Say exactly what you mean: proof language and the real line
Most Putnam points are lost before any mathematics happens — they are lost in sentences. A grader reading 'clearly the sequence converges' writes a zero next to it as fast as a grader reading a correct limit argument writes a ten. This week installs the sentence-level discipline everything else stands on, and pairs it with the one structure that discipline was invented for: the real numbers.
A proof is a machine whose parts are quantifiers. If you can say $\forall \varepsilon > 0\ \exists N$ and mean it, completeness of $\mathbb{R}$ stops being a slogan and becomes a tool you can point at a problem.
Key ideas — the week on one card
Quantifier order is meaning: $\forall \varepsilon\, \exists N$ (with $N$ depending on $\varepsilon$) is convergence; $\exists N\, \forall \varepsilon$ is a different and far stronger claim.
Completeness: every nonempty set of reals bounded above has a least upper bound — the axiom every limit argument secretly cites.
The epsilon-of-room lemma: if $a \le b + \varepsilon$ for EVERY $\varepsilon > 0$, then $a \le b$ — it closes most analysis proofs.
Induction earns full marks only with a named predicate $P(n)$, a verified base, and a step that visibly uses the hypothesis.
1Quantifiers are the load-bearing walls
Read these two sentences slowly: 'for every $\varepsilon > 0$ there is an $N$' and 'there is an $N$ such that for every $\varepsilon > 0$'. The first defines convergence. The second says the sequence is eventually constant at its limit — a wildly stronger claim. One swap of order, and you have proved something false. Putnam graders check quantifier order the way airport security checks passports.
The practical drill: before writing any proof this week, write its skeleton first — only quantifiers and connective words, no mathematics. 'For all... there exists... such that for all... .' If the skeleton is wrong, no amount of algebra inside it will save the proof. This habit costs thirty seconds and prevents the most common fatal slip in the archive's graded solutions.
Contrapositive deserves the same respect. 'If $f$ is unbounded then $f$ is not continuous on $[a,b]$' is proved most cleanly as 'if $f$ is continuous on $[a,b]$ then $f$ is bounded' — same statement, but now every tool about continuity is pointed forward instead of backward. When a direct proof feels like swimming upstream, write the contrapositive skeleton and check which direction the tools face.
∀ε ∃N, drawn: the band is ε, the cutoff is N, and the order of those two symbols is the whole definition.
Worked example
Negate precisely, then decide which is true for $f(x) = x^2$ on $\mathbb{R}$: (a) $\forall M\ \exists x : f(x) > M$, or its negation (b).
Nudge 1
Negation is mechanical before it is clever: flip each quantifier, then negate the inner inequality.
Reveal step 1
Mechanical negation: flip every quantifier, negate the core. $\neg(\forall M\ \exists x : f(x) > M)$ becomes $\exists M\ \forall x : f(x) \le M$ — 'some ceiling $M$ bounds $f$ everywhere'.
Nudge 2
To prove $\forall M\, \exists x$, treat $M$ as handed to you by an adversary — your job is one explicit $x$ built from $M$.
Reveal step 2
Now test (a) against $f(x) = x^2$: given any $M$, choose $x = \sqrt{|M|} + 1$. Then $x^2 = |M| + 2\sqrt{|M|} + 1 > M$. The witness is explicit — this is what 'there exists' obligates you to produce.
Nudge 3
Which of the two statements did that recipe just prove? Name the winner and say why the other fails.
Reveal step 3
So (a) is true: $x^2$ is unbounded above. Notice the proof of an $\forall\exists$ statement is a recipe: someone hands you $M$, you hand back $x$. If your 'proof' picks $x$ before seeing $M$, you proved the wrong statement.
Answer
(a) is true; its negation (b) asserts a global ceiling $\exists M\ \forall x : x^2 \le M$, which fails at $x = \sqrt{|M|}+1$.
Pitfall. Writing 'let $\varepsilon$ be small' or 'take $N$ large enough'. Graders read these as 'I don't know the dependence'. Always name the dependence: $N = \lceil 1/\varepsilon \rceil$ beats 'large enough' every single time.
2Completeness: the axiom that does all the work
Everything that separates $\mathbb{R}$ from $\mathbb{Q}$ is one axiom: every nonempty set bounded above has a least upper bound. The set $\{x \in \mathbb{Q} : x^2 < 2\}$ is bounded above in $\mathbb{Q}$ and has no least upper bound there — the hole at $\sqrt{2}$ is exactly what completeness fills. When a Putnam problem says 'show a maximum exists' or 'show the sequence converges', it is almost always asking you to find the right set and take its supremum.
