Example
To prove that every integer that is a perfect cube is a multiple of 9, or is 1 more than a multiple of 9, or is 1 less than a multiple of 9.
Proof:
Each cube number is the cube of some integer n. This integer n is either a multiple of 3, or 1 more or 1 less than a multiple of 3. So these 3 cases are exhaustive:
- Case 1: If n = 3p, then n3 = 27p3, which is a multiple of 9.
- Case 2: If n = 3p + 1, then n3 = 27p3 + 27p2 + 9p + 1, which is 1 more than a multiple of 9. For instance, if n = 4 then n3 = 64 = 9x7 + 1.
- Case 3: If n = 3p − 1, then n3 = 27p3 − 27p2 + 9p − 1, which is 1 less than a multiple of 9. For instance, if n = 5 then n3 = 125 = 9×14 − 1.
Read more about this topic: Proof By Exhaustion
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