Created at: 2025-05-11
Proving by case means dividing a problem into all the possible branches and then solving each branch individually.
In other words, "divide and conquer".
Proposition:
n is an integer, then n² + n + 6 is even.Fact:
n is an integer, it is either odd or even.Definitions:
n is even if n = 2k where k is another integer.n is odd if n = 2k + 1 where k is another integer.Branch 1 n is even:
n² + n + 6, then
(2k)² + 2k + 6, then
4k² + 2k + 6, then
2(2k² + k + 3), then
2x where x = (2k² + k + 3)
By fact (2) x is an integer.
By definition (a), we have that n² + n + 6 results in an even number when
n is even.
Branch 2 n is odd:
n² + n + 6, then
(2k + 1)² + 2k + 1 + 6, then
4k² + 4k + 1 + 2k + 1 + 6, then
4k² + 6k + 8, then
2(2k² + 3k + 4), then
2x where x (2k² + 3k + 4)
By fact (2) x is an integer.
By definition (a), we have that n² + n + 6 results in an even number when
n is odd.