Appendix E — Proof Techniques

For math majors, and for anyone who wants to read the "Formal" rigor tier of this book with confidence.

Calculus is where most students first meet the demand to prove — not merely compute. The Math Major Sidebars scattered through this book all assume the toolkit below. This appendix is a standalone reference: read it once now, and return to it whenever a Formal-tier proof leaves you wondering "but why is that step allowed?"

We organize by technique. Each section gives the logical shape of the method and then a worked example drawn from calculus itself, so you see the tool used on the material you actually care about.

1. What a Proof Is

A proof is a finite chain of statements, each one either an accepted axiom, a previously established result, or a logical consequence of earlier statements, ending at the claim you set out to establish. Nothing more, nothing less. A proof is not persuasion; it is deduction. A picture can convince you; only an argument can prove.

Most theorems have the form hypothesis $\implies$ conclusion: "If $H$, then $C$." The hypothesis $H$ is what you are given (and may freely assume); the conclusion $C$ is what you must reach. Confusing the two is the single most common beginner error. You are never allowed to assume $C$ and work backward to $H$ — that proves the converse, a different (and often false) statement.

Many calculus statements are quantified. Two symbols do almost all the work:

  • $\forall$ ("for all") — the claim must hold for every object of the stated type. To prove $\forall x,\, P(x)$, you fix an arbitrary $x$ and prove $P(x)$ using nothing special about it.
  • $\exists$ ("there exists") — at least one object works. To prove $\exists x,\, P(x)$, you must produce one (or prove one must exist).

The order of quantifiers matters enormously. Compare:

$$\forall \varepsilon > 0,\ \exists \delta > 0 : \cdots \qquad \text{versus} \qquad \exists \delta > 0,\ \forall \varepsilon > 0 : \cdots$$

The first (the real definition of a limit) lets $\delta$ depend on $\varepsilon$. The second demands one $\delta$ that works for all $\varepsilon$ at once — a far stronger, usually impossible, demand. Reading quantifiers left to right, in order, is the discipline the entire ε–δ machinery (Chapter 3) rests on.

To negate a quantified statement, flip each quantifier and negate the inside: the negation of $\forall x,\, P(x)$ is $\exists x,\, \neg P(x)$. This is the engine behind every proof by contradiction below.

2. Direct Proof

The default. Assume the hypothesis, apply definitions and known theorems step by step, arrive at the conclusion.

Worked example — Differentiability implies continuity (Ch. 5).

Why we care. This justifies treating "smooth" functions as well-behaved: differentiability is a stronger condition than continuity, so every derivative computation silently assumes continuity for free.

Claim. If $f$ is differentiable at $a$, then $f$ is continuous at $a$.

Key Idea. Continuity at $a$ means $\lim_{x \to a} f(x) = f(a)$, equivalently $\lim_{x \to a}\big(f(x) - f(a)\big) = 0$. We manufacture that difference out of the difference quotient, which we are given converges.

Formal Proof. Assume $f'(a)$ exists, i.e. $\displaystyle f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ is a finite number. For $x \neq a$ write

$$f(x) - f(a) = \frac{f(x) - f(a)}{x - a} \cdot (x - a).$$

Take the limit as $x \to a$. By the product law for limits,

$$\lim_{x \to a}\big(f(x) - f(a)\big) = \left(\lim_{x \to a}\frac{f(x) - f(a)}{x - a}\right)\left(\lim_{x \to a}(x - a)\right) = f'(a) \cdot 0 = 0.$$

Hence $\lim_{x \to a} f(x) = f(a)$, which is exactly continuity at $a$. $\blacksquare$

What it means. The converse fails — $f(x) = |x|$ is continuous but not differentiable at $0$ — so "differentiable" sits strictly inside "continuous." (See §9 on why you must never assume the converse.)

3. Proof by Contradiction

To prove statement $S$, assume $\neg S$ and derive an absurdity (a statement and its negation both holding). Since a true premise cannot lead to a contradiction, $\neg S$ must be false, so $S$ is true. This is reductio ad absurdum.

Worked example A — $\sqrt 2$ is irrational.

Key Idea. Rationality forces a parity contradiction in a fraction already in lowest terms.