The supremum has a two-line user manual, and both lines matter. First: $\sup S$ is an upper bound ($x \le \sup S$ for all $x \in S$). Second: nothing smaller is ($\forall \varepsilon > 0\ \exists x \in S : x > \sup S - \varepsilon$). The second line is the one that does work in proofs — it manufactures elements of $S$ arbitrarily close to the top, on demand. Most sup arguments are just: invoke line two, get a witness, run the hypothesis on it.
The Archimedean property — for any real $x$ there is a natural $n > x$ — looks too obvious to state, but it is the bridge between 'analysis' and 'actual numbers'. It is what lets you convert '$a \le b + \varepsilon$ for every $\varepsilon > 0$' into '$a \le b$', the single most reused micro-lemma in the entire subject. Prove that conversion once this week, cleanly, and you will use it forty times by December.
Worked example
Let $a, b \in \mathbb{R}$ satisfy $a \le b + \varepsilon$ for every $\varepsilon > 0$. Prove $a \le b$.
Nudge 1
Contrapositive: assume $a > b$ and produce a single $\varepsilon$ that breaks the hypothesis.
Reveal step 1
Skeleton first: we prove the contrapositive. Suppose $a > b$; we must exhibit one $\varepsilon > 0$ with $a > b + \varepsilon$.
Nudge 2
You own the positive quantity $a - b$; scale it so $b + \varepsilon$ lands strictly below $a$.
Reveal step 2
Choose $\varepsilon = \frac{a - b}{2}$. Since $a > b$, this $\varepsilon$ is positive — the hypothesis of the game is satisfied.
Nudge 3
Compute $b + \varepsilon$ explicitly — the midpoint of $a$ and $b$ sits below $a$.
Reveal step 3
Then $b + \varepsilon = b + \frac{a-b}{2} = \frac{a+b}{2} < \frac{a+a}{2} = a$. So $a > b + \varepsilon$ for this explicit $\varepsilon$, contradicting the assumption. Hence $a \le b$.
Answer
Contrapositive with the explicit witness $\varepsilon = (a-b)/2$; the midpoint $(a+b)/2$ separates $b + \varepsilon$ from $a$.
Pitfall. Concluding $a < b$ instead of $a \le b$. The hypothesis is compatible with $a = b$ (take any $\varepsilon$: $b \le b + \varepsilon$ holds). Strict versus non-strict is not a typo-level detail on this exam — it is frequently the entire problem.
3Induction that graders trust
Induction fails on the Putnam in one of two ways: the base case is checked for the wrong index, or the inductive step silently assumes more than the hypothesis grants. The fix is bureaucratic and unglamorous — state $P(n)$ as a displayed sentence before anything else. Not 'we induct on $n$', but: 'let $P(n)$ be the statement that $\dots$'. Once $P(n)$ has a body, checking $P(1)$ and proving $P(n) \Rightarrow P(n+1)$ become mechanical.
Strong induction is not a different technique; it is ordinary induction on the statement $Q(n) = P(1) \wedge \cdots \wedge P(n)$. Knowing this dissolves the fake dilemma of 'which induction do I use' — use whichever hypothesis your step actually needs, and say which. Recurrences that reach back more than one term ($a_{n+1}$ in terms of $a_n$ and $a_{n-1}$) need strong induction and TWO base cases; forgetting the second base case is a classic silent zero.
Worked example
Prove that $2^n > n^2$ for every integer $n \ge 5$.
Nudge 1
Check $n = 4$ as well as $n = 5$ — knowing where the claim FAILS tells you why the base is 5.
Reveal step 1
State it: let $P(n)$ be '$2^n > n^2$'. Base: $P(5)$ says $32 > 25$ — true. (Check: $P(4)$ is $16 > 16$, false — so $n \ge 5$ is not decoration; the base index is forced.)
Nudge 2
Multiply the hypothesis by 2 and reduce the goal to comparing $2n^2$ with $(n+1)^2$.
Reveal step 2
Step: assume $2^n > n^2$ for some $n \ge 5$. Then $2^{n+1} = 2 \cdot 2^n > 2n^2$. It remains to show $2n^2 \ge (n+1)^2$ for $n \ge 5$.
Nudge 3
Write $2n^2 - (n+1)^2$ as a shifted square and bound it using $n \ge 5$.
Reveal step 3
$2n^2 - (n+1)^2 = n^2 - 2n - 1 = (n-1)^2 - 2 \ge 4^2 - 2 = 14 > 0$ for $n \ge 5$. Chain the inequalities: $2^{n+1} > 2n^2 \ge (n+1)^2$, which is $P(n+1)$.
