computing sets

**wonderstrike** · February 24th 2009, 11:23 AM

Can someone help me out with the following question?

A= {1,3,5,6} and B={3,5} and C={a,b,c}

Compute:

a. A-B = {1,6}
b. B X C ( X meaning Cartesian Product) {(3,a),(3,b),(3,c),(5,a),(5,b),(5,c)}
c. A $A\cap B$ (B-A)
d. A $A\Delta B$ {1,6}

can anyone help me with 3.

**Plato** · February 24th 2009, 12:27 PM

Originally Posted by wonderstrike

A= {1,3,5,6} and B={3,5} and C={a,b,c}
Compute:
a. A-B
b. B X C ( X meaning Cartesian Product) * this one confuses the heck outta me
c. A Intersection (B-A)
d. A symmetric difference B
ps, how do I do the upside down U and symetric difference signs?

[tex]A\cap B[/tex] gives $A\cap B$
[tex]A\cup B[/tex] gives $A\cup B$
[tex]A \times B[/tex] gives $A \times B$
[tex]A\Delta B[/tex] gives $A\Delta B$ symmetric difference.

**wonderstrike** · February 24th 2009, 03:42 PM

Can I have help with c.

the way I understand is i need to do (B-A) 1st.

Well B-A is leaving me with nothing...

**Plato** · February 24th 2009, 03:48 PM

$A \cap \left( {B\backslash A} \right) = A \cap \left( {B \cap A^c } \right) = \emptyset$

[tex]A^c[/matH] is the complement of A, all elements not in A.

**wonderstrike** · February 24th 2009, 04:09 PM

Originally Posted by Plato

$A \cap \left( {B\backslash A} \right) = A \cap \left( {B \cap A^c } \right) = \emptyset$

$A^c$ is the complement of A, all elements not in A.

Right.. so $\emptyset$ means that they are disjoint right?

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