A student stares at the claim "if n is divisible by 6, then n is divisible by 3." They know it is true. They can check it for n = 6, 12, 18, 24. But when they try to write the proof, they get stuck — and the reason they get stuck is almost always the same. They start doing algebra too early, on words that are not yet equations.
Algebra operates on equations, not on English. "n is divisible by 6" is English. You cannot factor it, substitute it, or divide both sides of it. You first have to translate it into an equation: "n = 6k for some integer k." Now n is a symbol you can push around.
The single most reliable habit in direct-proof writing is this: before you touch algebra, unfold every defined term into its equation form. Write the definition out. Every term. No shortcuts.
What "unfolding" means
Unfolding a definition is taking a term with a one-word English label and replacing it with the algebraic condition the definition specifies. A defined term is a little compressed package; unfolding is the act of unzipping it.
A short catalogue of the most common unfoldings you will meet:
- "n is even" unfolds to "n = 2k for some integer k."
- "n is odd" unfolds to "n = 2k + 1 for some integer k."
- "a divides b" (written a \mid b) unfolds to "b = aq for some integer q."
- "A \subseteq B" unfolds to "for every x, x \in A \implies x \in B."
- "n is a multiple of 6" unfolds to "n = 6m for some integer m."
- "x is rational" unfolds to "x = p/q for some integers p, q with q \ne 0."
Why: algebra is a manipulation of symbols bound by equations. Until the defined term is in equation form, there is no symbol for algebra to grab. The English sentence "n is divisible by 6" does not contain any variable that algebra can touch — but n = 6k does.
The two-column move
A clean way to do this in practice is to set up two columns before you write any algebra.
Left column: the English claim. Right column: the equation form of every defined term.
For "if n is divisible by 6, then n is divisible by 3," the columns look like:
| English (hypothesis) | Equation form |
|---|---|
| n is divisible by 6 | n = 6k for some k \in \mathbb{Z} |
| English (conclusion) | Equation form |
|---|---|
| n is divisible by 3 | n = 3m for some m \in \mathbb{Z} |
Now the proof is almost writing itself. You have n = 6k. You want n = 3m. Substitute: n = 6k = 3 \cdot (2k). Take m = 2k; it is an integer (product of integers); done.
Without the two columns, students often try to "prove n is divisible by 3" by waving at the word divisible — and get nowhere, because there is no equation to manipulate.
What happens when you skip the unfolding
Here is a common failed attempt at "if n is even, then n^2 is even."
"If n is even, then n times n is even because even times anything is even."
This is not a proof. It invokes a fact — "even times anything is even" — without unfolding what even means. The fact is true, but using it this way hides the real structure, and on a slightly different problem this style falls apart.
Here is the same proof, with the unfolding done first:
- Hypothesis: n is even. Unfold: n = 2k for some integer k.
- Conclusion: n^2 is even. Unfold (target): n^2 = 2m for some integer m.
- Algebra: n^2 = (2k)^2 = 4k^2 = 2(2k^2). Let m = 2k^2; m \in \mathbb{Z}; done. \blacksquare
The second version is three lines and every line is a concrete manipulation. The unfolding is what made the algebra available.
The visualisation: words versus equations
Below is a slider that switches between "words" mode and "equations" mode for the same claim. Watch how much algebra becomes possible once the translation happens.
Unfolding compound hypotheses
When the hypothesis has multiple defined terms, unfold them all — do not pick favourites.
"If a \mid b and b \mid c, then a \mid c."
- Unfold a \mid b: b = aq for some integer q.
- Unfold b \mid c: c = br for some integer r.
- Unfold the conclusion a \mid c (target): c = as for some integer s.
- Bridge: c = br = (aq)r = a(qr). Take s = qr; it is an integer; done. \blacksquare
Notice that each defined term got its own variable (q, r, s). Never reuse the same letter for two different unfoldings — that is a source of silent mistakes. The right move is: fresh letter for every fresh existential.
Why a fresh letter each time: when you unfold "a \mid b" as "b = aq," the q is an integer specific to this particular pair (a, b). When you unfold "b \mid c" as "c = bq," the new q is a different integer — specific to (b, c). If you reuse the letter, you are silently claiming the same integer divides both pairs, which is not something your hypothesis gave you.
The rule of thumb
If, at any point in writing a direct proof, you are about to do algebra and you are unsure what symbol to push where — stop. Scroll back up to the hypothesis. Ask: "have I unfolded every defined term into an equation?" If there is even one defined term still in its English form, unfold it. Add a line to the proof: "Since X is [definition], X = \dots for some \dots."
This single discipline resolves most cases of proof paralysis. The English version of a claim is the advertisement; the equation version is the product. Work with the product.
The short summary
- A direct proof is algebra on equations — it cannot act on English terms.
- Before you attempt any manipulation, unfold every defined term into its equation form.
- Use a fresh integer letter for each unfolding to avoid silent collisions.
- Unfold the conclusion too — it tells you the target shape to aim for.
- If you are stuck mid-proof, the fix is almost always another unfolding, not a cleverer manipulation.
The discipline is the opposite of cleverness: it is slow, mechanical, and boring. That is exactly why it works — a mechanical step is a step you never forget to take.
Related: Mathematical Proof — Direct Proof · First Move on Every 'If P Then Q' · Same Variable in P and Q — Substitution and Algebra · State Every Definition or Assume Reader Knows?