Parity Rules in Proofs: How Even and Odd Numbers Help You Prove Things

Parity is a fancy word for something simple: whether an integer is even or odd. Parity arguments are one of the most beginner-friendly proof tools because they are concrete, reliable, and often lead to quick contradictions.

This post gives you the core parity rules, shows how to use them in proofs, and walks through a classic example (proving an equation has no rational solution).

The two definitions that power everything

An integer n is:

• Even if n = 2k for some integer k.

• Odd if n = 2k + 1 for some integer k.

That’s it. Every parity fact comes from these two forms.

The parity rules you should memorize

Here are the rules you will use constantly.

Addition and subtraction

• even + even = even

• odd + odd = even

• even + odd = odd

• (same rules for subtraction)

Multiplication

• even · anything = even

• odd · odd = odd

• A product is odd only if every factor is odd.

That last bullet is extremely useful in proofs: If xy is odd, then both x and y must be odd.

Powers

• If n is even, then n², n³, n⁴, ... are all even.

• If n is odd, then n², n³, n⁴, ... are all odd.

So n and nᵏ always have the same parity for positive integers k.

How parity gets used in proofs

Parity is often used in proof by contradiction:

1. Assume the opposite of what you want to prove.

2. Use algebra to create an equation involving integers.

3. Apply parity rules to show something impossible happens (like a number being both odd and even).

4. Conclude your assumption must be false.

The most common contradiction is this one:

• You prove a certain integer must be even, and also must be odd.

Another common contradiction (especially with rational numbers) is:

• You end up proving both numerator and denominator are even, which contradicts the fraction being in lowest terms.

The classic setup: “Assume a rational solution exists”

When a problem says “Show there is no rational solution,” your default template is:

1. Assume x is rational.

2. Write x = a / b where a, b ∈ ℤ, b > 0, and gcd(a, b) = 1 (lowest terms).

3. Substitute into the equation.

4. Multiply to clear denominators so you get an integer equation.

5. Use parity rules to force a contradiction.

Why “lowest terms” matters: If you ever prove a and b are both even, then gcd(a, b) ≥ 2, which contradicts “lowest terms.”

Example: How parity forces a contradiction

Suppose we are asked to show that x³ + x² = 1 has no rational solution.

Step 1: Assume x is rational

Let x = a / b

in lowest terms, meaning gcd(a, b) = 1.

Step 2: Substitute and clear denominators

(a / b)³ + (a / b)² = 1 (a³ / b³) + (a² / b²) = 1

Multiply by b³: a³ + a²b = b³

Factor: a²(a + b) = b³

Now we have a pure integer relationship.

Step 3: Parity punchline (the idea)

• If b is even, then b³ is even, so a²(a + b) is even. If a were odd, then a² would be odd and a + b would be odd (odd + even), so a²(a + b) would be odd · odd = odd, a contradiction. Hence a must be even, making both a and b even, which contradicts lowest terms.

• If b is odd, then b³ is odd, so a²(a + b) is odd. A product is odd only if each factor is odd, so a² must be odd and a + b must be odd. Since a² is odd, a must be odd. But now a and b are both odd, which makes a + b even (odd + odd = even), contradicting that a + b is odd.

Either way, contradiction. So no rational solution exists.

The key takeaway: parity is what traps you. You split into even/odd cases, and both cases fail.

Quick “proof moves” to practice

Here are three mini facts that show up a lot:

Move 1: “If a square is even, the number is even.”

If n² is even, then n must be even. Reason: If n were odd, then n² would be odd.

Move 2: “If a product is odd, each factor is odd.”

If xy is odd, then x is odd and y is odd.

Move 3: “Odd + odd = even” is a contradiction generator

If you ever need a + b to be odd while also knowing a and b are odd, you’re done.

Mini practice (with answers)

Problem A

Show that n² is odd if and only if n is odd.

Answer idea:

• If n is odd, write n = 2k + 1), then square it.

• If n² is odd, n cannot be even, so n must be odd.

Problem B

Show that a + b is even if and only if a and b have the same parity (both even or both odd).

Answer idea: List the two parity cases and use the addition rules.

Closing thought

Parity arguments are popular in intro proofs because they are:

• easy to apply,

• easy to explain,

• and surprisingly powerful for ruling out solutions.

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