CSC 2511 Fall 2010

Hi folks -- My starting thought on the puzzle is that while the rows are increasing, each path has

Yea On Mon, Oct 11, 2010 at 11:48 PM, gage saber <gage...@gmail.com> wrote: I'm not sure

My first homework was lost before it was graded. Did anyone get my assignment stapled to theirs? It

I agree... seeing also that (a^2 + b^2 is even) is not reflexive therefore not an equivalence. On Thu

The binary relation R is symmetric on N. Proof The binary relation R is symmetric on N if and only if

ill meet you there after work tomorrow. I'll try to make it there by 2pm. On Sep 15, 2010, at 10:

3|x+2y is not anti-symmetric. By example: x = 5 and y = 2; then 3|(5+2(2)) and 3|(2+2(5)) but a not

The set will be composed of any combination where both a and b are both odd or both even.

Much better! Thank you!

Yeah I just posted my answer in response to anne. Sorry i didn't take a look at what you wrote :P

P | Q | P v !(P ^ Q) ---+---+-------------- 0) T | T | TF T 1) T | F | TT F 2) F | T | TT F 3) F | F

Lol I was going to say. =) On Mon, Sep 6, 2010 at 7:16 PM, Anne S <anne....@gmail.com> wrote:

T = (1 + 5^.5)/2 T = 1 / (T - 1) -> T(T -1) = 1 -> T^2 - T - 1 = 0 -> T = (1 + 5^.5)/2 Yea

G is a planer graph -> G can be colored with at most 4 colors Converse G can be colored with at

3 + 5(2)^.5 = (a + b(2)^.5)^2 = (a + b(2)^.5)(a + b(2)^.5) = a^2 + 2ab(2)^.5 + 2b^2 = (a^2 + 2b^2) +

yeah... I guess that if you look at the implication as being a^2-b^2 odd then a even, b odd OR a odd,

Cool. Feel free to agree or disagree with the solutions I present and contribute your own. =) On Mon,