SOLUTION: How do I solve the following sets? A= {1,2,3,4,5} B= {2,4,6} C= {1,3,5} A∩(B∩C) AU(BUC)

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Question 447130: How do I solve the following sets?
A= {1,2,3,4,5} B= {2,4,6} C= {1,3,5}
A∩(B∩C)
AU(BUC)

Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
A= {1,2,3,4,5} B= {2,4,6} C= {1,3,5}
A∩(B∩C)
AU(BUC)
-------------------------------

A∩(B∩C)

We substitute the sets for the capital letters
which represent them:

{1,2,3,4,5}∩({2,4,6}∩{1,3,5})


Work inside the parentheses first:

The symbol "∩" (intersection) means to take only 
the elements that are in common to the two sets 
on each side of the ∩, and put them into one set.

There are no elements that are in common to the two
sets {2,4,6} and the set {1,3,5}, so therefore 
we replace the whole parentheses by the empty set
{ } or ∅. I'll use ∅. So now we have 

{1,2,3,4,5}∩∅

Again the symbol "∩" (intersection) means to take only 
the elements that are in common to the two sets 
on each side of the ∩, and put them into one set.
 
There are no elements that are in common to the two
sets {1,2,3,4,5,6} and the empty set ∅, so therefore 
we replace it by the empty set { } or ∅. So now
we have just

∅.

That's the answer.

--------------------------------------------------- 

AU(BUC)

{1,2,3,4,5}U({2,4,6}U{1,3,5})

Work inside the parentheses first:

The symbol "U" (union) means to take all the 
elements that are in either one or both of the 
two sets on each side of the U, and put them 
into one set.  So therefore we replace the whole 
parentheses by the set {1,2,3,4,5,6}. Notice that
it's a good idea to arrange them in numerical order
if the elements are numbers, and alphabetical order
if they are letters. So now we have:

{1,2,3,4,5}U{1,2,3,4,5,6}

Again the symbol "U" (union) means to take all the 
elements that are in either one or both of the 
two sets on each side of the U, and put them 
into one set.  So therefore we replace the above
by the set {1,2,3,4,5,6}. So we have just this:

{1,2,3,4,5,6}.

That's the answer.

Edwin

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