# (A intersect B)' proof

1. Nov 16, 2008

### kathrynag

1. The problem statement, all variables and given/known data
I need to prove or disprove:
(AintersectB)'=A'intersectB'

2. Relevant equations

3. The attempt at a solution
Let x$$\in$$(A$$\cap$$B)'
Then x$$\notin$$A$$\cap$$B
x$$\notin$$A or x$$\notin$$B
Then x$$\in$$A' or x$$\in$$B'
x$$\in$$A'$$\cup$$B'

2. Nov 16, 2008

### VeeEight

x is not an element of A intersection B means that x is not a common element of A and B. So you want to look at how you went from line 2 to line 3. You might want to try some examples to see what is happening, like A = (0,1), B = [0,1] (as a subset of the reals).

3. Nov 16, 2008

### kathrynag

Ok I typed line 2 and 3 wrong
Then x$$\notin$$A$$\cup$$B
x$$\notin$$A or x$$\notin$$B

4. Nov 16, 2008

### kathrynag

Let A={1,2,3}
B={3,4} universe={1,2,3,4,5,6}
(A intersect B)={3}
(A intersect B)'={1,2,4,5,6}

A'={4,5,6}
B'={1,2,5,6}
A'intersectB'={5,6}

5. Nov 16, 2008

### kittybobo1

You might want to check your 'ors' and 'ands'. If you have the union of (1,2,3) U (4,5,6), then given any x, that would mean that x is in (1,2,3) or (4,5,6), right? What about for intersections? Your proof is almost right..

6. Nov 16, 2008

### kathrynag

So, x$$\notin$$A$$\cap$$B.
This implies x$$\notin$${3}
So,x$$\in$${1,2,4,5,6}
Would this mean x$$\in$$A' or x$$\in$$B'?

7. Nov 16, 2008

### kathrynag

Ok so x$$\notin$$A and x$$\notin$$B.
Then x$$\in$$A' and x$$\in$$B'.
x$$\in$$A'$$\cup$$B'

8. Nov 16, 2008

### kittybobo1

well, no, you had it right using the 'or's. I just wasn't sure if moving from line 2 to line 3 you understood what you were doing or if you were formulating it to make your answer right.

9. Nov 16, 2008

### kathrynag

Ok, then did I go wrong somewhere in my proof? Should I have said x$$\notin$$A and x$$\notin$$B. Then if I went on from there, I could get a right conclusion?

10. Nov 17, 2008

### natives

If you dont care what method..You can use VENN DIAGRAMS!

11. Nov 17, 2008

This counterexample should tell you that the statement $$(A \cap B)' = A' \cap B'$$ is not true!

Last edited: Nov 17, 2008
12. Nov 17, 2008

### cristo

Staff Emeritus
A Venn diagram is not a rigorous mathematical proof.