[rove each of the fpllowing statements about sets, A,B and C?

Question

1..AnB=BnA
2..(AnB)nC=An(BnC)
3..An(BuC)=(AnB)u(AnC)
4..(A\B)\C=A\(BuC)

Answer 1

For each statement you have to prove that
1) left hand side contains the right hand side and
2) right hand side contains the left hand side.
It is not enough to show only one of the above statements. You must show both.

For #1
Step 1: Show that AnB is contained in BnA
Let x be in AnB, then by definition of intersection, x is in A and x is in B.
Since x is in A and x is in B, then x is in B and x is in A.
Thus x is in BnA
Step 2: Show that BnA is contained in AnB
Let x be in BnA, then by definition, x is in B and x is in A.
Since x is in B and x is in A, then x is in A and x is in B.
Thus, x is in AnB.
Therefore, it has been shown that AnB = BnA
Note: You have to do this for the other problems.
1) Start out by letting x be a member of the right hand side and show that it is also a member of the left hand side.
2) Let x be a member of the right hand side and show that x is a member of the left hand side.
If you can do step 1, then step 2 should be pretty easy. You're basically going in the reverse order.

Answer 2

n => intersect => common terms
u => union => all terms

So,
1..AnB=BnA can be proven by drawing the Venn diagram of 2 intersecting circles, and shading the intersection "oval".

2..(AnB)nC=An(BnC), similarly, can be proven by drawing 3 intersecting circles with one on top and two below, and shading the intersection.

3..An(BuC)=(AnB)u(AnC)
Draw the same linking of 3 circles as in question 2.
BuC implies that B and C are considered (shaded).
An(BuC) would be the region where A intersects the B/C region.

AnB would be like in question 1 and similarly for AnC.
The union of these two shaded regions is the same as that on An(BuC).

4..(A\B)\C=A\(BuC)
Errr.... I'm not sure what "\" means in Sets. Sorry.

Answer 3

Draw Venn Diagrams.

Answer 4

1. Suppose that there exists d, such that it is a member of AnB, then by definition d is an element of A and d is an alament of B. Or, in reverse, d is an element of B and an element of A. Therefore, it is an element of BnA.

Prove the others similarly.

Answer 5

Karl is a "neocon" and is more effective into political maneuvering or campaign administration, I trust Rick Perry's fact the fed desires to bypass and Ben's regulations are very undesirable for the economic device

Answer 6

k i am nt sure but tryin!
1. a intersection b ie d common thng in a and b will obviously will b equal to the common thng in b and a.
2.?
3.?
4.?
gosh man now i realised hw poor i am in maths
y dnt ya consult ne 11th maths book
its all der all dos de morgens laws.....blah blah!
tc

Answer 7

Like "ag iitkgp" said, draw Venn Diagrams, it makes is easier to see what you are doing.

Answer 8

damm easy use simple reasoing or venn idagram

Answer 9

Who cares,

Answer 10

I have absolutely no idea!!!!!!!