121 reputation
8
bio website entangled-logic.com
location Delta, Canada
age 35
visits member for 3 years, 5 months
seen Nov 29 '12 at 4:58

I am currently an Undergraduate Student of Simon Fraser University, Double Majoring in Mathematics and Computer Science. I have recently returned to school after fifteen years working in Industrial Production Engineering and Trades--CNC etc. I am involving myself in areas of Scientific Computing and working on a few of my own projects!


Sep
1
comment Series used in proof of irrationality of $ e^y $ for every rational $ y $
Thank you for pointing that out!
Sep
1
comment Series used in proof of irrationality of $ e^y $ for every rational $ y $
x-1 was my typo, thanks!
Apr
20
comment Boolean Expression Orders of Operation
thanks you for the heads up!
Apr
20
comment Boolean Expression Orders of Operation
This and your comments answer the question, thank you!
Apr
20
comment Boolean Expression Orders of Operation
ahhhh yes that is a problem, thank you!
Apr
20
comment Boolean Expression Orders of Operation
Very sorry about my attention to detail on this post!
Apr
20
comment Boolean Expression Orders of Operation
I hadn't written in all the steps properly. Given your correction(edited in), which I omitted in my original solution, I still get the same final answer. :/
Apr
20
comment Boolean Expression Orders of Operation
I forgot to make it F' = and also take the complement while I was copying it over, yes. I edited it in thank you!
Mar
12
comment Proof: Cartesian Product of Two Sets is a Set ZF
I thought as much, thanks for the reply! :)
Mar
12
comment Proof: Cartesian Product of Two Sets is a Set ZF
I'm not sure if by 'exists' you are meaning the same as 'is a set'. I'm not proving the existence of the Cartesian Product but that in fact it is not a 'proper class' if it is the product of two sets.
Mar
12
comment Proof: Cartesian Product of Two Sets is a Set ZF
Basic Set Theory by Azriel Levy
Mar
12
comment Proof: Cartesian Product of Two Sets is a Set ZF
This would still require using Power Set as below to finalize the proof!
Mar
12
comment Proof: Cartesian Product of Two Sets is a Set ZF
I guess the follow through is proving by replacement that if y is in t then y cannot be in s given the pair y(in replacement) = {s,t}. This said then for all s = A, for all t = B, A union B is a set by a proof already given 5.19 iii). This seems a bit convoluted to me and I will have to reevaluate the steps.
Mar
12
comment Proof: Cartesian Product of Two Sets is a Set ZF
I think I should be able to derive from this the answer that the text is suggesting I find. Thank you for the proof.