r \Rightarrow q$ , $s \Rightarrow q$ and $t \Rightarrow q$ separately, and concludes $p \Rightarrow q$ .
Indirect proof proves an equivalent statement instead of the original: proof by contradiction assumes the negation and derives an absurdity; proof by contrapositive proves $\lnot q \Rightarrow \lnot p$ instead of $p \Rightarrow q$ (the two are logically identical).
Proof by counterexample is how a general claim ("for every $n$ ...", "every continuous function is...") gets disproved: a single case where the claim fails is sufficient, because it is a direct proof of the negation of the claim.
The contrapositive of "if $p$ then $q$ " is "if not $q$ then not $p$ " — logically equivalent to the original. The converse, "if $q$ then $p$ ", is a different statement altogether and need not be true even when the original is.
Checking a statement for many values of $n$ , however many, is never a proof for all $n$ — the classic warning example is $2^{2^n}+1$ , prime for $n = 0,1,2,3,4$ but composite at $n = 5$ .
Key idea: a mathematical proof is a finite, gapless chain of logically justified steps from hypothesis to conclusion, built either directly (straightforward deduction, induction, or exhaustive cases) or indirectly (contradiction, contrapositive, or a single decisive counterexample), and no amount of numerical pattern-checking can substitute for one.

The core insight

A proof is a finite chain of logically justified steps from hypothesis to conclusion — built directly (straightforward deduction, induction over the natural numbers, or exhaustive cases) or indirectly (contradiction, the logically equivalent contrapositive, or one counterexample that kills a false generalisation) — and no finite amount of checking examples can ever replace it.

What this chapter covers

This chapter is a short, self-contained tour of how mathematics proves things, organised the way NCERT organises it. We open with what a proof actually is — a chain of deductions, each link justified by a definition, axiom, or earlier theorem — and why "many examples agree" is not the same as "proved for all cases." We then work through the three direct methods: the straightforward chain of implications (the method you already use whenever you solve an equation "step by reason"), the principle of mathematical induction (used whenever a claim is indexed by a natural number $n$ ), and proof by cases, used when the hypothesis naturally splits into finitely many exhaustive alternatives. We move to the three indirect methods: proof by contradiction (assume the statement false, reach an absurdity), proof by contrapositive (prove the logically equivalent "not $q$ implies not $p$ " instead of " $p$ implies $q$ "), and proof by counterexample (the only way to disprove a false general claim). Three full derivations work induction, contradiction, and cases end to end on genuine NCERT theorems — a matrix power formula, the infinitude of primes, and a triangle side-length identity. The worked examples apply the straightforward chain and the contrapositive to Class 12 staples, finishing with a higher-order example on the danger of trusting a pattern too early. We close with the traps examiners flag most often, a short quiz, and an exam-style cheat sheet.

The core ideas

What is a proof

A mathematical statement is a sentence that is either true or false, never both, never neither — for example " $x^2 - 5x + 6 = 0$ " (true or false depending on $x$ ) or "the function $f(x) = 2x+5$ is one-one" (true, once you check it). A proof of a statement is a sequence of statements, each one justified by a definition, an axiom, or a proposition already established, connected by the rules of deduction. Symbolically, to prove " $p \Rightarrow q$ " (read " $p$ implies $q$ "), you exhibit a chain

p \Rightarrow r \Rightarrow s \Rightarrow \cdots \Rightarrow q,

where every single arrow is licensed by something already known to be true. Because each link in the chain is individually justified, the whole chain is justified, and so " $p \Rightarrow q$ " is true. This is the deductive method, and it is the only method that yields a proof — no number of instances where $p$ happens to hold and $q$ happens to follow can substitute for exhibiting that chain. NCERT distinguishes two broad strategies for building the chain. A direct proof starts from the hypothesis $p$ itself (or an assumption you plan to discharge) and reasons forward to $q$ . An indirect proof instead proves a logically equivalent statement — the contrapositive $\lnot q \Rightarrow \lnot p$ , or it derives a contradiction from assuming $\lnot q$ (or $\lnot(p \Rightarrow q)$ ) is true. Each of these has, within it, several distinct techniques, mapped out below.

