- #1
press
- 11
- 0
Homework Statement
Use math induction and cases to prove that every integer is even or odd.
The Attempt at a Solution
Let p(n): n is even or is odd.
Let n be an integer. Assume p(n) is true.
Then n is even or n is odd.
Case 1: Assume n is even. Then n = 2k for some integer k. So n + 1 = 2k + 1. Thus n + 1 is odd.
So, n + 1 is odd or n + 1 is even.
Hence, p(n + 1) is true.
Case 2: Assume n is odd.Then n = 2k + 1 for some integer k. So n + 1 = 2(k + 1). Thus n + 1 is even.
So, n + 1 is even or n + 1 is odd.
Hence, p(n + 1) is true.
In either case, p(n + 1) is true. So, p(n) => p(n + 1)
-------------------------------------------------------------
I copied it from the back of my book. My question is how do we know n + 1 might also be even in Case 1 before we did Case 2? Is there a logical reason for that or is it just stylistic? I am inclined to re-write this proof like this below:
...
Case 1: Assume n is even. Then n = 2k for some integer k. So n + 1 = 2k + 1. Thus n + 1 is odd.
Hence, p(n + 1) is true
Case 2: Assume n is odd.Then n = 2k + 1 for some integer k. So n + 1 = 2(k + 1). Thus n + 1 is even.
Hence, p(n + 1) is true
So, n + 1 is even or n + 1 is odd.
In either case, p(n + 1) is true. So, p(n) => p(n + 1)
------------------------------------------------------
I am reasoning like this:
p or q
Case 1: Assume p
...
Therefore r
Case 2: Assume q
...
Therefore, s.
Therefore r or s.
------------------------------------------
Please, tell me where I am tripping up.
Thanks.
Last edited: