(A intersect B)' proof

[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\capB)'
Then x\notinA\capB
x\notinA or x\notinB
Then x\inA' or x\inB'
x\inA'\cupB'

[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\notinA\cupB
x\notinA or x\notinB

[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]

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\capB)'
Then x\notinA\capB
x\notinA or x\notinB
Then x\inA' or x\inB'
x\inA'\cupB'

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\notinA\capB.
This implies x\notin{3}
So,x\in{1,2,4,5,6}
Would this mean x\inA' or x\inB'?

[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\capB)'
Then x\notinA\capB
x\notinA or x\notinB
Then x\inA' or x\inB'
x\inA'\cupB'

Ok so x\notinA and x\notinB.
Then x\inA' and x\inB'.
x\inA'\cupB'

[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\notinA and x\notinB. Then if I went on from there, I could get a right conclusion?

[natives]

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

A Venn diagram is not a rigorous mathematical proof.