Venn Diagram problem (Set Theory)

[Ascleipus]

Homework Statement

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)?

Homework Equations

none^^

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

 

[LCKurtz]

Homework Statement

The universal set u=40
Set A = 20
Set B=17

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

n(A∩B) = 1/2n(A'∩B')

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

What is the value of n(A∩B)?

Homework Equations

none^^

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

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

 

[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?

 

[LCKurtz]

"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?

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.

 

[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 won't 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

 

[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

 

[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.

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 won't 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

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.

 

[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 =(

 

[LCKurtz]

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 =(

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?

 

[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.

 

[vela]

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

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.

 

[Ascleipus]

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.

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?

 

[vela]

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.

 

[ArcanaNoir]

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

 

[Ascleipus]

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

You should be able to put something in for S.

Now you can solve.

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

 

[ArcanaNoir]

S is 40, the universe

(a'nb') is everything 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...

 

[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?

 

[ArcanaNoir]

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

 

[vela]

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?

 

[Ascleipus]

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

 

[ArcanaNoir]

a+anb=20
b+anb=17
anb=(1/2)(40-(a+b+anb))

anb=20-(1/2)a-(1/2)b-(1/2)(anb)

+(1/2)(anb) +(1/2)(anb)


(3/2)(anb)=20-(1/2)a-(1/2)b multiply by 2

3(anb)=40-a-b


Now, since a+anb=20 and b+anb=17:
a=20-anb
b=17-anb

Now substitute into 3(anb)=40-a-b

3(anb)=40-(20-anb)-(17-anb)

Can you solve for anb now?

 

[ArcanaNoir]

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

You should have dumped a'nb' completely by substituting in (S-(a+b+anb)) where S was the universe (40)

 

[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?

I guess if you are going to just ignore my suggestions there's no point in continuing to try to help you.

 

[Ascleipus]

+(1/2)(anb) +(1/2)(anb)

where did these come from?

 

[ArcanaNoir]

I was just adding to both sides to get the anb all on one side. Sorry the spacing didn't come out right.

 

[Ascleipus]

I guess if you are going to just ignore my suggestions there's no point in continuing to try to help you.

I have been considering your suggestions all day in fact but i have not been able to produce the result, sorry if it seems like i have not, i do not want to be given the answer i assure you

 

[Ascleipus]

so a∩b=3?

 

[Ascleipus]

but how did you derive the equation a'∩b'=(1/2)(40-(a+b+a∩b))? I would understand if the 1/2 wasn't there, but if the 1/2 isn't there then the whole method doesn't work

 

[LCKurtz]

I think the notation in these posts is terrible. Equations like a∩b=3 or a+anb=20 just add to the confusion. Even though the idea of the proof is there in principle, it could be much clearer.

As I suggested before let x be the number of elements in AnB. Draw a Venn diagram like this:

Fill in the other colors with the proper expression in x then use the fact that the number of elements in the intersection is half the number in the yellow.

 

[Ascleipus]

This means that the yellow is 2x and the red is 17-x
and then i assume i use simultaneous equations to solve it, do tell me if I'm on the right track, but please no answer^^

 

[LCKurtz]

This means that the yellow is 2x and the red is 17-x
and then i assume i use simultaneous equations to solve it, do tell me if I'm on the right track, but please no answer^^

Don't use 2x for the yellow. Use the fact that it all adds up to 40 to figure out how many are in the yellow.

 

[Ascleipus]

but in order to figure out the yellow i need the intersection don't I?
40-(37-x) must be more appropriate then if i am to use that it adds up to 40

 

[Ascleipus]

it seemed reasonable to do 40-(37-x)=(37-x)/2

if the yellow is 40-(37-x) and the union of the two sets is 37-x if divide that by two then i should have a functioning equation, right?

however that just solves x as being 10.333...

 

[ArcanaNoir]

so a∩b=3?

Yes!

but how did you derive the equation a'∩b'=(1/2)(40-(a+b+a∩b))? I would understand if the 1/2 wasn't there, but if the 1/2 isn't there then the whole method doesn't work

a'nb' does not equal (1/2)(40-(a+b+a∩b))

a'nb' does equal (40-(a+b+a∩b))

Since anb=(1/2)a'nb' ,

anb= (1/2)(40-(a+b+a∩b))

 

[Ascleipus]

of course! thanks so much after all that i miss that 1 detail haha that's 2 problems you've helped me with, thank you to everyone else that has attempted to help me as well much appreciated everyone! physics forums is brilliant, hope i can someday help others like you have helped me :)