Formal Proof. Suppose, for contradiction, that $\sqrt 2 \in \mathbb{Q}$. Then $\sqrt 2 = p/q$ with $p, q \in \mathbb{Z}$, $q \neq 0$, and the fraction in lowest terms (so $p, q$ share no common factor). Squaring, $2 = p^2/q^2$, i.e. $p^2 = 2q^2$. Then $p^2$ is even, so $p$ is even (an odd number squared is odd), say $p = 2k$. Substituting, $4k^2 = 2q^2$, so $q^2 = 2k^2$, making $q^2$ — hence $q$ — even as well. But then $p$ and $q$ share the factor $2$, contradicting "lowest terms." Therefore $\sqrt 2$ is irrational. $\blacksquare$

Worked example B — A sequence cannot have two distinct limits.

Why we care. This is what lets us write "the limit," with a definite article. It is also the prototypical analysis contradiction proof, reused throughout Chapters 21–22.

Claim. If $a_n \to L$ and $a_n \to M$, then $L = M$.

Key Idea. If $L \neq M$ the two limits would eventually force every term to live inside two disjoint neighborhoods at once — impossible.

Formal Proof. Suppose $L \neq M$, and set $\varepsilon = \tfrac{1}{2}|L - M| > 0$. Because $a_n \to L$, there is $N_1$ with $|a_n - L| < \varepsilon$ for all $n > N_1$. Because $a_n \to M$, there is $N_2$ with $|a_n - M| < \varepsilon$ for all $n > N_2$. For any $n > \max(N_1, N_2)$, the triangle inequality gives

$$|L - M| \le |L - a_n| + |a_n - M| < \varepsilon + \varepsilon = 2\varepsilon = |L - M|.$$

So $|L - M| < |L - M|$, an absurdity. Hence $L = M$. $\blacksquare$

Warning. Contradiction is powerful but easy to overuse. If a direct proof is available, prefer it: contradiction proofs are harder to read because the reader must hold a false assumption in mind throughout. Reach for contradiction when the conclusion is a negation, a uniqueness, or an irrationality/impossibility claim.

4. Proof by Contrapositive

The statement $H \implies C$ is logically equivalent to its contrapositive $\neg C \implies \neg H$ (not to be confused with the converse $C \implies H$, which is not equivalent). When the negations are easier to manipulate, prove the contrapositive instead.

Worked example. Show: if $f$ is not continuous at $a$, then $f$ is not differentiable at $a$.

This is the contrapositive of "differentiable $\implies$ continuous" (§2). Because we already proved that direct implication, the contrapositive holds automatically — no new work. This illustrates the practical payoff: proving one direction gives you the contrapositive for free, and you may invoke whichever phrasing is more convenient in a later argument.

A cleaner standalone example: prove that if $n^2$ is even then $n$ is even. The direct route is awkward (what is the square root of an even number?). The contrapositive — "if $n$ is odd then $n^2$ is odd" — is immediate: $n = 2k+1 \Rightarrow n^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$, odd. Done. (We quietly used this fact inside the $\sqrt2$ proof of §3.)

5. Proof by Induction

To prove $P(n)$ holds for all integers $n \ge n_0$:

  1. Base case — prove $P(n_0)$ directly.
  2. Inductive step — assume $P(k)$ (the inductive hypothesis) for an arbitrary $k \ge n_0$, and prove $P(k+1)$.

The two together topple the whole infinite row of dominoes. Induction is the right tool whenever a claim is indexed by a natural number: a finite sum, a recursively defined object, or a derivative rule for integer powers.

Worked example A — The closed form $\displaystyle\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$.

Base case ($n=1$): the left side is $1$; the right side is $\frac{1\cdot 2}{2} = 1$. ✓

Inductive step. Assume $\sum_{i=1}^{k} i = \frac{k(k+1)}{2}$. Then

$$\sum_{i=1}^{k+1} i = \left(\sum_{i=1}^{k} i\right) + (k+1) = \frac{k(k+1)}{2} + (k+1) = (k+1)\!\left(\frac{k}{2} + 1\right) = \frac{(k+1)(k+2)}{2},$$

which is the formula with $n = k+1$. By induction it holds for all $n \ge 1$. $\blacksquare$ (This identity underlies the exact evaluation of Riemann sums in Chapter 13.)

Worked example B — The power rule for positive integers (Ch. 5).

Claim. For every integer $n \ge 1$, $\dfrac{d}{dx}\,x^n = n x^{n-1}$.

Key Idea. Bootstrap from $n$ to $n+1$ using the product rule, which we treat as already established.

