(A intersect B)' proof

[kathrynag]

Homework Statement

I need to prove or disprove:
(AintersectB)'=A'intersectB'

Homework Equations

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'

 

[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).

 

[kathrynag]

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).

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

 

[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}

 

[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..

 

[kathrynag]

Homework Statement

I need to prove or disprove:
(AintersectB)'=A'intersectB'

Homework Equations

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'

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}

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'?

 

[kathrynag]

Homework Statement

I need to prove or disprove:
(AintersectB)'=A'intersectB'

Homework Equations

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'

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

 

[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.

 

[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?

 

[natives]

If you don't care what method..You can use VENN DIAGRAMS!

 

[pizzasky]

Let A = {1,2,3} , B = {3,4} , universe = {1,2,3,4,5,6}

$$A \cap B$$ = {3}
$$(A \cap B)'$$ = {1,2,4,5,6}

A' = {4,5,6}
B' = {1,2,5,6}
$$A' \cap B'$$ = {5,6}

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

 

[cristo]

If you don't care what method..You can use VENN DIAGRAMS!

A Venn diagram is not a rigorous mathematical proof.