'No', 'Never', 'Irrational', 'Infinite', 'Does Not Exist' — Spotting the Contradiction Signal Words

Picking a proof technique should not be a long deliberation. For a huge fraction of proof problems, the wording of the claim itself tells you which technique to reach for. If you can learn to spot a small list of keyword patterns, you can skip the agonising and jump straight to the right first attempt.

This article is about five such keywords — no, never, irrational, infinite, does not exist — and why each of them is a near-guarantee that proof by contradiction is the cleanest route.

The five keyword signals

Read the claim once. If any of these words (or close synonyms) appears prominently in the conclusion, your first move is almost always contradiction.

Keyword	Example claim	Why contradiction is natural
no	"There is no rational r with r^2 = 2."	Negates existence — assume one exists, find the wall.
never	"n^2 is never* 2 \pmod 4."*	Universal negation — assume some n breaks it, derive impossibility.
irrational	"\log_2 3 is irrational."	Irrational = not rational — a negation in disguise.
infinite(ly many)	"There are infinitely many* primes."*	Assume the list is finite, construct one not on the list.
does not exist	"There does not exist* a largest integer."*	Direct negation of existence — classic contradiction template.

Each entry in the right column is the same shape: assume the opposite, follow the algebra, hit a wall. That uniformity is why these claims cluster under one technique.

Why each signal works

"No" and "does not exist" are negations of existence. Direct proof demands that you produce something, but a non-existence claim has no object to produce. Instead, you assume the object exists, use its properties to derive equations, and eventually show the equations cannot be simultaneously satisfied.

"Never" is a universal quantifier over a negation — "for all n, n^2 is not 2 \pmod 4." You cannot check infinitely many cases directly. But if you assume some n does satisfy the forbidden property, you get a single, concrete n to argue about, and the algebra on that n usually yields an impossibility.

"Irrational" is literally defined as "not rational." Whenever you see this word, mentally rewrite the claim: "there is no pair of integers (p, q) with q \neq 0 and the given number = p/q." Now you have an existence-negation — and the recipe snaps into place.

"Infinite(ly many)" looks different because it seems positive — it asserts existence, not absence. But the standard contradiction move flips it: "Suppose there are only finitely many," list them, manipulate them, and produce one that is not on the list. The finite-list assumption gives you a finite set to manipulate algebraically; direct exhibition of an infinite sequence is possible but usually harder.

Why the signals cluster: each one names a situation where the "thing being asserted" is the absence of something, or the infinitude of something. Direct proof needs a concrete handle to manipulate; contradiction hands you one by flipping the claim. The flip is worth memorising because it is mechanical — no creativity needed to set up the first line.

The keyword scanner

Drag the dot below to cycle through sample claims. The readout identifies the keyword and names the first-try technique.

Five common signal words, five claims, one recommended technique. The scanner turns keyword recognition into reflex: read the claim, pattern-match on the signal word, and move straight to *"Assume, for contradiction, ..."* as your first line.

A subtle case: "at most" and "at least"

Beyond the five explicit keywords, phrases with bounds often hide the same logic:

"At most finitely many n satisfy P" → negate to "infinitely many n satisfy P" → contradiction.
"At least two distinct primes divide n" → direct (constructive) usually wins here, because you can exhibit the primes.
"Exactly one x satisfies ..." (uniqueness) → assume two, derive contradiction.

The test is simple: does the claim forbid something or count something? If it forbids or negates, contradiction. If it produces or counts, direct is often cleaner.

Where the signal can mislead

Keyword recognition is a heuristic, not a law. Two situations where a signal word appears but direct proof still wins:

Constructive non-existence proofs. "There is no polynomial of degree n with more than n distinct roots" — you can prove this directly using the factor theorem without assuming the negation. The keyword "no" pushes you toward contradiction, but a clean direct argument exists.
Infinitely many, constructively. "There are infinitely many even numbers" — trivially direct: exhibit 2, 4, 6, \ldots. Contradiction here would be absurd. The keyword "infinitely many" is overridden by an obvious constructive sequence.

When the keyword signal points one way but a short direct proof is visible, take the direct proof. The keyword is a default, not a rule.

Why the signal still wins most of the time: in school-level and JEE-level problems, the claims most worth proving with non-trivial techniques are exactly the ones where direct constructions are not available. Textbook authors choose contradiction-friendly examples precisely because they illustrate the technique. When a signal word appears in a homework problem, the odds strongly favour contradiction being the intended path.

Spotting the signal in three JEE-style claims

Claim 1: "Prove that there is no integer n such that n^2 \equiv 2 \pmod 4."

Signal: no. Rewrite: "Assume some integer n satisfies n^2 \equiv 2 \pmod 4." Check cases: n even → n^2 \equiv 0; n odd → n^2 \equiv 1. Neither is 2. Contradiction. \blacksquare

Claim 2: "Prove that \sqrt{6} is irrational."

Signal: irrational. Rewrite: "Assume \sqrt{6} = p/q in lowest terms." Square: 6q^2 = p^2. Then 2 \mid p^2, so 2 \mid p; write p = 2k, so 6q^2 = 4k^2, giving 3q^2 = 2k^2. Now 2 \mid 3q^2, and since \gcd(2,3)=1, 2 \mid q^2, so 2 \mid q. Contradiction with lowest terms. \blacksquare

Claim 3: "Show that a largest prime does not exist."

Signal: does not exist. Rewrite: "Assume P is the largest prime." Form N = 2 \cdot 3 \cdot 5 \cdots P + 1. Any prime dividing N is greater than P. Contradiction. \blacksquare

The reflex in short

Read the claim.
Scan for keywords: no, never, irrational, infinite(ly), does not exist, or hidden negations (at most, unique).
If any match — default first move is "Suppose, for the sake of contradiction, that ...".
Proceed to the algebra; wait for the wall.

The 30 seconds you spend scanning for signal words are the most productive 30 seconds in any proof-writing session. They convert technique selection from a cognitive bottleneck into a reflex, freeing your thinking for the parts of the proof that actually require thinking.