Proving Sum of Odds is Even

Intro to Proofs · Axiom Academy

EXAMPLE Proving Sum of Odds is Even Construct a rigorous direct proof using the algebraic representation of odd integers. Prove that the sum of any two odd integers is an even integer. We'll build the proof one step at a time — at each step, choose the move that keeps the argument rigorous. Excellent work! You've constructed a complete direct proof. Here's what made it rigorous: Direct proof structure: We proved a universal claim by working with arbitrary odd integers and showing the property holds in general — not just for examples. Algebraic representation: Any odd integer can be written as 2k + 1 for some integer k . Capturing parity algebraically is what makes the argument work. Distinct variables: Using 2m + 1 and 2n + 1 (not the same variable twice) keeps the two integers independent and the proof fully general. Reveal the form 2k : Factoring the 2 out of 2m + 2n + 2 = 2(m + n + 1) exposes exactly the definition of an even number. Closure: Because integers are closed under addition, m + n + 1 is an integer, so the sum really does have the form . This technique — represent numbers algebraically, then manipulate to reveal the property you want — is the backbone of parity and number-theory proofs. You'll reuse it for sums, products, and proofs about even, odd, and divisible integers.

This is the written version of the interactive lesson above. See the full Intro to Proofs course.