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| bio | website | entangled-logic.com |
|---|---|---|
| location | Delta, Canada | |
| age | 33 | |
| visits | member for | 2 years, 2 months |
| seen | Nov 29 '12 at 4:58 | |
| stats | profile views | 114 |
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!
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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. :/ |
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Apr 20 |
revised |
Boolean Expression Orders of Operation added 78 characters in body |
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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! |
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Apr 20 |
revised |
Boolean Expression Orders of Operation edited body; added 1 characters in body |
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Apr 20 |
asked | Boolean Expression Orders of Operation |
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Mar 12 |
comment |
Proof: Cartesian Product of Two Sets is a Set ZF I thought as much, thanks for the reply! :) |
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Mar 12 |
accepted | Proof: Cartesian Product of Two Sets is a Set ZF |
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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. |
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Mar 12 |
awarded | Supporter |
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Mar 12 |
comment |
Proof: Cartesian Product of Two Sets is a Set ZF Basic Set Theory by Azriel Levy |
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Mar 12 |
awarded | Scholar |
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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! |
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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. |
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Mar 12 |
awarded | Editor |
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Mar 12 |
revised |
Proof: Cartesian Product of Two Sets is a Set ZF deleted 4 characters in body |
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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. |
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Mar 12 |
awarded | Student |
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Mar 12 |
asked | Proof: Cartesian Product of Two Sets is a Set ZF |
