Thread: Floors and Ceiling proof (Discrete Mathematics)

1. Floors and Ceiling proof (Discrete Mathematics)

∀a ∈ Z, if a is odd then ⌊a/2⌋ is odd or ⌈a/2⌉ is odd.

Hi,

I'm having a little trouble proving this statement.
I supposed a is an integer that is odd.
Then a has the form 2k+1 for some integer k.
Then a/2 is k + 1/2.
Now I'm not sure how to show that the floor or ceiling is odd.

I was thinking of saying that since k is an integer, once it is added with a strictly rational number(1/2), it would result in a non-integer, and thus be a number between two integers. Then I could say that since every consecutive integer is of opposite parity, then when you round down(floor) or round up(ceiling), it would result in 2 consecutive integers, and one of them must be odd.

Would that be sufficient?

2. Originally Posted by Whiz
∀a ∈ Z, if a is odd then ⌊a/2⌋ is odd or ⌈a/2⌉ is odd.

Hi,

I'm having a little trouble proving this statement.
I supposed a is an integer that is odd.
Then a has the form 2k+1 for some integer k.
Then a/2 is k + 1/2.
Now I'm not sure how to show that the floor or ceiling is odd.

I was thinking of saying that since k is an integer, once it is added with a strictly rational number(1/2), it would result in a non-integer, and thus be a number between two integers. Then I could say that since every consecutive integer is of opposite parity, then when you round down(floor) or round up(ceiling), it would result in 2 consecutive integers, and one of them must be odd.

Would that be sufficient?
Yup, that is exactly right.

3. Okay I wasn't sure of that one since my other questions were of a very different style.

Thank you!