Base case ($n=1$): $\frac{d}{dx}\,x = 1 = 1\cdot x^{0}$. ✓

Inductive step. Assume $\frac{d}{dx}\,x^k = k x^{k-1}$. Write $x^{k+1} = x \cdot x^k$ and apply the product rule:

$$\frac{d}{dx}\,x^{k+1} = \frac{d}{dx}\big(x \cdot x^{k}\big) = (1)\cdot x^{k} + x \cdot \big(k x^{k-1}\big) = x^k + k x^{k} = (k+1)x^{k}.$$

Since $(k+1)x^{(k+1)-1} = (k+1)x^k$, the formula holds for $n = k+1$. By induction it holds for all $n \ge 1$. $\blacksquare$

Math Major Sidebar. The rational and real-exponent versions of the power rule are not proved by induction — there is no "next" real number. They require logarithmic differentiation or the chain rule applied to $x^r = e^{r \ln x}$. Induction is a tool for the integers only; do not stretch it past its domain.

A variant, strong induction, lets the inductive step assume $P(j)$ for all $j \le k$, not just $P(k)$. It is logically equivalent to ordinary induction and is convenient when $P(k+1)$ depends on several earlier cases.

6. The ε–δ Method — The Signature Calculus Proof

Every informal "gets arbitrarily close" in calculus is made precise by ε–δ. This is the technique that separates calculus the computation from calculus the theory, introduced formally in Chapter 3.

Definition. $\displaystyle\lim_{x \to a} f(x) = L$ means:

$$\forall\, \varepsilon > 0,\ \exists\, \delta > 0 : \quad 0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon.$$

Read it as a game. The adversary names a tolerance $\varepsilon > 0$ on the output. You must respond with a tolerance $\delta > 0$ on the input such that staying within $\delta$ of $a$ (but not at $a$ — note $0 < |x-a|$) guarantees staying within $\varepsilon$ of $L$. If you can always win, the limit holds.

The method. Proofs come in two stages, and you should keep them visually separate:

  1. Scratch work (find $\delta$). Start from the target $|f(x) - L| < \varepsilon$ and algebraically extract a factor of $|x - a|$, bounding the rest. This tells you what $\delta$ to pick. This stage is exploratory and does not belong in the final proof.
  2. The proof (verify $\delta$ works). State your $\delta$ up front, then show the implication holds.

Worked example A — a linear limit. Prove $\displaystyle\lim_{x \to 2}(3x + 1) = 7$.

Scratch: $|(3x+1) - 7| = |3x - 6| = 3|x - 2|$. To force this below $\varepsilon$, demand $|x - 2| < \varepsilon/3$.

Proof. Let $\varepsilon > 0$. Choose $\delta = \varepsilon/3$. If $0 < |x - 2| < \delta$, then $|(3x+1) - 7| = 3|x - 2| < 3\delta = \varepsilon$. $\blacksquare$

Worked example B — a nonlinear limit (the case beginners must master). Prove $\displaystyle\lim_{x \to 3} x^2 = 9$.

Scratch: $|x^2 - 9| = |x - 3|\,|x + 3|$. The factor $|x + 3|$ is not constant, so we tame it first by agreeing in advance that $\delta \le 1$. Then $|x - 3| < 1$ gives $2 < x < 4$, so $|x + 3| < 7$. Now $|x^2 - 9| < 7|x - 3|$, and to beat $\varepsilon$ we want $|x - 3| < \varepsilon/7$. We need both constraints, so take $\delta = \min(1, \varepsilon/7)$.

Proof. Let $\varepsilon > 0$. Choose $\delta = \min(1, \varepsilon/7)$. Suppose $0 < |x - 3| < \delta$. Since $\delta \le 1$ we have $|x - 3| < 1$, hence $|x + 3| = |(x-3) + 6| \le |x - 3| + 6 < 7$. Since also $|x - 3| < \varepsilon/7$,

$$|x^2 - 9| = |x - 3|\,|x + 3| < \frac{\varepsilon}{7}\cdot 7 = \varepsilon. \qquad \blacksquare$$

The "restrict $\delta \le 1$ to bound the wild factor" maneuver is the workhorse of every polynomial and rational ε–δ proof.

The ε–N analogue for sequences (Ch. 21–22). Sequences have no continuous input variable, so $\delta$ on the input is replaced by an index threshold $N$:

$$\lim_{n \to \infty} a_n = L \iff \forall\, \varepsilon > 0,\ \exists\, N \in \mathbb{N} : \quad n > N \implies |a_n - L| < \varepsilon.$$

Worked example. Prove $\displaystyle\lim_{n \to \infty} \frac{1}{n} = 0$.

