Overview

On this page, we introduce how to write proofs clearly. This article is based on a handout by Evan Chen on proof-writing and mathematical English: https://web.evanchen.cc/handouts/english/english.pdf

Why learn proof writing?

The common question that is asked frequently is "Why do we care so much about proof-writing? Isn't it okay as long as I can get a 7?"

Evan Chen has a nice answer to this question: see https://web.evanchen.cc/faq-contest.html#C-24

The main idea of proof writing is that everyone who reads the proof should be able to understand why that statement is true. And understanding the 'why' aspect is crucial. without that I can claim anything is true. If proof writing is not there, I can do ridiculous things. For example, I can say that x>xx>x. It seems ridiculous and that is what will happen without proof writing. In fact, you cannot get a 7 without a proper written proof*. Some tips while writing proofs are:

  • Clearly state what you are proving and define all variables.
  • Write in complete sentences and justify each step.
  • Identify and highlight the key invariant, idea, or construction early.

Core Concepts

Induction

To prove a statement for all integers n≥1n \ge 1, establish a base case and then show that truth for n=kn = k implies truth for n=k+1n = k+1. However we cannot use this for statements which requires the state of multiple variables such as a degree 2 or more recursive sequence (Where strong induction will be used).

Example

Claim.

13+23+⋯+n3=(1+2+⋯+n)2.1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2.

Proof.

Base Case (n=1n = 1):

13=12,1^3 = 1^2,

so the statement holds.

Inductive Step: Assume the identity holds for n=kn = k, i.e.,

13+23+⋯+k3=(1+2+⋯+k)2.1^3 + 2^3 + \cdots + k^3 = (1 + 2 + \cdots + k)^2.

We must prove it holds for n=k+1n = k+1:

13+23+⋯+k3+(k+1)3=(1+2+⋯+k+(k+1))2.1^3 + 2^3 + \cdots + k^3 + (k+1)^3 = (1 + 2 + \cdots + k + (k+1))^2.

Using the inductive hypothesis, it suffices to show:

(1+2+⋯+k+(k+1))2−(1+2+⋯+k)2=(k+1)3.(1 + 2 + \cdots + k + (k+1))^2 - (1 + 2 + \cdots + k)^2 = (k+1)^3.

Apply the difference of squares:

A2−B2=(A−B)(A+B),A^2 - B^2 = (A - B)(A + B),

where

A=1+2+⋯+k+(k+1),B=1+2+⋯+k.A = 1 + 2 + \cdots + k + (k+1), \quad B = 1 + 2 + \cdots + k.

Then

A−B=k+1,A+B=2(1+2+⋯+k)+(k+1).A - B = k+1, \quad A + B = 2(1 + 2 + \cdots + k) + (k+1).

Using

1+2+⋯+k=k(k+1)2,1 + 2 + \cdots + k = \frac{k(k+1)}{2},

we compute:

A+B=k(k+1)+(k+1)=(k+1)2.A + B = k(k+1) + (k+1) = (k+1)^2.

Thus,

A2−B2=(k+1)(k+1)2=(k+1)3.A^2 - B^2 = (k+1)(k+1)^2 = (k+1)^3.

This completes the inductive step.

Therefore, the statement holds for all n≥1n \ge 1. □\square

Logical Chain

Ensure your argument follows a clear logical progression:

X  ⟹  Y  ⟹  Z.X \implies Y \implies Z.

Avoid gaps in reasoning—every implication must be justified. Avoid circular proofs, where you justify a statement with the statement itself.

Example

Claim.

a2+b2=c2a^2 + b^2 = c^2

Proof.

sin⁡(x)2+cos⁡(x)2=1\sin(x)^2 + \cos(x)^2 = 1

Now let the hypotenuse be represented by c, the opposite by b and the adjacent by a. Then:

sin⁡(x)=bccos⁡(x)=ac  ⟹  (bc)2+(ac)2=1\sin(x) = \frac{b}{c}\quad\cos(x) = \frac{a}{c} \implies (\frac{b}{c})^2+(\frac{a}{c})^2=1

Hence,

(bc)2+(ac)2=b2c2+a2c2=1(\frac{b}{c})^2+(\frac{a}{c})^2=\frac{b^2}{c^2}+\frac{a^2}{c^2}=1
b2c2+a2c2=1  ⟹  a2+b2=c2\frac{b^2}{c^2}+\frac{a^2}{c^2}=1 \implies \boxed {a^2+b^2=c^2}

Holds for all a,b,c∈Ra,b,c \in \mathbb{R}

Note: This example is of a cirular proof and the proof is wrong

Structure

Good structure is essential for clarity:

  • Begin by stating what you are proving.
  • Present the main idea early.
  • Use a Claim–Proof format when appropriate.
  • Center important computations and separate them from text.
  • Guide the reader through your reasoning.