Answer
$P(5)$ holds and $P(n) \Rightarrow P(n+1)$ via $2n^2 \ge (n+1)^2 \iff (n-1)^2 \ge 2$, true from $n \ge 5$ (indeed from $n \ge 3$).
Pitfall. Proving the step for all $n$ but checking the base at $n = 1$ when the statement is false there. The base case and the step must cover the SAME range — graders check the seam between them first.
Before you open the gates
Write the quantifier skeleton before the mathematics — thirty seconds, catches the most common fatal slip.
Every 'there exists' in your proof must come with a named witness; every dependence ($N$ on $\varepsilon$) must be explicit.
When a maximum or limit 'should' exist, name the set, argue it is nonempty and bounded, and take the sup — then use the $\sup S - \varepsilon$ line to extract witnesses.
State $P(n)$ as a full sentence before inducting; make the base indices match where the step starts working.
Reread every strict inequality in your final draft and ask: does equality actually break, or did I just want it to?
Check yourself
1. Which statement is equivalent to '$(a_n)$ does NOT converge to $L$'?
Negation flips every quantifier of $\forall\varepsilon\,\exists N\,\forall n$: some bad $\varepsilon$ works against every $N$ — infinitely many terms stay $\varepsilon$-far from $L$.Repair: revisit §1 Quantifiers are the load-bearing walls ↑
2. $S = \{1 - \tfrac{1}{n} : n \in \mathbb{N}\}$. Which is true?
Each rung: attempt cold, one hint if stuck, worked resolution only after a real try.
Rung 1 (direct application). Write the formal negation of the statement $\exists N\ \forall n \ge N : a_n > 0$, then decide which of the two — the statement or its negation — holds for $a_n = (-1)^n / n$.
One hint
Negate left to right: each $\exists$ becomes $\forall$, each $\forall$ becomes $\exists$, and the inner predicate flips.
Worked resolution
The negation is $\forall N\ \exists n \ge N : a_n \le 0$. For $a_n = (-1)^n/n$: given any $N$, pick an odd $n \ge N$; then $a_n = -1/n < 0$. So the negation holds and the original statement is false — the sequence keeps returning below zero no matter how far out you go. [Source: Drills the negation discipline from this week's Hammack logic chapters.]
✓ rung 1 done
Rung 2 (the axiom at work). Let $S = \{x \in \mathbb{R} : x^2 < 2\}$ and $s = \sup S$. Prove that $s^2 = 2$ — completeness hands you $s$; your job is to rule out $s^2 < 2$ and $s^2 > 2$.
One hint
If $s^2 < 2$, the single choice $h = \tfrac{2 - s^2}{2s+1}$ (which is $< 1$ because $s \ge 1$) makes $(s+h)^2 < 2$, contradicting 'upper bound'. If $s^2 > 2$, then $h = \tfrac{s^2-2}{2s}$ makes $s - h$ a POSITIVE smaller upper bound — handle $x \le 0$ and $x > 0$ separately when you compare against it.
Worked resolution
$S$ is nonempty ($1 \in S$) and bounded above (if $x \ge 2$ then $x^2 \ge 4$, so every element is $< 2$); by completeness $s$ exists, and $s \ge 1$. If $s^2 < 2$: take $h = \tfrac{2 - s^2}{2s + 1} > 0$; since $s \ge 1$ gives $2 - s^2 \le 1 < 2s + 1$, we have $h < 1$. Then $(s+h)^2 = s^2 + h(2s + h) < s^2 + h(2s + 1) = s^2 + (2 - s^2) = 2$ — the strict step is $h < 1$ — so $s + h \in S$, contradicting that $s$ is an upper bound. If $s^2 > 2$: take $h = \tfrac{s^2 - 2}{2s} > 0$ and note $s - h = \tfrac{s^2 + 2}{2s} > 0$. Then $(s-h)^2 = s^2 - (s^2 - 2) + h^2 = 2 + h^2 > 2$. Every $x \in S$ satisfies $x < s - h$: for $x \le 0$ this is immediate since $s - h > 0$, and for $x > 0$, $x^2 < 2 < (s-h)^2$ with both sides positive gives $x < s - h$. So $s - h$ is an upper bound smaller than $s$ — contradicting leastness. Hence $s^2 = 2$. [Source: This is the classic supremum argument behind Rudin Ch. 1's existence of roots — the week's assigned spine.]
✓ rung 2 done
Rung 3 (Putnam-shaped induction). Prove that for every integer $n \ge 1$: $\dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \cdots + \dfrac{1}{\sqrt{n}} < 2\sqrt{n}$.
