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theEric
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Math proof help, please! Relations and subsets involved. I don't see the solution!
My math book proved that if sets \[A, B,C,D\] existed such that \[A \subseteq C \]and\[B \subseteq D\] then \[(A \times B) \subseteq (C \times D)\]
I need to show that the converse of that is false. I think the converse is\[(A \times B) \subseteq (C \times D)\]implies\[A \subseteq C \]and\[B \subseteq D\].
But I can't understand how it is false for some group of sets \[A,B,C,D\].
 2 years ago
 2 years ago
theEric Group Title
Math proof help, please! Relations and subsets involved. I don't see the solution! My math book proved that if sets \[A, B,C,D\] existed such that \[A \subseteq C \]and\[B \subseteq D\] then \[(A \times B) \subseteq (C \times D)\] I need to show that the converse of that is false. I think the converse is\[(A \times B) \subseteq (C \times D)\]implies\[A \subseteq C \]and\[B \subseteq D\]. But I can't understand how it is false for some group of sets \[A,B,C,D\].
 2 years ago
 2 years ago

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zugzwang Group TitleBest ResponseYou've already chosen the best response.1
Use null sets :)
 2 years ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Alright, thanks.. I never went into writing a proof on that, but I thought about it and didn't see it working... Thank you!
 2 years ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.1
No problem :)
 2 years ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Ah!! I think I finally see it.. Thanks for making me concentrate more on the null set issue!
 2 years ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.1
Got it? :)
 2 years ago
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