Proof Mastery: Exposition, Lemmas & Scratchpad Strategy

Study Sheet

Proof Mastery: Exposition, Lemmas & Scratchpad Strategy

How a 7 becomes a full-score proof: structure, lemmas, and what to do with the first ninety minutes

Rigorous Exposition

Concept
Structure the reader can follow

State what you will prove. Define every symbol. Announce the method ("we induct on nn", "suppose for contradiction"). Prove claims in the order they are used. Handle degenerate cases explicitly. End with the conclusion in the words of the problem.

Graders read fast and grade what is on the page. A solution whose steps are correct but scattered across the page, with variables introduced mid-equation and cases handled "similarly", receives partial credit for the parts the grader can verify. A weaker idea written in order, with every case, often outscores a stronger idea written as scratch work. Structure is not decoration; it is the grading unit.

Tip
Forbidden phrases

"Clearly", "obviously", "it is easy to see", "similarly" (unless the symmetry is genuine and stated), "by a well-known theorem" (name it), "WLOG" (say what symmetry justifies it). Each hides exactly the step a grader will test.

Concept
Proof by picture is not a proof

A diagram supports the exposition and must not carry any logical weight. Every configuration claim ("PP lies between BB and CC", "the circles meet at two points") is either proved or stated as an assumption that is handled in a separate case.

Geometry solutions lose the most points here. The fix is a habit: after drawing, list every "obvious" fact from the picture and mark each as proved, assumed, or case-split. Angles should be directed or the configuration should be fixed by an explicit assumption with the other case addressed.

Lemma Generation

Concept
Break the problem into claims

Before solving, write the two or three intermediate facts that would make the problem easy ("Lemma 1: ff is injective. Lemma 2: f(0)=0f(0) = 0. Then…"). Prove them separately, each with its own short proof, then assemble.

A massive problem is a sequence of small ones. Lemma generation is the skill of guessing the right small ones — usually by asking "what property of the object would make the conclusion one line?" and then "can I prove that property alone?" Each lemma proved is partial credit even if the assembly fails, and the structure keeps the write-up readable.

Example
A functional-equation skeleton

Find all f:Q→Qf: \mathbb Q \to \mathbb Q with f(x+f(y))=f(x)+yf(x + f(y)) = f(x) + y. Lemma 1: ff is injective — if f(y1)=f(y2)f(y_1) = f(y_2) the equation gives f(x)+y1=f(x)+y2f(x) + y_1 = f(x) + y_2. Lemma 2: f(f(y))=f(0)+yf(f(y)) = f(0) + y (set x=0x = 0), so ff is surjective. Lemma 3: f(0)=0f(0) = 0 — pick y0y_0 with f(y0)=0f(y_0) = 0; then f(x)=f(x)+y0f(x) = f(x) + y_0 gives y0=0y_0 = 0. Assembly: now f(f(y))=yf(f(y)) = y; substituting y→f(y)y \to f(y) in the original equation gives f(x+y)=f(x)+f(y)f(x + y) = f(x) + f(y), so ff is additive on Q\mathbb Q, hence f(x)=cxf(x) = cx; the involution forces c2=1c^2 = 1. Verification: both f(x)=xf(x) = x and f(x)=−xf(x) = -x satisfy the equation. Three lemmas, one assembly, one check — each a separate paragraph in the write-up.

Side note
Where it goes wrong

A lemma must be proved for the objects in the problem, not for the ones in your head. A common false step in functional equations: “ff is injective, hence surjective.” That holds for maps of a finite set to itself and fails on infinite sets: f(n)=2nf(n) = 2n on Z\mathbb Z is injective and misses every odd number. In the skeleton above, surjectivity came from a separate identity, f(f(y))=f(0)+yf(f(y)) = f(0) + y, whose right side takes every value — that is the kind of reason a lemma needs.

Scratchpad Strategy

Concept
The first ninety minutes

Read all three problems. Spend twenty minutes on small cases of each: compute n=1,2,3,4n = 1, 2, 3, 4; draw the configuration accurately with a compass; test the inequality at equality and at extremes. Guess the answer or the key structure. Only then choose which problem to write.

Small cases are not a warm-up; they are the method. The answer to "find all nn" is visible at n≤6n \le 6; the equality case of an inequality tells you which classical inequality to use; the pattern in a recurrence suggests the invariant. Organize the scratch work: one problem per page, guesses boxed, dead ends crossed out with a note on WHY they died (so you do not retry them).

Concept
Guess the structure, then prove it

Once small cases suggest "the answer is nn even" or "ff must be linear", the proof splits into two halves: show the guessed objects work (usually easy) and show nothing else does (the real problem). Write both; the easy half is points on the table.

A common failure: proving that the answer works and forgetting to prove nothing else does, or the reverse. "Find all" means both directions, and the write-up should have visible headers for each — "Verification" and "Uniqueness" — so the grader and you can see that both were done.

Tip
Time budget for a 4.5-hour session

Read + small cases on all three: 30 min. Full attack on the most accessible: 90 min including write-up. Second problem: 90 min. Last 30 min: reread both write-ups as a hostile grader, fix gaps, add missing cases, box the answers. Problem 3 only with time left AFTER the reread.

Try it
Try it: guess from small cases

A sequence has a1=1a_1 = 1 and an+1=an+gcd⁡(n,an)a_{n+1} = a_n + \gcd(n, a_n). Compute a2,…,a8a_2, \ldots, a_8 and guess a structure worth proving.

Show the answer

a2=1+gcd⁡(1,1)=2a_2 = 1 + \gcd(1, 1) = 2, a3=2+gcd⁡(2,2)=4a_3 = 2 + \gcd(2, 2) = 4, a4=4+gcd⁡(3,4)=5a_4 = 4 + \gcd(3, 4) = 5, a5=5+gcd⁡(4,5)=6a_5 = 5 + \gcd(4, 5) = 6, a6=6+gcd⁡(5,6)=7a_6 = 6 + \gcd(5, 6) = 7, a7=7+gcd⁡(6,7)=8a_7 = 7 + \gcd(6, 7) = 8, a8=8+gcd⁡(7,8)=9a_8 = 8 + \gcd(7, 8) = 9. Guess: an=n+1a_n = n + 1 for n≥3n \ge 3; the gcd is always 11 because an=n+1a_n = n + 1 and gcd⁡(n,n+1)=1\gcd(n, n + 1) = 1 — a one-line induction. (The real problem varies the recurrence so the jumps land on primes; the scratch method is the same.)