Anonymous

# Prove |S U T| = |S| + |T| - |S ∩ T|?

Set theory

Prove |S U T| = |S| + |T| - |S ∩ T|

Relevance

asumming |S| = a and

|T| = b

and

|S ∩ T| = c

if they have no common member then c=0

if x belong to S U T it means

case1) x belong to S only or

case2) x belong to T only or

case3) x belong to both

case 1 is if x belong to S but not T, and the number of them are

a-c

case 2 : b-c

case3 :c

a-c + b -c + c = a + b -c

• I am not sure what class this is, so the language might be different, but here would be my proof.:

Assume that there exists a set S and a set T. By rule of Sets, the union of sets S and T is created by the items that are both unique to sets S and T as well as the items shared. The union is then comprised of the addition of sets S and T minus the intercetion, since the addition includes twice the shared items.

Hope this helps you.

• Draw venn diagrams. I'm not sure how you might go about it algebraically though, it seems pretty self evident as it is xD

S U T is the set of all elements without repetition

S + T is the set of all elements with repetition

S ∩ T is the set of all elements which repeat.

(the set of all elements with repetition) - (the set of all repetitions) = (the set of all elements without repetition).

• Two equations:

|S U T| = |S| + |T \ S|

|T| = |T \ S| + |S ∩ T|

If you need to prove these, they're not hard.

Then

|T \ S| = |T| - |S ∩ T|

so

|S U T| = |S| + |T| - |S ∩ T|