One hint
Induct. The whole problem is the one inequality $2\sqrt{n} + \tfrac{1}{\sqrt{n+1}} \le 2\sqrt{n+1}$ — rationalize $\sqrt{n+1} - \sqrt{n}$ and it falls open.
Worked resolution
Base $n = 1$: $1 < 2$. Step: assume the sum to $n$ is $< 2\sqrt{n}$. Then the sum to $n+1$ is $< 2\sqrt{n} + \tfrac{1}{\sqrt{n+1}}$, so it suffices to show $2\sqrt{n} + \tfrac{1}{\sqrt{n+1}} \le 2\sqrt{n+1}$, i.e. $\tfrac{1}{\sqrt{n+1}} \le 2(\sqrt{n+1} - \sqrt{n})$. Rationalizing, $2(\sqrt{n+1} - \sqrt{n}) = \tfrac{2}{\sqrt{n+1} + \sqrt{n}}$, and $\sqrt{n+1} + \sqrt{n} \le 2\sqrt{n+1}$ gives $\tfrac{2}{\sqrt{n+1} + \sqrt{n}} \ge \tfrac{2}{2\sqrt{n+1}} = \tfrac{1}{\sqrt{n+1}}$. Chain closed; induction complete. The lesson: make the inductive step an honest inequality chain a grader can audit line by line. [Source: A standard strengthened-induction exercise in the spirit of the week's Hammack induction assignment.]
✓ rung 3 done
Prove it — constructed response
Let $A$ and $B$ be nonempty sets of real numbers, each bounded above, and let $A + B = \{a + b : a \in A,\ b \in B\}$. Prove that $\sup(A+B) = \sup A + \sup B$. Every quantifier in the right order; every use of completeness named.
Self-grade against the rubric — completion requires the judgment, not the text
Model proof (compare AFTER grading yourself)
Let $\alpha = \sup A$ and $\beta = \sup B$; both exist by completeness since $A, B$ are nonempty and bounded above. (Upper bound.) For every $a \in A$ and $b \in B$ we have $a \le \alpha$ and $b \le \beta$, so $a + b \le \alpha + \beta$. Thus $A+B$ is nonempty, bounded above by $\alpha + \beta$, and $s := \sup(A+B)$ exists with $s \le \alpha + \beta$. (Leastness.) Let $\varepsilon > 0$. Since $\alpha - \varepsilon/2 < \alpha$ is not an upper bound of $A$, there is $a \in A$ with $a > \alpha - \varepsilon/2$; likewise $b \in B$ with $b > \beta - \varepsilon/2$. Then $s \ge a + b > \alpha + \beta - \varepsilon$. Since this holds for every $\varepsilon > 0$, the $\varepsilon$-slack lemma from this week's section 2 gives $s \ge \alpha + \beta$. Combining, $s = \alpha + \beta$. $\blacksquare$
The gates
Track — Proof Writing (2 hrs/day)
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- Read: Velleman *How to Prove It* Ch. 1–3 — sentential logic, quantifiers, proof structure
- Read: Hammack *Book of Proof* Ch. 4–6 — direct, contrapositive, contradiction (free online)
- Watch: Michael Penn Proof Writing playlist — videos 1–4
- Do: Velleman Ch. 1 ex 1, 3, 5, 7, 9 · Ch. 2 ex 1, 2, 4, 6 · Ch. 3 ex 1, 3, 5, 7
- Do: Hammack Ch. 4 ex 1, 3, 5, 7, 9 · Ch. 5 ex 1, 3, 5 · Ch. 6 ex 1, 3
- Do: PnB section 1.1–1.2 probs 1–6
sources & assignments (8)
SourceVelleman *How to Prove It* — Velleman *How to Prove It* Ch. 1–3 — sentential logic, quantifiers, proof structure
SourceHammack *Book of Proof* — Hammack *Book of Proof* Ch. 4–6 — direct, contrapositive, contradiction (free online)
SourceMichael Penn — Proof Writing (playlist) — Proof-writing lecture series — direct playlist companion to this gate's reading. Watch-for: begins with set-theory foundations (sets, Cartesian products, operations) before the proof-technique videos.
SourceTao — Solving Mathematical Problems — Tao *Solving Mathematical Problems* Ch. 2 §2.1–2.3 (number theory) + Ch. 3 (algebra/analysis) — read, then reconstruct each worked example from a blank page
SourceMichael Penn — Abstract Algebra (playlist) — Intuition companion for “Track — Proof Writing (2 hrs/day)” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises. Watch-for: this playlist opens with set-theory / proof-writing videos — skip to the 'Abstract Algebra |' entries for the group/ring/field content this gate needs.