Proof. Let $\varepsilon > 0$. By the Archimedean property there is an integer $N$ with $N > 1/\varepsilon$. If $n > N$, then $\left|\frac{1}{n} - 0\right| = \frac{1}{n} < \frac{1}{N} < \varepsilon$. $\blacksquare$

The grammar is identical to ε–δ — name the tolerance, exhibit the threshold, verify the inequality — which is why mastering one transfers directly to the other.

7. "If and Only If," and Uniqueness Proofs

Biconditionals. A statement $P \iff Q$ ("$P$ if and only if $Q$") asserts two implications. You must prove both:

  • ($\Rightarrow$) $P \implies Q$, and
  • ($\Leftarrow$) $Q \implies P$.

Label the two halves explicitly. A frequent error is proving only the direction that comes easily and declaring victory. For instance, "$f$ has a horizontal tangent at $c$ $\iff$ $f'(c) = 0$" needs both: that a horizontal tangent forces $f'(c)=0$, and that $f'(c)=0$ produces a horizontal tangent. (Fermat's theorem, used in Chapter 9, supplies one direction for interior extrema — and the converse famously fails at $f(x)=x^3$, $c=0$, a reminder that the two directions are genuinely independent.)

Uniqueness. To prove "there is a unique object with property $P$," prove two things:

  1. Existence — at least one such object exists (often constructive, §8).
  2. Uniqueness — if $u$ and $v$ both have property $P$, then $u = v$.

The uniqueness half is almost always run by assuming two candidates and forcing them equal. We already did exactly this for limits in §3 (Worked Example B): existence is the limit definition; uniqueness is the contradiction argument. The same two-step pattern proves uniqueness of antiderivatives up to a constant (Ch. 14): if $F' = G' = f$ on an interval, then $(F - G)' = 0$, and the Mean Value Theorem (Ch. 9) forces $F - G$ to be constant.

8. Counterexamples

To disprove a universal claim $\forall x,\, P(x)$, you need exactly one $x$ where $P$ fails. One counterexample beats any amount of supporting evidence. Constructing sharp counterexamples is as much a part of analysis as constructing proofs — they map the precise boundary of a theorem.

Worked example — $a_n \to 0$ does not imply $\sum a_n$ converges (Ch. 22).

The "$n$th term test" says: if $\sum a_n$ converges, then $a_n \to 0$. Students routinely assume the converse. It is false, and the harmonic series is the canonical counterexample. Take $a_n = 1/n$. Then $a_n \to 0$, yet $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges. Group the terms:

$$1 + \tfrac12 + \underbrace{\left(\tfrac13 + \tfrac14\right)}_{>\,\frac12} + \underbrace{\left(\tfrac15 + \cdots + \tfrac18\right)}_{>\,\frac12} + \underbrace{\left(\tfrac19 + \cdots + \tfrac1{16}\right)}_{>\,\frac12} + \cdots$$

Each parenthesized block of $2^{k-1}$ terms sums to more than $\frac12$ (every term in a block exceeds the last, which is $\frac{1}{2^k}$, and there are $2^{k-1}$ of them, totaling $\frac12$). Infinitely many blocks each exceeding $\frac12$ push the partial sums past every bound, so the series diverges. The terms shrinking to zero is necessary for convergence but nowhere near sufficient. $\blacksquare$

The Key Insight. A counterexample does not just refute a claim; it locates the missing hypothesis. The harmonic series shows that "$a_n \to 0$" must be supplemented (by a rate of decay — see the integral and comparison tests) before convergence follows.

Keep a mental zoo of standard counterexamples: $|x|$ (continuous, not differentiable), $x^3$ ($f'=0$ at a non-extremum), the harmonic series (terms $\to 0$, series diverges), and $\sin(1/x)$ (a limit that fails to exist by oscillation).

9. Common Logical Pitfalls

These are the traps that turn a confident argument into a wrong one. Learn to spot them in your own writing.

  • Affirming the consequent. From $H \implies C$ and the fact that $C$ holds, concluding $H$. Invalid — this is mistaking a statement for its converse. "Differentiable $\Rightarrow$ continuous; $f$ is continuous; therefore $f$ is differentiable" is exactly this fallacy ($|x|$ at $0$ refutes it).