Completeness Checklist

Many otherwise strong solutions lose points due to missing cases. Always verify:

  • Did I prove all parts of the problem?

  • Did I prove both directions (if required)?

    • Especially for “find all” problems.
  • Are all conditions of the theorems I used satisfied?

  • Did I address edge cases (e.g., parallel lines in geometry)?

  • Did I explicitly justify equalities and constructions?

Common Pitfalls

  • Failing to prove all required cases
  • Using unjustified or illogical steps
  • Poor structure or unclear organization
  • Usage of circular reasoning

mathematical argument

In this section we talk about mathematical proofs; you can't get a 77 if you're mathematically wrong.

"Find all" type problem is always two-part problem

Some olympiad problems look like this: "Find all nn such that the condition holds." If you think the answer set is SS, then we need to prove the following:

  • For n∈Sn\in S prove that the condition holds.
  • For n∉Sn\notin S, prove that nn doesn't satisfy the condition.

Notice that we have to prove both of two statements; we need to write it down even if it's trivial! It is a common mistake to forget writing "this clearly works" and losing 11 point in a contest. Thus, we recommend writing the solution in the following form:

  1. Start by describing all the solutions you found, which looks like "The answer is n∈{1,2,4,12}n\in \{1, 2, 4, 12\}".
  2. Show that these solutions you stated works. You must write this even if it's trivial, because it's the part of the problem.
  3. Prove that those are the only ones that satisfy the condition. Again, you must write this part even if it's trivial, since it's a part of the problem.

Note that sometimes part 2 and 3 can be swapped, though you must write both. By skipping the trivial part, the grader cannot tell whether you know it is trivial or whether you do not know the proof and therefore did not write it.

optimization problem is always two-part problem

Optimization problem works similarly as above; suppose a problem with: "find the minimum/maximum of mm" and you've got the answer MM, then we need to prove that:

  • There exists a situation where m=Mm=M.
  • For all possible configurations we must have that MM is the minimum/maximum of mm.

Thus, we recommend writing proofs as follows:

  1. Describe your answer by writing "The answer is m=(something)m=\text{(something)}"
  2. Show that the bound is achieved: "which is achieved when blah blah".
  3. Show that we always have m≤Mm\le M (or m≥Mm\ge M).

How much do I need to write?

Answer: Include enough that grader will know that you know the solution. For a general advise:

  • If you're not sure a lemma you're using is not well known, it is safer to write down a complete statement of a lemma and the outline of a proof of it.
  • Be careful of the definition! Wrong claims or typo's can be fixed quite easily by a grader, but if you write a definition wrong there's no way a grader can fix it. In particular, include diagram a if you need.

style argument

In this section we talk about styles of writing: make the paper more readable. The advantage of doing this is:

  • The grader can understand your proof easily, and prevent misunderstanding.
  • It's more easier to review your proofs during contests (or even in practice).
  • Writing formal proof can make you understand a problem and the idea well.

Emphasize important claims, lemmas

Olympiad problems usually have a starting point, and few claims, that will lead to a conclusion. Thus, emphasizing each claim is powerful since graders can know outline of a solution before actually reading small sentences (If you watch YouTube, it's just like a timestamp on long videos). More precisely:

  • Isolated claims/proofs should be in new line, with whitespace before and after.
  • When you do casework, always split cases into separate paragraph or bullet points.
  • Write in backwards manner; instead of writing "blah blah, thus p=3p=3", write "claim: p=3p=3, proof".
  • Even sometimes using colors or cover with box, e.g. p=3\boxed{p=3} is helpful.