ProblemsVelleman — Velleman Ch. 1 ex 1, 3, 5, 7, 9 · Ch. 2 ex 1, 2, 4, 6 · Ch. 3 ex 1, 3, 5, 7Putnam
ProblemsHammack — Hammack Ch. 4 ex 1, 3, 5, 7, 9 · Ch. 5 ex 1, 3, 5 · Ch. 6 ex 1, 3Putnam
ProblemsPnB — PnB section 1.1–1.2 probs 1–6Putnam
Track A — Analysis: Rudin Ch. 1 (1.5 hrs/day)
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- Read: Rudin *PMA* Ch. 1 (~26 pp.) — ordered fields, completeness axiom, ℝ and ℂ construction
- Watch: Bright Side of Mathematics Real Analysis — videos 1–2
- Do: Rudin Ch. 1 ex 1, 3, 5, 8, 10, 13, 16
- Do: PnB section 3.1 intro (pp. 104–108) — read only; preview series vocabulary
SourceMichael Penn — Proof Writing (playlist) — Watch ONE video matching this gate's proof type (direct/contrapositive/contradiction/induction), then redo one of this gate's exercises in that video's style. Output: the rewritten proof. Watch-for: begins with set-theory foundations (sets, Cartesian products, operations) before the proof-technique videos.
SourceMichael Penn — Putnam Exam Solutions (playlist) — Intuition companion for “Track A — Analysis: Rudin Ch. 1 (1.5 hrs/day)” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
Putnam Contact — deconstruct the game early (30–45 min)
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Keeps Week 1 tied to the actual contest without turning proof-foundation week into an archive grind.
Ritual: Do not score this as a solve. The win condition is recognizing the proof engine and writing the first-move sentence.
sources & assignments (3)
SourcePutnam archive — official-style early A1 contact — Use Putnam 2017 A1 as the solved model and Putnam 2018 A1 as the first timed contact. Mark claim, first move, key lemma, and final check.
ProblemsPutnam archive deconstruction — Putnam 1969 B1 — read a complete solution; label claim, first move, key lemma, final check.Putnam
ProblemsPutnam archive timed contact — Putnam 1973 B1 — 15-minute attempt; write one sentence: “The first move was ___ because ___.”Putnam
Problem-Solving Reps — weekly homework
—
Standing weekly homework — the problem-solving book stack, every week. Solve ALL listed; write ONE full clean solution (the rest may stay scratch). Up to 2 due re-solves from your review queue surface first.
- Reconstruct: pick 1 Rudin exercise → close notes → rewrite from memory → write one pattern note: "When I see ___, first move is ___."
- Start: proof error log (logic gap / skipped definition / false claim)
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Ritual: pick 1 Rudin exercise → close notes → rewrite from memory → write one pattern note: "When I see ___, first move is ___."
sources & assignments (3)
SourceVelleman *How to Prove It* — Week reference — this gate applies this week's lead material (“Track — Proof Writing (2 hrs/day)”). If an attempt stalls, the repair source is here.
SourceMichael Penn — Proof Writing (playlist) — Week reference — this gate applies this week's lead material (“Track — Proof Writing (2 hrs/day)”). If an attempt stalls, the repair source is here. Watch-for: begins with set-theory foundations (sets, Cartesian products, operations) before the proof-technique videos.
SourceMichael Penn — Abstract Algebra (playlist) — Intuition companion for “Reflect: reconstruct and log the pattern” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises. Watch-for: this playlist opens with set-theory / proof-writing videos — skip to the 'Abstract Algebra |' entries for the group/ring/field content this gate needs.
Source: Miguel A. Lerma, Northwestern Putnam team training problems (2023). Local PDF: /library/Lerma/putnam-training-2023.pdf. Hints + full solutions are in the PDF's later parts.
ProblemsLerma Training 2023 §1 Induction — Work problems 1.1–1.17 (4–6 per session). Strong vs ordinary induction; pick the right inductive quantity. Self-check against the PDF's hints/solutions parts.Putnam
ProblemsLerma Training 2023 §10 Telescoping — Work problems 10.1–10.7 (4–6 per session). Rewrite terms as differences so the sum collapses. Self-check against the PDF's hints/solutions parts.Putnam
ProblemsLerma Training 2023 §14 Miscellany — Work problems 14.1–14.17 (4–6 per session). Mixed-technique problems — name the governing idea before computing. Self-check against the PDF's hints/solutions parts.Putnam
Exit contract — Week 1
Verification remaining
reading progress…
Carry-forward repairs
reading queue…
Next week opens with
W2 · Methods + AIME Fluency + Analysis: Metric Spaces the exam follows the final taper week