Figure

A branching map of proof methods showing direct proof splitting into straightforward chain induction and cases, and indirect proof splitting into contradiction contrapositive and counterexample

Direct proof: the straightforward chain

The simplest and most familiar method is the straightforward approach: a chain of arguments that leads directly from what is given or assumed, using axioms, definitions, or already-proved theorems, to what is to be proved, using rules of logic. Every time you solve $x^2 - 5x + 6 = 0$ by factoring and write " $\Rightarrow$ " between lines, you are executing this method — each arrow is a deduction, and the reason for it (an identity, a property of real numbers, a previously proved rule) can be written in brackets alongside. NCERT's own reading of Example 1 spells this out explicitly: if $p$ is " $x^2-5x+6=0$ " and $q$ is " $x=3$ or $x=2$ ", the chain $p \Rightarrow r \Rightarrow s \Rightarrow \cdots \Rightarrow q$ is the proof that $p \Rightarrow q$ .

Direct proof: mathematical induction

Many statements are not single claims but a whole family indexed by a natural number $n$ — "for every $n \in \mathbb{N}$ , ...". Checking such a statement for $n = 1, 2, 3, \ldots, 100$ never proves it for $n = 101$ ; a method that handles all natural numbers at once is needed. That method is the principle of mathematical induction, resting on an axiom about $\mathbb{N}$ itself: for a subset $S \subseteq \mathbb{N}$ , if $1 \in S$ , and $k \in S$ implies $k+1 \in S$ for every $k$ , then $S = \mathbb{N}$ . In proof practice: to show $P(n)$ holds for every $n \geq j$ (usually $j=1$ ), prove two things — the base case $P(j)$ is true, and the inductive step, that $P(k)$ true implies $P(k+1)$ true for every $k \geq j$ . Together these force $P(n)$ true for every $n \geq j$ : the base case starts things off, and the inductive step, applied repeatedly, reaches every subsequent $n$ .

Direct proof: cases (exhaustion)

Sometimes the hypothesis $p$ of " $p \Rightarrow q$ " cannot be attacked in one uniform argument, but it can be split into a finite number of exhaustive alternatives — $p = r \lor s \lor t$ (read " $r$ or $s$ or $t$ "), where every instance of $p$ falls into at least one of $r$ , $s$ , $t$ . If each of the conditionals $r \Rightarrow q$ , $s \Rightarrow q$ and $t \Rightarrow q$ can be proved separately, then $(r \lor s \lor t) \Rightarrow q$ is proved, and since $p = r \lor s \lor t$ , so is $p \Rightarrow q$ . This is proof by cases, sometimes called proof by exhaustion; it is convenient only when the number of genuinely distinct cases is small (a triangle's third angle, for instance, is acute, obtuse, or a right angle — exactly three exhaustive alternatives).

Indirect proof: contradiction

Instead of proving $p \Rightarrow q$ head-on, proof by contradiction (reductio ad absurdum) assumes the statement is false and shows this assumption leads somewhere impossible. Concretely, you assume $\lnot q$ (or assume $p$ true and $q$ false) and, using valid deductions, arrive at a conclusion that contradicts something already known to be true — often the very assumption you started with, or an established fact such as "every natural number greater than 1 has a prime divisor." Since a true chain of deduction cannot produce a false conclusion from a true one, the only thing that can have gone wrong is the initial assumption; hence $\lnot q$ is false, i.e. $q$ is true.

Indirect proof: contrapositive

The contrapositive of "if $p$ then $q$ " is "if not $q$ then not $p$ ", formed by swapping the hypothesis and conclusion and negating both. The two statements are logically equivalent — proving one proves the other — so whenever $p \Rightarrow q$ is awkward to attack directly, you are free to prove $\lnot q \Rightarrow \lnot p$ instead. This must not be confused with the converse, "if $q$ then $p$ " (swap only, no negation), which is generally a different statement, not automatically true just because the original is.

Indirect proof: counterexample

A general claim of the form "for every $n$ , $P(n)$ holds" is disproved the moment a single $n$ is exhibited for which $P(n)$ fails. Such an $n$ is a counterexample. This works because disproving " $p \Rightarrow q$ for all cases" is logically the same as proving " $\lnot(p \Rightarrow q)$ for some case" — and exhibiting one concrete case where $p$ holds but $q$ fails is itself a direct proof of that negation. No matter how many cases confirm a general claim, that evidence is never a proof; but a single disconfirming case is always a complete disproof.

Figure

A domino chain showing the base case toppling the first domino and each domino toppling the next to represent the inductive step reaching every natural number

Derivations

Derivation AMathematical Induction Proves a Matrix Power Formula

NCERT's Example 3 asks for a proof that if $A = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$ , then $A^n = \begin{pmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{pmatrix}$ for every positive integer $n$ . Direct computation cannot settle this once and for all, since it asserts something about every natural number $n$ simultaneously — you cannot multiply $A$ by itself infinitely many times and check each answer by hand. The principle of mathematical induction is built for statements of exactly this shape, and working through this proof shows both halves of the method — the base case that starts the induction, and the inductive step that carries truth from any one $n$ to the next — together covering all of $\mathbb{N}$ in one finite argument.

