Division Into Cases Examples
17. The product of any two consecutive integers is even.
Proof
Let n and n+1 any two be consecutive integers. There are two cases, n is even or is n is odd. We
need to show n(n  1) is even in both cases.
Case 1. n is even. Let k be such that n  2k . Then
n(n  1)  (2k )(2k  1)
 2  k (2k  1)
 2t
Where t  k (2k  1) is an integer. Therefore, by definition we have that n(n  1) is even.
Case 2. n is odd. Let k be such that n  2k 1 . Then
n(n  1)  (2k  1)(2k  1  1)
 (2k  1)(2k  2)
 2(2k  1)( k  1)
 2t
where t  (2k  1)(k  1) is an integer. Therefore, by definition n(n  1) is even.
28a. The product of any three consecutive integers is divisible by 3.
Proof
Let n, n+1, n+2 be any three consecutive integers. There are three cases, n  3k or n  3k  1 or
n  3k  2 where k is some integer. We need to show the product n  n  1 n  2 is divisible by
3 in each of these cases.
Case 1: n  3k for some k. Then
 n  n  1 n  2    3k  3k  1 3k  2 
 3  k  3k  1 3k  2  
 3m
where m  k  3k  1 3k  2 is an integer. Hence by definition n  n  1 n  2 is divisible by 3 .
Case 2: n  3k  1 for some k. Then
n  n  1 n  2    3k  1 3k  2  3k  3
  3k  1 3k  2   3   k  1
 3   3k  1 3k  2  k  1
 3m
where m   3k  1 3k  2 k  1 is an integer. Hence by definition n  n  1 n  2 is divisible
by 3 .
Case 3: n  3k  1 for some k. Then
n  n  1 n  2    3k  2  3k  3 3k  4 
  3k  2   3   k  1 3k  4 
 3   3k  2  k  1 3k  4 
 3m
where m   3k  2 k  1 3k  4 is an integer. Hence by definition n  n  1 n  2 is divisible
by 3 .