  • Assuming what you want to prove (begging the question / circular reasoning). Using the conclusion, or something equivalent to it, as a step. In ε–δ proofs this sneaks in as "assume $|f(x)-L| < \varepsilon$" at the start — but that is the goal, not a given. You may only assume $0 < |x - a| < \delta$.

  • Dividing by zero (or by a possibly-zero quantity). Cancelling $(x - a)$ from both sides of an equation is legal only when $x \neq a$ — which, fortunately, the "$0 < |x - a|$" in the limit definition guarantees. Outside that protection, every cancellation needs an explicit "and this factor is nonzero because…".

  • Confusing necessary and sufficient. "$f'(c) = 0$" is necessary for an interior extremum but not sufficient (Ch. 9). Swapping the two is the structural cousin of affirming the consequent.

  • Quantifier swap. Silently exchanging $\forall \varepsilon\,\exists\delta$ for $\exists\delta\,\forall\varepsilon$ — the difference between pointwise and uniform continuity, and a genuine error even experts must guard against.

  • Proof by example. A single case verifying $P$ never proves $\forall x,\,P(x)$ (though, by §8, a single case can disprove it). Diagrams persuade; they do not prove.

10. Reading Proofs Actively, and Attacking "Prove That…" Problems

Reading. A proof is not prose to skim; it is a program to run in your head. As you read each line, ask: Which rule justifies this step? Where was each hypothesis used? What would break if I dropped it? Keep paper beside you and re-derive every "clearly" and "it follows that." If you cannot reconstruct a step, you have not read it — you have only looked at it. When a proof finishes, restate its strategy in one sentence ("contradiction via parity," "ε–δ with the factor bounded by restricting $\delta \le 1$"). That sentence is what you will actually remember.

Writing — a workflow for "Prove that $H \implies C$":

  1. Translate. Write $H$ and $C$ in precise symbols, expanding every definition (especially limits, continuity, convergence) into its quantified form. Half of all calculus proofs are won at this step: once "continuous at $a$" becomes "$\forall \varepsilon\,\exists\delta\,\dots$", the next move is usually forced.
  2. Choose a technique. Negation in the conclusion or a uniqueness/impossibility claim → contradiction or contrapositive. A statement indexed by $n \in \mathbb{N}$ → induction. A limit → ε–δ / ε–N. A biconditional → split into two implications.
  3. Do scratch work separately. For ε–δ, find $\delta$ by working backward from the target inequality — but never present that backward reasoning as the proof.
  4. Write forward. The final proof flows from hypothesis to conclusion, each step justified, no gaps. State your $\delta$, $N$, or chosen object before you use it.
  5. Audit against §9. Did you divide by something possibly zero? Assume the conclusion? Affirm a converse? Swap quantifiers?
  6. End deliberately. Close with $\blacksquare$ and a one-line statement of what was shown.

Math Major Sidebar. When stuck, try to prove it is false — attempt a counterexample. Either you find one (and the claim was wrong, or you misread the hypotheses), or the obstruction you keep hitting reveals exactly which hypothesis makes the theorem true. This adversarial back-and-forth, due in spirit to Pólya, is how working mathematicians actually find proofs.

Standard Calculus Proofs Worth Mastering

These recur in any rigorous treatment; each is an exercise in one or more techniques above.

  1. Differentiability $\implies$ continuity — direct (Ch. 5; §2 here).
  2. The power rule for positive integers — induction (Ch. 5; §5 here).
  3. Uniqueness of the limit of a sequence — contradiction (Ch. 21; §3 here).
  4. Divergence of the harmonic series — counterexample / grouping (Ch. 22; §8 here).
  5. Rolle's Theorem, then the Mean Value Theorem from it — existence via the Extreme Value Theorem (Ch. 9).
  6. Fundamental Theorem of Calculus, Part 1, from the accumulation function — direct, via ε and the squeeze (Ch. 14).
  7. The geometric series formula and the ratio test — algebra plus ε–N (Ch. 22).
  8. Taylor's theorem with the Lagrange remainder — repeated MVT (Ch. 23).

Where to Go Deeper

  • Velleman, How to Prove It — the definitive first book on proof structure, quantifiers, and logic; start here if §1 felt new.
  • Spivak, Calculus — calculus done as analysis; the starred exercises are ε–δ proofs and reward exactly the skills above.
  • Abbott, Understanding Analysis — the cleanest modern account of the ε–δ and ε–N world this appendix lives in.
  • Pólya, How to Solve It — heuristics for finding a proof, not just checking one; pairs with §10.