When the Hypothesis Is a Negation, Direct Proof Stalls — Switch

A direct proof begins by using the hypothesis. You write it down, unpack it with a definition, and let the algebra do the rest. That recipe works beautifully when the hypothesis is a positive statement — n is even, a divides b, the function is continuous at x_0. You write n = 2k or b = aq and you are off.

It stops working when the hypothesis is a negation. "n is not even." "There is no integer x such that x^2 = 2." "a does not divide b." Try to unpack those with a definition and you hit a wall — the definitions tell you what "even" and "divides" are, not what they are not. Your first step is missing, and sitting longer does not conjure one.

The pattern is worth learning to spot in one glance, because the fix is almost mechanical: pivot out of direct into one of the two techniques designed for negative starting material.

The shape to recognise

Before you write anything, read the hypothesis and ask: is it shaped like a positive assertion I can unpack, or like a denial?

Positive assertions, which direct proof handles:

n is even. (Definition unpacks to n = 2k.)
p is prime. (Definition unpacks to "only divisors are 1 and p.")
A \subseteq B. (Definition unpacks to "if x \in A then x \in B.")
\lim_{x \to a} f(x) = L. (Definition unpacks to an \varepsilon–\delta statement.)

Denials, which usually stall direct proof:

n is not divisible by 3.
p is irrational. (i.e. not of the form a/b.)
A \not\subseteq B.
f is not continuous at a.
There is no x with x^2 = 2.

Why the second list stalls: a denial has no canonical unpacking in the forward direction. "n is not divisible by 3" gives you infinitely many possibilities (n leaves remainder 1, or remainder 2) and no single equation to manipulate. You cannot write "n = \text{something}" because there is no single "something" that denial corresponds to.

The pivot — two options

When the hypothesis is a negation, you have two standard workarounds. Both restart the proof from a positive assumption.

Option A — Contrapositive. Replace "P \Rightarrow Q" with its logically equivalent form "\lnot Q \Rightarrow \lnot P." You assume \lnot Q — which is often a clean positive statement if Q itself was a negation — and derive \lnot P. See The Contrapositive of p → q for the full technique.

Option B — Contradiction. Assume both P and \lnot Q together, and derive a contradiction. This lets you use the negative hypothesis alongside the negation of the conclusion, which often gives you two working premises instead of one stuck one. See Proof by Contradiction.

Which one to pick? The rule of thumb:

If Q is itself a negation, \lnot Q is positive — contrapositive is usually cleanest.
If Q is a positive claim and P is the only negation, contradiction gives you the positive P to work with, plus the extra premise \lnot Q.

A recognition flowchart

case:

Four candidate proofs. Case 1 has a clean direct path. Case 2 has denials on both sides — the contrapositive flips them to positives. Case 3 has a negated conclusion — contrapositive again. Case 4 pairs a positive hypothesis with an internally contradictory target — contradiction is the natural method.

A 60-second worked pivot

Claim. If n^2 is not divisible by 3, then n is not divisible by 3.

Direct attempt. Assume n^2 is not divisible by 3. What does that mean? That there is no integer q with n^2 = 3q. This is an infinite family of non-equations, and you cannot plug it into algebra. The direct path has no first step.

Pivot. Both sides are negations, so the contrapositive turns both into positives: if n is divisible by 3, then n^2 is divisible by 3.

Direct proof of the contrapositive. Assume n is divisible by 3. Then n = 3k for some integer k. Hence n^2 = 9k^2 = 3(3k^2), and 3k^2 is an integer, so n^2 is divisible by 3. \blacksquare

The original statement follows by the contrapositive equivalence. The unusable hypothesis was the only obstacle; once flipped, the proof is three lines.

A diagnostic checklist

Before you start writing, read P and Q out loud and tick:

Does P contain the word "not", "no", "irrational", "never", "does not", or a \neg symbol? If yes, direct proof is at risk.
Does Q contain a negation? If yes, contrapositive is probably cleaner than contradiction.
Is P universal ("for all") and Q an existence-denial ("there is no")? Contradiction is almost always the technique.
Is there a definition that unpacks P into an equation? If no, direct proof has no first step regardless of what Q looks like.

If three of these tick toward "pivot," you save yourself from staring at the page. Write "By contrapositive, it suffices to prove: if \lnot Q, then \lnot P" or "Suppose for contradiction that P and \lnot Q" and restart.

The mental model

Think of direct proof as a forward chain. It needs a start link — a positive hypothesis with a definition that converts it to an equation. A negation provides no start link, so the chain has nowhere to hook in. Contrapositive and contradiction each supply a different start link:

Contrapositive supplies a new start link by negating the conclusion.
Contradiction supplies a new start link by adding \lnot Q as an extra premise, which is often the concrete positive fact you were missing.

Learning to recognise a missing start link — in a single read of the statement — is the single largest speedup you can add to your proof workflow. It is the difference between twenty minutes of staring and thirty seconds of pivoting.