When Is It Really a Contradiction (vs Just Something Weird)?

You are midway through a proof by contradiction. You assumed the opposite of what you want to prove. You ran some algebra. Now you are staring at something that looks off — an unexpected fraction, a negative number where you expected a positive one, a formula that does not match what your intuition said. Is this the contradiction? Can you stop now?

Usually: no, not yet. Weirdness and contradiction are different things, and the difference is the whole point of the method. This article lays down the sharp test and a two-part checklist for when your proof is actually finished.

The sharp definition

A contradiction is a statement of the logical form "P and \lnot P" — a single claim and its explicit negation, both asserted to be true at once. Equivalently: a statement that is false under every possible interpretation, like "2 = 3" or "x is even and x is odd" or "\gcd(p, q) = 1 and \gcd(p, q) \geq 2."

Anything weaker than this is not a contradiction, however uncomfortable it feels.

Why the strict form matters: proof by contradiction works because the logical principle from a false assumption, anything follows is never actually triggered — you rule out the assumption precisely because it would force a statement that cannot be true under any interpretation. A "surprising" or "ugly" intermediate is still internally consistent; it does not let you conclude the assumption was wrong. Only an outright logical impossibility does.

The two-part checklist

Before declaring a proof finished, apply this check to the line that looks contradictory:

Check 1 — Can you write it as "P and \lnot P"? Pair the offending line with some earlier line (often the assumption itself) so that the two, taken together, form a literal "X is true and X is false." If you cannot produce this pairing in one sentence each, you do not have a contradiction yet.

Check 2 — Is the claim false in all cases, not just the one at hand? "This specific number turned out to be negative" is not a contradiction unless you had already asserted it was positive. "This fraction is not in lowest terms" is not a contradiction unless you had asserted (or were granted) that it was. The collision must be with something fixed, not with a preference.

If both checks pass, the proof is done. If either fails, you are looking at a surprising intermediate step, not a contradiction — keep going.

Three examples: real contradiction, fake contradiction, unclear

Example A — real contradiction. In the \sqrt{2} proof you derive that p and q are both even, and you had assumed \gcd(p, q) = 1.

Check 1: pair them. "\gcd(p, q) = 1" (assumed) and "\gcd(p, q) \geq 2" (consequence of both being even). That is P and \lnot P.
Check 2: false under all interpretations. There is no way for a pair of integers to both have gcd 1 and share a factor of 2. Period.

Both checks pass. Real contradiction.

Example B — fake contradiction. You are proving that a certain equation has no real solutions. You assumed a solution x exists and, after algebra, derived x^2 = 7.

Check 1: try to pair. x^2 = 7 does not contradict anything you assumed — you never said x^2 \neq 7.
Check 2: false always? No: x = \sqrt{7} makes it true. This is simply an equation with a real solution.

No contradiction. Worse: this intermediate step actually undermines your claim — the original equation did have a real solution, and you should stop trying to prove otherwise.

Example C — unclear until re-framed. In a proof you derive x = 0 and you had written earlier "let x be a positive integer."

Check 1: pair. "x > 0" (from the setup) and "x = 0" (derived). These together form P and \lnot P — specifically "x > 0 is true and x > 0 is false."
Check 2: false always? Yes — no positive integer is zero.

Both pass. Real contradiction, but only after you explicitly quote the earlier "let x be positive" line. If you had not written that, x = 0 would just be a number.

The trap: "weird" vs "impossible"

The frequent student move is to treat any unexpected result as a win. You set out to prove a claim, the algebra hands you something that does not match your intuition, and you declare victory. This is almost always wrong. Intuition is calibrated on familiar cases; the algebra might be revealing a true fact you did not know. Before you stop, ask whether the weird thing actually denies a specific, explicit earlier statement.

Here is a checklist you can actually run in your head:

Point to the earlier line that the new line contradicts. If you cannot point to one, there is no contradiction yet.
Read the two lines aloud one after the other: "X is true. X is false." If both sentences are syntactically natural and mutually exclusive, you are done.
If the earlier line was itself an assumption (blue), great — the contradiction kills the assumption, which is the win you wanted. If it was some other fact (like a definition or a lemma), you still have a contradiction, but check that the assumption is really responsible for it.

When the contradiction is delayed

Sometimes the algebra produces a weird-looking result that seems like it might be a contradiction but is actually "halfway there." The classic example: in the \sqrt{2} proof, deriving "p is even" is not the contradiction on its own. It is only a consequence of the assumption; it does not yet violate anything. You must push further — substitute p = 2k, derive "q is even," and then combine with \gcd(p, q) = 1 — to get the real contradiction.

Why this intermediate deserves patience: the contradiction is the sentence where the derivation directly collides with the set-up. "p is even" is an ordinary fact about p, not a collision with anything you assumed. Only the paired statement "both p and q share a factor of 2" is inconsistent with \gcd(p, q) = 1. Students who stop at "p is even" think they are done; they have merely filled in half the form.

Spot the real contradiction

You are trying to prove that \log_2 3 is irrational. Suppose, for contradiction, that \log_2 3 = p/q with p, q positive integers and \gcd(p, q) = 1. Then 2^{p/q} = 3, so 2^p = 3^q.

Is this the contradiction? Apply the checklist.

Check 1: can you pair 2^p = 3^q with an earlier line as P and \lnot P? You never asserted 2^p \neq 3^q. So no immediate pairing.
Check 2: is 2^p = 3^q false in all cases? Yes, for positive integers p, q — but only once you invoke the theorem that the only way 2^a = 3^b is a = b = 0, which is a consequence of unique prime factorisation.

So the contradiction is real, but it requires one more citation: the fundamental theorem of arithmetic. Without it, 2^p = 3^q is just an equation. With it, it directly conflicts with p, q \geq 1.

Result: contradiction confirmed — after the theorem is brought in. You would note this step as "which is impossible by unique prime factorisation."

The habit to build: after any algebraic derivation that feels like a contradiction, pause and run the two-part check. If both parts pass and you can quote the earlier line being denied, the proof is done. If either fails, keep going — the algebra has more to reveal.