Even + Even = Even, Animated: Watch 2k and 2m Group Into 2(k + m)

The claim "even plus even equals even" is so obvious that your instinct is to wave it through without proof. But the reason the direct proof of this innocent-looking statement is important is not its content — it is its shape. Every proof about even and odd numbers, every proof about divisibility, every proof about numbers of a certain form uses this exact template: unpack the hypothesis with the definition, do one algebraic rearrangement, repack the conclusion with the definition.

Watch the rearrangement happen. Below, two stacks of paired blocks ("pairs" = twos) sit side by side. You drag a dot, and the stacks slide together, a bracket appears around them, and the factor of 2 is pulled out front. The visual is the proof.

The animation

k (green pairs): m (blue pairs): stage:

Two coloured stacks of pairs. Green holds $k$ pairs ($= 2k = a$); blue holds $m$ pairs ($= 2m = b$). Press Play (or drag the stage slider): the stacks slide together, the count collapses to $k+m$ pairs, and a bracket pulls the $2$ out front. The sum is a multiple of $2$.

The proof in three lines

Here is the written version the animation illustrates.

Claim. The sum of two even integers is even.

Proof.

a + b \;=\; 2k + 2m \;=\; 2(k + m).

Since k and m are integers, k + m is an integer (integers are closed under addition). Therefore a + b = 2 \times (\text{integer}), which means a + b is even. \blacksquare

The three algebraic equalities are the stages of the animation:

a + b = 2k + 2m — stage 2, the stacks joined but not yet regrouped.
2k + 2m = 2(k + m) — stages 3 and 4, the regroup and factor.
2(k + m) is a multiple of 2 — the stage-4 declaration.

Why this short proof is the whole idea: the word "even" is defined as "a multiple of 2." To prove something is even, you must write it in that form. The algebra 2k + 2m = 2(k + m) does exactly that — it packages the sum as 2 \cdot (\text{something integer}). The definition of even is the check: does this expression match "2 \times integer"? Yes. Done.

The same template for "even \times odd = even"

Once you have seen the shape, the variant proofs write themselves.

Claim. The product of an even integer and an odd integer is even.

Proof. Let a = 2k (even) and b = 2m + 1 (odd). Then

a b \;=\; 2k (2m + 1) \;=\; 2 \cdot \bigl(k(2m + 1)\bigr).

Since k(2m + 1) is an integer, ab is 2 \times (\text{integer}), hence even. \blacksquare

Same scaffolding: unpack both hypotheses with the definition, multiply, factor out the 2, re-pack with the definition. The only thing that changed is the specific algebra in the middle.

What the animation hides (and why that matters)

The animation shows specific numbers of pairs — say, k = 3 and m = 3. A student might reasonably ask: "But how do I know this works for k = 100, m = 7?"

The answer: the animation is just a picture of one case. The argument itself — 2k + 2m = 2(k + m) — is algebra with variables, not specific numbers. Whatever values k and m take, the equality 2k + 2m = 2(k + m) is an instance of the distributive law. The distributive law holds for all integers, not just the ones in the picture. So the proof covers every choice of k and m simultaneously.

This is the same point made visually in 100 Examples vs One Direct Proof: the proof argues about the form of the integers, not about specific values.

A visual worked example — with numbers

Take a = 6, b = 10. Then k = 3, m = 5. The animation would show a green stack of 3 pairs and a blue stack of 5 pairs. Join them: 8 pairs total. The sum 6 + 10 = 16 = 2 \times 8. Even, confirmed.

Now take a = 4, b = 8. Then k = 2, m = 4. Green: 2 pairs. Blue: 4 pairs. Joined: 6 pairs. Sum 4 + 8 = 12 = 2 \times 6. Even.

Whatever k and m are, the merged stack has k + m pairs — and a stack of pairs is, by definition, an even number.

The odd + odd case (and why it gives even, not odd)

A natural next question: what about odd + odd? The animation shape changes slightly.

Let a = 2k + 1 and b = 2m + 1. Each stack now has an extra single block beyond the pairs.

a + b = (2k + 1) + (2m + 1) = 2k + 2m + 2 = 2(k + m + 1).

The two extra singles combine into a new pair, bumping the pair count from k + m to k + m + 1. Sum: 2 \times (k + m + 1) — still even.

So: odd + odd = even. Even + even = even. Only even + odd (one pair-stack, one pair-stack-plus-one) gives an unmatched single block, making the sum odd.

The ethos — pictures as proof scaffolding

The animation is not the proof. The proof is the algebra 2k + 2m = 2(k + m). But the animation is the mental scaffolding that makes the algebra feel obvious, and that scaffolding is what you carry with you to new problems. When you encounter "show that three consecutive integers sum to a multiple of 3," you reach for the same mental move — unpack each integer with its definition-in-form, add, factor, re-pack.