Parity (Even/Odd) Math Example 2

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Example 2

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

Solution

  1. 1
    Let the two consecutive integers be nn and n+1n+1.
  2. 2
    Their sum: n+(n+1)=2n+1n + (n+1) = 2n + 1.
  3. 3
    2n2n is always even (divisible by 22), so 2n+12n + 1 is always odd.
  4. 4
    Therefore the sum of any two consecutive integers is odd.

Answer

Sum =2n+1= 2n + 1, which is always odd.
Using algebra with a general integer nn lets us prove the parity result for all possible pairs at once. The form 2n+12n+1 is the definition of an odd number: any number of the form 2k+12k+1 for integer kk is odd.

About Parity (Even/Odd)

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

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