Other general advice

  • Display long equation; rather than writing something like : ∡ARC=∡ARQ+∡QRC=∡APQ+∡QXC=∡APQ+∡QAP=∡AQP=∡AQD=∡ACD=∡ABC\measuredangle ARC =\measuredangle ARQ+\measuredangle QRC =\measuredangle APQ+\measuredangle QXC =\measuredangle APQ+\measuredangle QAP =\measuredangle AQP=\measuredangle AQD=\measuredangle ACD=\measuredangle ABC. Instead write:
∡ARC=∡ARQ+∡QRC=∡APQ+∡QXC=∡APQ+∡QAP=∡AQP=∡AQD=∡ACD=∡ABC\begin{array}{rll} \measuredangle ARC & = \measuredangle ARQ + \measuredangle QRC \\ & = \measuredangle APQ + \measuredangle QXC \\ & = \measuredangle APQ + \measuredangle QAP \\ & = \measuredangle AQP \\ & = \measuredangle AQD \\ & = \measuredangle ACD \\ & = \measuredangle ABC \end{array}
  • Include more space than you really think.
  • Include diagrams if you can.
  • Write more! If a definition of an object you gave was complicated, consider writing an example. If you're unsure that the explanation you gave was incomplete, include more information. There's no reason not to.
  • It's useful to add sentences like "The key idea is XX" or "important claim is YY".

In general, be kind to the readers.

During a contest

Here's a short section about writing in the contest.

  • Before writing a solution, give a short outline to it and confirm that each step is right. Also, note what lemmas to state before the proof of a problem.
  • Write diagrams largely, possibly using the whole one page.
  • Leave more space! In practice with computer-written solution, you can insert text or copy-paste it anytime. In hand-written solution, you can't do this, thus leaving space will help you.

Example

Question. How many prime numbers pp are there such that 29p+129^p+1 is a multiple of p?

Answer. There are 33 such prime numbers

Case 1:

Now let us assume that p≠29p\ne29. Then we observe that the Greatest Common divisor of pp and 2929 is 1.

The key idea here is to useFermat′s  Little  Theorem\boxed{Fermat's\;Little\;Theorem} to get:

29p−1≡1 (mod p)  ⟹  29p≡29 (mod p)29^{p-1} \equiv 1\:(mod\:p) \implies 29^p \equiv 29\:(mod\:p)

Hence we need:

29p+1≡30≡0 (mod p)  ⟹  p∣3029^p+1\equiv30\equiv0\:(mod\:p) \implies p\mid30

p=2,3 and 5p=2,3\:and\:5 satisfies the condition which is of 3 primes in total.

Case 2:

For p=29p=29, we know that

p∣29pp\mid29^p

Thus

p∤29p+1p\nmid 29^p+1

And we are done

Practice Problems

StatusSourceProblem NameDifficultyTags
USAMOVery Hard
Show TagsAlgebra, Proof
USAMOVery Hard
Show TagsNumber Theory, Proof
Berkeley Math Circle: Monthly Contest 6Hard
Show TagsColoring schemes, extremal arguments, Logic
Berkeley Math Circle: Monthly Contest 2Hard
Show TagsLogic
Berkeley Math Circle Monthly Contest 1Hard
Show TagsAlgorithms, Logic
Berkeley Math CircleHard
Show TagsLogic
Berkeley Math Circle Monthly Contest #1Hard
Show TagsInduction / smoothing, Logic
Berkeley Math CircleHard
Show TagsLogic, Pigeonhole principle
Harvard-MIT Mathematics TournamentHard
Show TagsColoring schemes, extremal arguments, Logic
Harvard-MIT Mathematics Tournament, Team Round AHard
Show TagsColoring schemes, extremal arguments, Logic
2nd Annual Harvard-MIT November TournamentHard
Show TagsLogic
Harvard-MIT Mathematics TournamentHard
Show TagsColoring schemes, extremal arguments, Logic
Harvard-MIT Mathematics TournamentHard
Show TagsColoring schemes, extremal arguments, Logic
HMICHard
Show TagsLogic
HMMTHard
Show TagsEnumeration with symmetry, Logic
HMMT NovemberHard
Show TagsColoring schemes, extremal arguments, Logic

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