Proving A x (BUC) = (A x B) U (B x C) in Modern Algebra

[OhyesOhno]

This problem involves modern algebra sets... particularly on cartesian products... so here's the problem:

Prove that A x (BUC) = (A x B) U (B x C)

Note that U here is 'union'

How do I prove that?? I know by the law of De Morgen that A U (BnC) = AUB n AUC but I don't know how to prove that one...

 

[nicksauce]

Mmmm I'm not so sure that this statement is true.

Suppose C = {1}, B={0}, A={37,38}.

Then Ax(BUC) = Ax{0,1} = {(37,0), (37,1), (38,0), (38,1)}. Meanwhile (AxB)U(BxC) = {(37,0),(38,0)}U{(0,1} = {(37,0),(38,0),(0,1)}

Did you happen to mean Ax(BUC) = (AxB)U(AxC) ?

 

[OhyesOhno]

Oh yeah, sorry about that. Ax(BUC) = (AxB)U(AxC) is what I meant. Sorry!

 

[Dick]

Pick an element (x,y) of the left side and show it's in the right side and conversely. If you describe the contents of the two sides of the equation in words, you'll find you are saying the same thing.