Parity (Even/Odd) Math Example 3

Follow the full solution, then compare it with the other examples linked below.

Example 3

medium
Prove that the sum of two odd numbers is always even.

Solution

  1. 1
    Let the two odd numbers be 2a+12a + 1 and 2b+12b + 1, where aa and bb are integers.
  2. 2
    Add them: (2a+1)+(2b+1)=2a+2b+2=2(a+b+1)(2a + 1) + (2b + 1) = 2a + 2b + 2 = 2(a + b + 1).
  3. 3
    Since a+b+1a + b + 1 is an integer, the sum 2(a+b+1)2(a + b + 1) is divisible by 2, so it is even.

Answer

The sum of two odd numbers is always even: (2a+1)+(2b+1)=2(a+b+1)(2a+1) + (2b+1) = 2(a+b+1).
Every odd number has the form 2k+12k+1. When you add two such numbers, the two '+1' parts combine to give an extra 2, making the total divisible by 2. This algebraic proof works for all odd numbers, not just specific examples.

About Parity (Even/Odd)

The classification of integers as even (evenly divisible by 2, with no remainder) or odd (not divisible by 2).

Learn more about Parity (Even/Odd) →

More Parity (Even/Odd) Examples

Example 1 easy

Without fully computing, determine the parity (odd or even) of [formula] and of [formula].

Example 2 medium

Prove that the sum of any two consecutive integers is always odd.

Example 4 easy

Classify each as odd or even without fully computing: (a) [formula], (b) [formula], (c) [formula].

Example 5 medium

In a group of [formula] people, each person shakes hands with every other person exactly once. Is th

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