# cartesian product proofs

• October 29th 2009, 09:45 AM
cartesian product proofs
A X (B U C)= (A X B) U (A X C)

proof:
Assume (x,y) is an element of A X (B U C). This means x is an element of A and y is an element of B or y is an element of C. Since (x,y) can be x as an element of A and y as an element of B, (x,y) is an element of A X B. Since (x,y) can also have x as an element of A and y as an element of C, (x,y) is an element of A X C.

A X (B-C)= (A X B) - (A X C)
not sure

A X (B intersect C)= (A X B) intersect (A X C)

• October 29th 2009, 10:31 AM
Plato
Quote:

A X (B-C)= (A X B) - (A X C)
not sure

Here is one. They all work about the same way.
$\begin{gathered}
(x,y) \in A \times (B\backslash C) \hfill \\
x \in A \wedge y \in B \wedge y \notin C \hfill \\
\left( {x \in A \wedge y \in B} \right) \wedge \left( {x \in A \wedge y \notin C} \right) \hfill \\
(x,y) \in A \times B \wedge (x,y) \notin A \times C \hfill \\
(x,y) \in \left( {A \times B} \right)\backslash \left( {A \times C} \right) \hfill \\
\end{gathered}$
• October 29th 2009, 10:40 AM
is a slash equal to the difference
also is the first 1right
• October 29th 2009, 02:05 PM
i dont understand
• October 29th 2009, 02:17 PM
Plato
Quote:

i dont understand

Maybe you should try to get some one-on-one help from your instructor.
• October 29th 2009, 04:42 PM
A X (B U C)= (A X B) U (A X C)
Proof:
(x,y) E A X (B U C).
x E A and y E (B U C)
X E A and (y E B U y E C)
(x E A and y E B) or (x E A and y E C)
(x,y) E A X B or (x,y) E A X C
(x,y) E (A X B) or (A X C)

and the argument can be reversed

the only thing I cant justify is how I go from step 3 to step 4 and how I go from step 4 to step 3.

E= element of
• October 29th 2009, 04:48 PM
cartesiian proof
Proof:
(x,y) E A X (B intersect C)
x E A and y E (B intersect C)
x E A and y E B and y E C
(x E A and y E B) and (x E A and y E C)
(x,y) E A X B and (x,y) E A X C
(x,y) E (A X B) intersect (A X C)

again i am only not sure how to justify going from step 3 to step 4 and how to go from step 4 to step 3

the argument reverses to prove they are equal

E= element of