# Homework Help: Venn Diagram problem (Set Theory)

1. Sep 3, 2011

### Ascleipus

1. The problem statement, all variables and given/known data
The universal set u=40
Set A = 20
Set B=17
n(A∩B) = 1/2n(A'∩B')

What is the value of n(A∩B)?
2. Relevant equations
none^^

3. The attempt at a solution
The first thing I thought was that because Set A and Set B add up to 37 there must be 3 remaining outside the two sets, and since n(A'∩B') = everything outside the 2 sets n(A'∩B') = 3 and since n(A∩B) = 1/2n(A'∩B') then n(A∩B) = 1.5 but a set can't have 0.5 of an element, and as such I am stuck, please help me solve this :)

Cheers

Last edited: Sep 3, 2011
2. Sep 3, 2011

### LCKurtz

Pretty carelessly written. Presumably you mean the number of members of each of those sets are the numbers given.

What, exactly, is in that denominator? Parentheses are important.

Remember n(AUB) = n(A) + n(B) - n(A∩B)

Last edited: Sep 3, 2011
3. Sep 3, 2011

### Ascleipus

"What, exactly is in that denominator? Parentheses are important."
terribly sorry, it's quite late, it's meant to say n(A∩B)=½n(A'∩B') - the question reads exactly like this

also you are correct it is the number of elements, though your tip, i'm sorry i should probably mention that i haven't seen how it is with having the ½ in front of the expression, though i assume it means ½ the number of elements in that set?

Last edited: Sep 3, 2011
4. Sep 3, 2011

### LCKurtz

Yes. And it should be written (1/2)n(A'∩B'). Here's another hint. Let x be the unknown number of elements in A∩B and express everything in terms of x. See if that helps.

5. Sep 3, 2011

### Ascleipus

I hate to disappoint you, I am terribly grateful that you are helping me with this, I just cannot seem to get the numbers to work out, I always end up with n(A∩B) as being equal to something .5 It may just be that i'm tired, i don't know but i really want to figure it out or i wont be able to sleep due to curiosity :rofl: i must be going about this an entirely different way, maybe if you can explain to me why my original method didn't work? because mathematically it seems like it should work except for the fact that the result makes no sense

6. Sep 3, 2011

### ArcanaNoir

Your original method didn't work because if A and B share any elements then A+B is less than 37 and so there could be more than 3 elements outside of AUB

7. Sep 3, 2011

### LCKurtz

Draw a Venn diagram and try the hint I gave you above. Fill each area with the expression for the number of elements it in terms of x.

8. Sep 4, 2011

### Ascleipus

Ughh, i just cannot seem to connect the dots, I am meant to be doing this using simultaneous equations correct?

Please don't leave me on my own now need this solved by monday =(

Last edited: Sep 4, 2011
9. Sep 4, 2011

### LCKurtz

Did you draw the Venn diagram like I suggested? A rectangular box with two intersecting circles and start with putting an x in the area where the circles intersect. Did you do that or are you waiting for me to do it for you?

10. Sep 4, 2011

### ArcanaNoir

Do draw the venn diagram. Label the part A-AnB as "a" (or whatever you like, this is just my preference) and label B-BnA as "b" and the part AnB as anb.

Then, a+anb=?
b+anb=?
and anb=(1/2)(S-(a+b+anb))

You should be able to put something in for S.

Now you can solve.

11. Sep 4, 2011

### vela

Staff Emeritus
To give a specific example to illustrate ArcanaNoir's point, say

U = {1, 2, ..., 40}
A = {1, 2, ..., 20}
B = {1, 2, ..., 17}

Then A ∪ B = {1, 2, ..., 20}, so there are 20 elements, namely {21, 22, ..., 40}, outside of A ∪ B, not 3.

Note that this is also a counterexample to n(A∩B) = 1/2 n(A'∩B') since

A' = {21, ..., 40}
B' = {18, ..., 40}
A' ∩ B' = {21, ..., 40}

so 1/2 n(A' ∩ B') = 10 but n(A ∩ B) = 17.

Last edited: Sep 4, 2011
12. Sep 4, 2011

### Ascleipus

but you can't exactly assume that set A contains the elements 1,2...20 and set B contains the elements 1,2...17, can you?

Last edited: Sep 4, 2011
13. Sep 4, 2011

### vela

Staff Emeritus
Right, a specific example doesn't prove a claim in general. On the other hand, a claim needs to hold for all cases, so if you can find even one case where it doesn't hold, you've disproved it.

For example, you reasoned that because n(A)=20, n(B)=17, and n(U)=40, there are exactly 3 elements outside of A ∪ B. I've given you an example showing that's not the case in general.

Similarly, you have made the claim n(A ∩ B) = 1/2 n(A' ∩ B'), but this relationship doesn't hold for the specific sets A and B from the example, which again means that claim can't be true in general.

14. Sep 4, 2011

### ArcanaNoir

I think that was just a counter example. Can you do the problem using the equations I set up?

15. Sep 4, 2011

### Ascleipus

what is this S, how did you form this equation?

16. Sep 4, 2011

### ArcanaNoir

S is 40, the universe

(a'nb') is everthing outside of A and B, which is the universe minus (a plus b plus anb)

Oh, sorry I should have said S was the universe, that's what my class uses, I forgot it wasn't used here....I'm very stressed by my probability homework...

17. Sep 4, 2011

### Ascleipus

it's no problem, hope your probability works out well :)

Though i will have to burden you some more, i do not see how knowing that
a+a∩b=20 and
b+b∩b=17 helps me just seems like i'm always missing this 1 key bit of information, it is simultaneous equations i am meant to be using correct?

18. Sep 4, 2011

### ArcanaNoir

You need those two plus the long one. Then you have three equations and three variables.

19. Sep 4, 2011

### vela

Staff Emeritus
Suppose A and B had no elements in common. Then n(A ∪ B) = 37, right? Now what if A and B had 1 element in common? A would have 19 elements plus the 1 in common with B while B would have 16 elements plus the 1 in common with A, so n(A ∪ B) = 19 + 16 + 1 = 36. What if A and B had two elements in common?

20. Sep 4, 2011

### Ascleipus

I only end up with equations with a'∩b' included =(