Question

Need help with a discrete math problem.

Answer 1

If \(\chi_A \cdot \chi_B=1\), then that means \(\chi_A=1\) and \(\chi_B=1\).
If \(\chi_A \cdot \chi_B=0\), then that means one of the following is true:
1. \(\chi_A=0\), \(\chi_B=1\)
2. \(\chi_A=1\), \(\chi_B=0\)
3. \(\chi_A=\chi_B=0\).

Answer 2

Now, if \(\chi_A(x)=1\) and \(\chi_B(x)=1\), then this means \(x \in A\) and \(x \in B\).
If \(\chi_A(x)=1\) and \(\chi_B(x)=0\), then \(x\in A\) and \(x \notin B\).
If you repeat this for the rest, you should be able to figure out the set with characteristic function \(\chi_A \cdot \chi_B\)

Answer 3

So does this mean XA*XB would only be in subset A?

Answer 4

@phi

Do you know how to do this problem? I don't understand the solution =\

Answer 5

Do you have a definition for characteristic function ? I am guessing it means "is a member of " ?

Answer 6

So I just looked it up and found this:

I(x) = { 1, if x belongs to B,
= { 0, if x doesnot belong to B. where x is a variable, which can take any letter from the alphabets.

Answer 7

ok. then Xa * Xb would be 1 only if the member is in both sets

Answer 8

which sounds like A intersect B

Answer 9

ok

Answer 10

so it asks what subset of S has characteristic function (a), (b), and (c). would that mean a is (a) characteristic function of the set A?

Answer 11

(a) is the characteristic function of \( A \cap B\)

Answer 12

I don't understand how you got that =\

Answer 13

Xa (y) is 1 if y is in set A
Xb (y) is 1 if y is in set B

Xa (y) * Xb (y) is 1 if y is in set A and in set B

by definition, if y is in A and in B it is in \( A \cap B\)

Answer 14

ohhhhhhhh makes sense!

Answer 15

so then then for part (b) if Xa + Xb - Xa(intersects)B = 1

Answer 16

I think this characteristic function is only allowed 0 or 1 as its output values

Xa(y) + Xb(y) - \( X_{A \cap B}\)(y)
y is in A but not in B = 1 + 0 - 0 = 1
y is in B but not in A = 0 +1 -0 = 1
y is in both A and B, then 1+1 - 1 = 1
y is not in A and not in B= 0 + 0 -0 =0

Answer 17

so I'm thinking something like A\B, B\A and A(union)B

Answer 18

It sure looks like A \( \cup \) B to me

the reason the characteristic function is so complicated is that if y is in both A and B, just using
Xa(y) + Xb(y) would give 2 , and that is not allowed.

Answer 19

so A\B and B\A wouldn't be answers?

Answer 20

what does A\B mean?

Answer 21

y is in A but not in B = 1 + 0 - 0 = 1

which is the Set A - the set B (A\B)

Answer 22

in A but not in B

Answer 23

oh wait, I don't think I understood it correctly

Answer 24

if you evaluate
Xa(y) + Xb(y)-\(X_{A \cap B}\)(y)

with y in B, you get a 1
but A\B should give you a 0 ?

Answer 25

ok, I think it makes sense now, so the answer would be AUB

Answer 26

because y is in both A and B

Answer 27

y is in one or the other or both

Answer 28

ah I see. Now it makes sense. :)

Answer 29

so for (c)

y is in A but not in B = 1 + 0 - 2 * 0 = 1
y is in B but not in A = 0 + 1 - 2 * 0 = 1
y is in both A and B, then 1 + 1 - 2 * 1 = 0
y is not in A and not in B = 0 + 0 - 2 * 0 = 0

Answer 30

looks good.

Answer 31

So the answer would be A(compliment)(intersects)B(compliment)?

Answer 32

A' means not in A ?
but y is in A but not in B = 1 + 0 - 2 * 0 = 1

Answer 33

oh right

Answer 34

hmm, I'm not sure

Answer 35

so when y is in both A and B it equals 0

Answer 36

I am thinking A \( \cup \) B - A \( \cap \) B

Answer 37

that would do it :O

Answer 38

Thank you so much for the help!

Answer 39

and for putting up with my stupidity :D

Answer 40

All makes sense now.