Step 1. Let $P(n)$ be the assertion $A^n = \begin{pmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{pmatrix}$ , and establish the base case $P(1)$ : directly, $A^1 = A = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$ , exactly the claimed formula with $n=1$ . So $P(1)$ is true.

Step 2. Assume the inductive hypothesis: suppose $P(k)$ is true for some particular $k \geq 1$ , i.e. $A^k = \begin{pmatrix} \cos k\theta & \sin k\theta \\ -\sin k\theta & \cos k\theta \end{pmatrix}$ . This is a conditional "if," to be discharged once the implication is proved, not an assertion for all $k$ .

Step 3. Show $P(k)$ true forces $P(k+1)$ true. By the definition of matrix powers, $A^{k+1} = A^k \cdot A$ . Substituting the inductive hypothesis for $A^k$ and the given form for $A$ , $A^{k+1} = \begin{pmatrix} \cos k\theta & \sin k\theta \\ -\sin k\theta & \cos k\theta \end{pmatrix}\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$ .

Step 4. Multiply entry by entry. The top-left entry $\cos k\theta\cos\theta - \sin k\theta\sin\theta$ collapses, by the cosine addition identity, to $\cos(k+1)\theta$ . The top-right entry $\cos k\theta\sin\theta + \sin k\theta\cos\theta$ collapses, by the sine addition identity, to $\sin(k+1)\theta$ . The bottom-left entry is $-\sin k\theta\cos\theta - \cos k\theta\sin\theta = -\sin(k+1)\theta$ , and the bottom-right is $-\sin k\theta\sin\theta + \cos k\theta\cos\theta = \cos(k+1)\theta$ . So $A^{k+1} = \begin{pmatrix} \cos(k+1)\theta & \sin(k+1)\theta \\ -\sin(k+1)\theta & \cos(k+1)\theta \end{pmatrix}$ , exactly $P(k+1)$ .

Step 5. Conclude by the axiom underlying mathematical induction. $P(1)$ is true (Step 1), and $P(k)$ true implies $P(k+1)$ true for every $k \geq 1$ (Steps 2 to 4). So the set of $n$ for which $P(n)$ holds must be all of $\mathbb{N}$ . Hence $A^n = \begin{pmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{pmatrix}$ for every positive integer $n$ .

Go deeper · Derivation B · 2 minProof by Contradiction Shows the Primes Never End

Derivation BProof by Contradiction Shows the Primes Never End

NCERT's Example 5 proves that the set of all prime numbers is infinite, and it is the textbook illustration of proof by contradiction because the direct route — somehow constructing infinitely many primes by an explicit formula — simply is not available. Instead the strategy is to assume the opposite of what we want, that the primes form a finite list, and to manufacture from that finite list a specific number that cannot possibly be on it, which collapses the assumption. Studying this proof carefully also shows how proof by contradiction and proof by cases often work together: the final contradiction here is reached only after splitting into two exhaustive alternatives.

Step 1. State the goal and its negation. Let the statement to prove be "the set $P$ of all prime numbers is infinite." Proof by contradiction begins by assuming the negation: assume $P$ is finite, so every prime can be listed exhaustively as $p_1, p_2, \ldots, p_k$ for some finite $k$ , with no prime existing outside this list.

Step 2. Construct a number designed to test the assumption. Define $N = (p_1 p_2 p_3 \cdots p_k) + 1$ . Since $N$ is strictly larger than every number in the list, $N$ itself is not one of $p_1, \ldots, p_k$ .

Step 3. Split into the two exhaustive cases for what $N$ could be: $N$ is either prime or composite, since every integer greater than 1 is one or the other (a fact already established in earlier classes).

Step 4. Examine the case " $N$ is prime." If $N$ is prime, then by its own construction $N$ is a prime number not equal to any of $p_1, \ldots, p_k$ — but the assumption in Step 1 was that this list contains every prime. This directly contradicts the assumption.

Step 5. Examine the case " $N$ is composite." If $N$ is composite, it must have some prime divisor. But dividing $N$ by any $p_i$ from the list leaves remainder $1$ , by the very construction $N = (p_1 p_2 \cdots p_k)+1$ , so none of $p_1, \ldots, p_k$ divides $N$ . Hence $N$ 's prime divisor is some prime not in the list — again contradicting the assumption that the list is exhaustive.

