# Thread: Help! Proof with Sets

1. ## Help! Proof with Sets

Let A, B, C be sets. Prove either using a membership table or other method that:
(A\B) x C = (AxC) \ (BxC)

2. ## Honestly

when it comes to sets first what I would do is draw three intersecting cirlces that represent the sets then test your hypothesis...that is the best way to visualize...from there it may become readily apparent how to do the problem

3. $\displaystyle (x,y) \in \left( {A\backslash B} \right) \times C\quad \Rightarrow \quad x \in A \wedge x \notin B \wedge y \in C$
$\displaystyle x \in A \wedge x \notin B \wedge y \in C \Rightarrow \quad \left( {x,y} \right) \in A \times C \wedge \left( {x,y} \right) \notin B \times C$
$\displaystyle \left( {x,y} \right) \in A \times C \wedge \left( {x,y} \right) \notin B \times C \Rightarrow \quad \left( {x,y} \right) \in \left[ {\left( {A \times C} \right)\backslash \left( {B \times C} \right)} \right]$

Can you turn that around to get both ways?

4. Do I literally just reverse what you showed me?
And then say that this proves them equal because (x,y) is on both (A\B)xC and (AxC)\(BxC) ?

5. Originally Posted by vballnyy2
Do I literally just reverse what you showed me?
And then say that this proves them equal because (x,y) is on both (A\B)xC and (AxC)\(BxC) ?
Not necessarily. See if you can do it or if you need a different proof to go backwards, make sure that it works from your beginning assumption of (AxC)\(BxC).

And to prove that two sets are equal, you prove that they are subsets of each other - not "because (x,y) is on both (A\B)xC and (AxC)\(BxC)" but maybe that's what you mean.

6. Originally Posted by vballnyy2
Do I literally just reverse what you showed me?
Not quite! It is a bit more tricky.
$\displaystyle \left( {x,y} \right) \notin B \times C \Rightarrow \quad x \notin B \vee y \notin C$
It really is a different proof.

7. Originally Posted by Plato
Not quite! It is a bit more tricky.
$\displaystyle \left( {x,y} \right) \notin B \times C \Rightarrow \quad x \notin B \vee y \notin C$
It really is a different proof.

OK I'm giving it a try... not sure if this makes sense:

So for the second part...

If (x,y) is in [(AxC)\(BxC)] then (x,y) is not in (BxC).
This means that x is not in B or y is not in C.
Since (x,y) is in (AxC) then y is in C, so x is not in B.
Therefore, x is in A, x is not in B, and y is in C,
so (x,y) is in (A\B) x C.
Therefore [(AxC)\(BxC)] is a subset of (A\B)xC (and from before vice versa), so the two sides are equal.

Does this make sense at all??

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# (A\B)xC=(AxC)\(BxC)

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