# Set theory homework - Theoretic reasoning

1. Jul 14, 2011

### loplol

1. The problem statement, all variables and given/known data

Prove where X and Y are both sets use theoretic reasoning

i) Z \ (X $\cap$ Y) = (Z \ X) $\cup$ (Z \ Y)
ii)(Y \ X) $\cup$ Z = (Y $\cup$ Z) \ (X \ Z)
iii) Z \ (Y \ X) = (X $\cap$Z) $\cup$(Z \ Y)

2. Relevant equations
\ = set difference

3. The attempt at a solution
i know you dont do other peoples work for them but i am very clueless on this subject a guiding hand with step by step help would be greatly appreciated

2. Jul 15, 2011

### gb7nash

What is "theoretic reasoning"?

3. Jul 15, 2011

### HallsofIvy

If you mean just regular "proofs", each of those is basically to prove "A= B" where A and B are sets. And the standard way to do that is to prove "$A\subset B$" and then "$B\subset A$". Finally, the way to prove $A\subset B$ is to start by saying "let x be a member of A" and prove, using the definitions of A and B, that x must be a member of B.

For example, you want to prove, in (i), that
$$Z\(A\cap B)= (Z\A)\cup (Z\B)$$
so you start by proving
$$Z\(A\cap B)\subset (Z\A)\cup (Z\B)$$

And you do that by saying "let x be in $Z\(A\cap B)$" which means, of course, that x is in Z but not in $A\cap B$. Saying that x is not in $A\cap B$ means x is not in both A and B. So there are two possibilities- x is in Z and A but not in b or x is in Z and B but not in A. Do those two possibilities as two cases:
1) x is in Z and A but not in B. Then, since x is in Z but not in B, x is in Z/B. Do you see why that means x must be in $(Z/A)\cup(Z/B)$?
2) x is in Z and B but not in A. Then, since x is in Z but not in A, ...

4. Jul 15, 2011

### loplol

ok i understand 1)

heres my attempt at 2)
(Y \ X) ∪ Z = (Y ∪ Z) \ (X \ Z)

let x be in (Y \ X) ∪ Z
-x is in Y and Z but not in X, so x is in X\Z
also (Y ∪ Z) \ X, because x is in Y and Z but not in X

hence (Y \ X) ∪ Z = (Y ∪ Z) \ (X \ Z)

THIS OK?

5. Jul 16, 2011

### HallsofIvy

No, but it is a start. That shows that
$$(Y\X)\cup Z\subset (Y\cup Z)\(X\Z)$$
You still need to show the other way.

6. Jul 16, 2011

### loplol

what do you mean when you say show the other way?

is the 'other way' this?
-x is in X and Z but not in Y, so x is in Y\Z

so overall
(Y \ X) ∪ Z = (Y ∪ Z) \ (X \ Z)

let x be in (Y \ X) ∪ Z
-x is in Y and Z but not in X, so x is in X\Z
-x is in X and Z but not in Y, so x is in Y\Z

also (Y ∪ Z) \ X, because x is in Y and Z but not in X

hence (Y \ X) ∪ Z = (Y ∪ Z) \ (X \ Z)

7. Jul 16, 2011

### gb7nash

A = B if and only if A ⊂ B and B ⊂ A.

You showed that (Y \ X) ∪ Z ⊂ (Y ∪ Z) \ (X \ Z). You still need to show that (Y ∪ Z) \ (X \ Z) ⊂ (Y \ X) ∪ Z.

8. Jul 17, 2011

### loplol

ok lol kinda obvious isnt it

ok heres a revised version of 2
(Y \ X) ∪ Z = (Y ∪ Z) \ (X \ Z)

let x be in (Y \ X) ∪ Z
-x is in Y and Z but not in X, so x is in X\Z
-x is in Y and Z but not in X, so (Y ∪ Z) \ X

let x be in (Y ∪ Z) \ (X \ Z)
-x is in Y and Z, so x is in Y∪Z
-x is in Z but not in X, so x is in A\C
-x is in Y and Z but not in X, so x is in (Y∪Z)\X

hence (Y \ X) ∪ Z = (Y ∪ Z) \ (X \ Z)

good?