Step 6. Conclude. Both exhaustive cases (Steps 4 and 5) contradict the assumption made in Step 1. Since assuming " $P$ is finite" leads to a contradiction in every possible case, the assumption must be false. Hence $P$ , the set of all prime numbers, is infinite.

Go deeper · Derivation C · 2 minProof by Cases Establishes the Projection Formula

Derivation CProof by Cases Establishes the Projection Formula

NCERT's Example 4 proves that in any triangle $ABC$ , the side $a$ satisfies $a = b\cos C + c\cos B$ — the projection formula, obtained by dropping a perpendicular from vertex $A$ to side $BC$ (extended if necessary) and reading off two right-triangle projections. This needs proof by cases rather than one uniform argument for a purely geometric reason: whether the foot of the perpendicular from $A$ lands inside segment $BC$ or outside it depends on whether angle $C$ is acute, obtuse, or exactly a right angle, and the algebra of the projections differs slightly in each configuration. Splitting the hypothesis " $ABC$ is a triangle" into these three exhaustive sub-cases, and proving the same conclusion in each, is proof by cases at work.

Step 1. Identify the hypothesis and set up the exhaustive split. Let $p$ be " $ABC$ is a triangle" and $q$ be " $a = b\cos C + c\cos B$ ". Since angle $C$ of any triangle is either acute, obtuse, or a right angle, and these three alternatives cover every possibility, split $p$ into $r$ : "angle $C$ is acute", $s$ : "angle $C$ is obtuse", and $t$ : "angle $C$ is a right angle", so that $p = r \lor s \lor t$ .

Step 2. Prove $r \Rightarrow q$ (angle $C$ acute). Drop perpendicular $AD$ from $A$ to $BC$ ; because $C$ is acute, $D$ falls between $B$ and $C$ , so $a = BC = BD + DC$ . In right triangle $ADB$ , $BD/AB = \cos B$ , so $BD = c\cos B$ . In right triangle $ADC$ , $DC/AC = \cos C$ , so $DC = b\cos C$ . Adding, $a = c\cos B + b\cos C$ , which is exactly $q$ .

Step 3. Prove $s \Rightarrow q$ (angle $C$ obtuse). Now $D$ falls outside segment $BC$ , beyond $C$ , so $a = BC = BD - DC$ . As before, in right triangle $ADB$ , $BD = c\cos B$ . But in right triangle $ADC$ , the angle at $C$ inside that right triangle is $\angle ACD = 180^\circ - C$ (the supplement, since $C$ is obtuse), so $DC/AC = \cos(180^\circ - C) = -\cos C$ , giving $DC = -b\cos C$ . Substituting, $a = c\cos B - (-b\cos C) = c\cos B + b\cos C$ , again exactly $q$ .

Step 4. Prove $t \Rightarrow q$ (angle $C$ a right angle). Here the perpendicular from $A$ meets $BC$ exactly at $C$ , so $D = C$ and triangle $ACB$ is itself right-angled at $C$ . Then $BC/AB = \cos B$ directly, so $a = c\cos B$ . Since $C = 90^\circ$ , $\cos C = 0$ , so $b\cos C = 0$ , and $a = c\cos B = c\cos B + 0 = c\cos B + b\cos C$ , which is exactly $q$ once more.

Step 5. Combine the three cases. Steps 2, 3 and 4 establish $r \Rightarrow q$ , $s \Rightarrow q$ and $t \Rightarrow q$ respectively. Since $p = r \lor s \lor t$ , it follows that $(r \lor s \lor t) \Rightarrow q$ , i.e. $p \Rightarrow q$ . Hence, for any triangle $ABC$ , $a = b\cos C + c\cos B$ .

Worked examples

Board level - solving a quadratic equation as a chain of justified deductions

Show, using the straightforward-chain method with reasons stated at each step, that if $x^2 - 5x + 6 = 0$ then $x = 3$ or $x = 2$ .

Step 1. Start from the given statement $p$ : $x^2 - 5x + 6 = 0$ . Factorise the left side using the identity $x^2 - 5x + 6 = x^2 - 3x - 2x + 6 = x(x-3) - 2(x-3) = (x-3)(x-2)$ , established by ordinary algebra. This replaces $p$ with the equivalent statement $r$ : $(x-3)(x-2) = 0$ .

