Direct Proof for Secondary Mathematics Students

The basic shape

A direct proof of “if A, then B” assumes A under the stated conditions and derives B using definitions, algebra and established properties. It does not begin by assuming B. It does not rely on a handful of favourable examples.

For Secondary Mathematics, a useful template is: define the arbitrary object, translate the given property into mathematics, transform it legally, and identify the required property in the final form.

Example 1: even plus even

Claim: the sum of two even integers is even.

Let the two even integers be 2a and 2b, where a and b are integers. Their sum is 2a + 2b = 2(a + b). Since a + b is an integer, the sum is two times an integer. Therefore the sum is even.

The variables are arbitrary. That is what lets the argument cover all pairs of even integers, rather than selected examples.

Example 2: odd plus odd

Let two odd integers be 2a + 1 and 2b + 1. Their sum is 2a + 2b + 2 = 2(a + b + 1). The expression in brackets is an integer, so the sum is even.

The key move is not sophisticated algebra. It is choosing a representation that contains the definition of oddness.

Example 3: a divisibility result

Claim: if an integer n is divisible by 6, then n² is divisible by 12.

Since n is divisible by 6, write n = 6k for some integer k. Then n² = 36k² = 12(3k²). Because 3k² is an integer, n² is divisible by 12.

This proves a sufficient condition. It does not claim that n being divisible by 6 is necessary for n² to be divisible by 12. That reverse direction is a separate question; see Necessary and Sufficient Conditions.

Example 4: consecutive integers

Claim: the product of two consecutive integers is even.

Let the consecutive integers be n and n + 1. Consecutive integers have opposite parity: one is even and the other odd. Therefore one factor is divisible by 2, so their product n(n + 1) is divisible by 2 and hence even.

An alternative proof splits into two cases: if n is even, the product is even; if n is odd, n + 1 is even, so the product is even. Both are direct arguments.

Example 5: an algebraic inequality

For real x, prove x² + 1 ≥ 2x.

For every real x, (x − 1)² ≥ 0 because a real square is nonnegative. Expanding gives x² − 2x + 1 ≥ 0. Rearranging gives x² + 1 ≥ 2x.

The proof starts from an established nonnegative quantity and transforms it into the desired inequality. Equality occurs when x − 1 = 0, so x = 1.

Direct proof in geometry

A direct geometry proof follows the same logic even though the representation changes. Begin with given lengths, angles, parallel lines or shape properties. Apply established angle, congruence, similarity or circle results. Each statement should have a reason that does not depend on the picture merely looking correct.

If two sides of a triangle are equal, for example, the corresponding base angles are equal by the isosceles-triangle theorem. If a transversal crosses parallel lines, the appropriate angle relationships follow from parallelism. The diagram organises the proof; the theorem licenses the step.

Choose the representation that contains the definition

• Even integer → 2k.
• Odd integer → 2k + 1.
• Multiple of m → mk.
• Consecutive integers → n and n + 1.
• Nonnegative square → u² ≥ 0 for real u.

Good representation often does most of the intellectual work. Once the definition is encoded, ordinary algebra can expose the conclusion.

Common direct-proof failures

Using a specific example. Writing 8 + 14 = 22 does not prove even plus even for arbitrary even integers.

Assuming the conclusion. If you begin “let the sum be even”, you have inserted the result you were asked to establish.

Losing the domain. If k must be an integer, say so. A divisibility proof depends on the multiplier being an integer.

Proving the converse. “If A then B” and “if B then A” are different claims. Check which direction the question asks for.

Practice

1. Prove that the sum of three consecutive integers is divisible by 3. Let them be n, n + 1 and n + 2. Their sum is 3n + 3 = 3(n + 1).

2. Prove that the square of an odd integer is odd. Let n = 2k + 1. Then n² = 4k² + 4k + 1 = 2(2k² + 2k) + 1.

3. Prove that if an integer is divisible by 8, it is divisible by 4. Write n = 8k = 4(2k), where 2k is an integer.

Continue

Start with What Counts as Proof? if you need the proof/evidence distinction. Use Counterexamples when a universal claim may be false, and return to the Secondary Mathematics Master Index.