Every direct proof opens with "Assume P." That two-word phrase grants you a licence to treat P as if it were true for the rest of the proof. But students new to proof-writing quickly hit a confusing question: "Does that mean I can assume specific examples of P? Does it mean I can assume things related to P? Can I assume a consequence of P that I already know is true?"
The answer has three parts and they are worth nailing down before you write your next proof.
Part 1 — You may treat P as a true fact for the rest of the argument
If P is "n is even," then after "Assume P" you can use "n is even" anywhere in the proof as if it were an established theorem. You can apply the definition of even to n: "by definition, n = 2k for some integer k." That is the canonical first move.
You cannot, however, pretend P said something stronger. If P is "n is even," you cannot assume "n is divisible by 4" — that would be a different (stronger) statement, and assuming it would mean you are proving something about multiples of 4, not about even numbers.
Why this strictness matters: a proof is supposed to work whatever specific objects satisfy the hypothesis. If you assume n is divisible by 4, your conclusion only applies to multiples of 4, not to all even numbers. The claim you proved is therefore weaker than the one you set out to prove — and the proof fails.
Part 2 — You may not assume specific examples of P
A trap students routinely fall into: after "Assume n is even," they say "let n = 6." That is not legal. Choosing n = 6 proves your statement only for n = 6, not for all even integers. One example is not a proof (see 100 Examples vs One Direct Proof).
The right move is to treat n as an arbitrary even integer — n = 2k where k is some unspecified integer. Every statement you derive must hold for every possible k, not one convenient choice. The letter k carries all the cases.
This has a useful phrasing: the variable n in "Assume n is even" is universally quantified. It stands for any integer satisfying the hypothesis. The proof must work uniformly across all such n.
Part 3 — You may use anything already proved, defined, or axiomatic
The assumption "P" does not stand alone. You bring with you:
- Every axiom of arithmetic. Commutativity, associativity, distributivity, closure under addition and multiplication.
- Every definition. Even, odd, prime, rational, limit — whatever definitions apply to the objects in the proof.
- Every previously proved theorem. If you proved "the sum of two even numbers is even" earlier, you may invoke it.
What you cannot invoke is something you have not yet established. Circular proofs — using the thing you are trying to prove as a step in the proof — are always illegal. If the claim is "if P then Q," you cannot use "Q" anywhere before you have derived it.
The "assume box" — a visual
A walked example of the common trap
Claim. For every even integer n, n^2 + 2n is even.
A common student attempt:
Assume n is even. Let n = 4. Then n^2 + 2n = 16 + 8 = 24, which is even. So the claim is true.
This is not a proof. The student has verified the claim for one even integer, not for every even integer. Dropping in n = 4 violates Part 2 — specific examples are not allowed.
The correct version:
Assume n is even. By definition, n = 2k for some integer k. Then
n^2 + 2n = (2k)^2 + 2(2k) = 4k^2 + 4k = 2(2k^2 + 2k).Since 2k^2 + 2k is an integer, n^2 + 2n is 2 \times (\text{integer}), hence even. \blacksquare
The variable k covers every even integer in one sweep. No specific value, no loss of generality.
A second walked example — the strengthening trap
Claim. If n is prime and n > 2, then n is odd.
A tempting move:
Assume n is prime and n > 2. Then n is one of 3, 5, 7, 11, \ldots — all of which are odd. So n is odd.
This assumes the conclusion by listing cases that are all odd — but the listing is itself what needs proof. It smuggles in the answer.
The correct move is a direct proof from the hypothesis:
Assume n is prime and n > 2. Suppose for contradiction n is even. Then n = 2k for some integer k. Since n > 2, k > 1. So n has divisors 1, 2, n (and possibly more). In particular 2 is a divisor of n different from 1 and n, so n is not prime — contradiction. Hence n is odd. \blacksquare
The hypothesis (prime, n > 2) was used; no specific values of n were plugged in.
What "Without Loss of Generality" is (and is not)
A phrase like "without loss of generality, assume a \le b" (abbreviated WLOG) looks like it breaks Part 2 — you are choosing a specific case. But WLOG is allowed only when the statement is symmetric in a and b, so the case a \le b covers the case a \ge b by just relabelling. You have not actually picked a specific example; you have just named which of two symmetric halves you are treating.
If the statement is not symmetric, WLOG is illegal. This is the fine print of Part 2 — symmetry is a genuine gain, but you must justify it.
The short summary
- Granted: exactly what the hypothesis P says — no more, no less.
- Not granted: specific examples, stronger statements, the conclusion Q itself.
- Always available: axioms, definitions, previously proved theorems.
The rule of thumb: whenever you feel the urge to drop in a specific value, ask "does the claim hold for every value satisfying the hypothesis, or only for this one?" If every value, you must work with the variable. If only this one, the claim is narrower than you think.
Related: Mathematical Proof — Direct Proof · 100 Examples Checked vs One Direct Proof · Forward-Chaining Tree · Proof by Contradiction