Step 2. Apply the previously established theorem "for real $a, b$ , $ab = 0 \Rightarrow a = 0$ or $b = 0$ " to statement $r$ , giving statement $s$ : $x - 3 = 0$ or $x - 2 = 0$ .

Step 3. Add the same quantity to both sides of each equation in $s$ (adding equal quantities on either side does not alter an equation): $x - 3 + 3 = 0 + 3$ or $x - 2 + 2 = 0 + 2$ , which simplifies using the additive identity of real numbers to $x = 3$ or $x = 2$ . This is statement $q$ , the conclusion. The chain $p \Rightarrow r \Rightarrow s \Rightarrow q$ proves that $x^2-5x+6=0$ implies $x=3$ or $x=2$ .

Board level - proving a matrix statement using the contrapositive

Prove: "if a square matrix $A$ is invertible, then $A$ is non-singular", using the method of proof by contrapositive.

Step 1. Write the statement symbolically as $p \Rightarrow q$ , where $p$ is "matrix $A$ is invertible" and $q$ is " $A$ is non-singular" (that is, $|A| \neq 0$ ). Instead of proving $p \Rightarrow q$ directly, form its contrapositive, $\lnot q \Rightarrow \lnot p$ : "if $A$ is not non-singular, then $A$ is not invertible."

Step 2. Assume $\lnot q$ : $A$ is not non-singular, which by definition means $A$ is singular, i.e. $|A| = 0$ .

Step 3. Recall the formula $A^{-1} = \dfrac{\operatorname{adj} A}{|A|}$ for the inverse of a square matrix. Since $|A| = 0$ (Step 2), this expression is undefined — division by zero is not permitted — so $A^{-1}$ does not exist, meaning $A$ is not invertible. This is exactly $\lnot p$ .

Step 4. Having proved $\lnot q \Rightarrow \lnot p$ , invoke the logical equivalence of a statement and its contrapositive: $\lnot q \Rightarrow \lnot p$ is true if and only if $p \Rightarrow q$ is true. Hence "if $A$ is invertible, then $A$ is non-singular" is proved.

Higher-order thinking (HOTS) - disproving a plausible-looking generalisation with a counterexample

Fermat once conjectured that $2^{2^n} + 1$ is prime for every natural number $n$ . Investigate this claim for $n = 0, 1, 2, 3, 4$ and determine, with a counterexample if needed, whether the generalisation "for every $n \in \mathbb{N}$ , $2^{2^n}+1$ is prime" is true.

Step 1. Compute the first several cases directly. For $n=1$ : $2^{2^1}+1 = 2^2+1 = 5$ , prime. For $n=2$ : $2^{2^2}+1 = 2^4+1 = 17$ , prime. For $n=3$ : $2^{2^3}+1 = 2^8+1 = 257$ , prime. Each check succeeds, and it is tempting — but not logically permitted — to treat this run of successes as a proof for all $n$ .

Step 2. Test the claim at $n=5$ : $2^{2^5}+1 = 2^{32}+1 = 4{,}294{,}967{,}297$ . Checking this number for factors (a computation, not a leap of faith) shows $4{,}294{,}967{,}297 = 641 \times 6{,}700{,}417$ , a genuine product of two integers greater than 1.

Step 3. Interpret the result logically. $4{,}294{,}967{,}297$ is composite, not prime, so the statement " $2^{2^n}+1$ is prime" fails at $n=5$ . By the method of proof by counterexample, exhibiting this one failing case is a complete, direct proof of the negation of the general claim — no further checking of any other $n$ is needed. The generalisation "for every $n \in \mathbb{N}$ , $2^{2^n}+1$ is prime" is false; $n=5$ is a counterexample. Timing note: about 3 to 4 minutes if $641$ is given or recalled as a known factor (verifying the product is quick); longer if you must discover it by trial division, which is not expected in an exam. The transferable skill is not the arithmetic but the logic: Step 1's three or four successes are evidence, not proof, and the moment a single failing case is found, the search stops — one clean counterexample closes the argument completely.

Common traps

The single most common error in this chapter is treating a run of confirming examples as though it were a proof — mistaking inductive evidence ("pattern-spotting") for the formal deductive method of mathematical induction, entirely different things despite the similar name. Three further slips recur. Students confuse the converse with the contrapositive — swapping hypothesis and conclusion without negating gives the converse, generally not equivalent to the original, whereas swapping and negating gives the contrapositive, which always is. Students state the inductive step incorrectly, writing "prove $P(k+1)$ is true" rather than "prove $P(k) \Rightarrow P(k+1)$ " — induction never asserts $P(k)$ true on its own, only that its truth would force $P(k+1)$ 's. And in proof by cases, students list cases that are not exhaustive, forgetting the right-angle case and covering only "acute or obtuse," leaving a logical gap the proof does not close.

