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#### bergausstein

##### Active member

- Jul 30, 2013

- 191

1. show that if $A\cap B\,=\,\emptyset$, then $\left(A\,x\,C\right)\cap \left(B\,x\,C\right)\,=\,\emptyset$ that is if A and B have no elements in common, then there can be no ordered pair $\left(x,c\right)$ which is in both AxB and BxC.

my solution,

$A\,x\,C\,=\,\{\left(a,c\right)|a\in A\,\wedge\,c\in C\}$

$B\,x\,C\,=\,\{\left(b,c\right)|b\in B\,\wedge\,c\in C\}$

since AxC and BxC have no elements in common $A\cap B\,=\,\emptyset$

2. explain why $\left(A\cap B\right)\,x\,C\,=\,\left(A\,x\,C \right)\cap \left(B\,x\,C\right)$, that is if an ordered pair (x,c) is in AxC and in BxC, then (x,c) must be in $\left(A\cap B\right)\,x\,C$ and conversely.

thanks!