Common trapNo finite number of confirming examples is ever a proof — one counterexample, on the other hand, is a complete disproof, forever

The Fermat-number example above is the canonical warning: $2^{2^n}+1$ is prime for $n=0,1,2,3,4$ — five confirmations in a row — and yet the claim "prime for every $n$ " is false, refuted at $n=5$ . The same lesson appears with "every continuous function is differentiable": $x^2$ , $e^x$ and $\sin x$ are all continuous and differentiable, which might suggest the pattern is universal, until $\phi(x) = |x|$ is checked — continuous everywhere but not differentiable at $x=0$ , a single counterexample that disproves the generalisation completely. Do not write "true for $n=1,2,3$ , hence true for all $n$ " on any proof; either complete the inductive step or search actively for a counterexample.

You should be able to…

State precisely what a mathematical proof is, and explain why a chain of confirmed examples is not one.
Distinguish direct proof from indirect proof, and name the three methods within each family covered in this chapter.
Write a straightforward-chain proof for a simple algebraic or functional statement, stating the reason for each deductive step.
State the principle of mathematical induction precisely, including both the base case and the inductive step, and apply it to prove a formula indexed by $n$ .
Recognise when a hypothesis splits naturally into exhaustive cases, and construct a proof by cases that covers every alternative.
Construct a proof by contradiction, correctly identifying the assumption being negated and the contradiction it produces.
Form the contrapositive of a conditional statement correctly, distinguish it from the converse, and use the equivalence to prove a statement indirectly.
Disprove a false general claim with a single, correctly justified counterexample.

Ask the tutor: Want to see how these same six proof methods apply to a claim of your own choosing, worked out step by step?

What exactly is the difference between a conjecture and a theorem?

A conjecture is a mathematical statement believed to be true, often on the strength of many confirming examples or a compelling pattern, but for which no proof yet exists. A theorem is a statement that has been proved — a finite, checkable chain of deductive steps has been exhibited linking accepted axioms and definitions to the conclusion. Fermat's guess that $2^{2^n}+1$ is always prime was a conjecture; once $n=5$ was checked and found composite, the conjecture was disproved rather than becoming a theorem. Many famous conjectures in mathematics (Goldbach's conjecture, for instance) remain unresolved for centuries precisely because a proof, or a counterexample, is far harder to produce than the pattern is to notice.

Why does proving the contrapositive count as proving the original statement?

Because "if p then q" and "if not q then not p" are logically equivalent — they are true or false together in every possible case, which can be checked with a truth table. If p is true and q is false, both the original statement and its contrapositive are false (a false implication has a true hypothesis and false conclusion); in every other combination of truth values, both statements come out true. Since the two statements never disagree, a valid proof of either one is automatically a valid proof of the other. This is genuinely useful when "not q implies not p" happens to be an easier chain to build than "p implies q" directly, as in the invertible-matrix example in this chapter.

In induction, why can we not just check n equals 1, 2, 3 and stop there?

Because checking finitely many values, however many, only tells you the statement holds for those specific values — it says nothing, by itself, about the infinitely many values you have not checked. Mathematical induction gets around this with a different kind of argument: instead of checking every $n$ one at a time (impossible, since $\mathbb{N}$ is infinite), you prove one base case and then prove that truth at any single value $k$ forces truth at the next value $k+1$ . That second fact is a single, finite proof that covers the "domino effect" from any $k$ to $k+1$ simultaneously, which is what lets a finite argument reach every natural number. Skipping the inductive step and relying only on checked cases is not induction at all — it is unproven pattern-matching.

Is proof by contradiction the same as proof by contrapositive?

They are related but distinct. Proof by contrapositive proves "if p then q" by proving the specific equivalent statement "if not q then not p" — you still end with a direct, forward chain of deduction, just starting from $\lnot q$ instead of $p$ . Proof by contradiction is more general: you assume the statement to be proved is false (which could mean assuming $p$ true and $q$ false, or assuming the negation of any statement at all, not necessarily of implication form) and derive any contradiction whatsoever — not necessarily $\lnot p$ , but a clash with any known fact, or even with the assumption itself, as in the infinitude-of-primes proof. Every contrapositive proof can be viewed as a contradiction proof, but not every contradiction proof follows the contrapositive's specific pattern.