Question

Show that: \[\overline{A \cap B} \equiv \overline A \cup \overline B\]

Answer 1

this has something to do with set theory

Answer 2

break the line, change the sign

Answer 3

like i said it's set theory so it's not that simple

Answer 4

it has to be proven

Answer 5

Assume: (A∩B)′⊈(A′∪B′)
∴∃x such that x∈(A∩B)′, x∉(A′∪B′)
[x∉(A′∪B′)]⟹(x∉A′∧x∉B′)
(x∈A∧x∈B)
[x∈(A∩B)]
[x∉(A∩B)′]
⊕ (contradiction)
∴(A∩B)′⊆(A′∪B′) … (1)Proposition: (A′∪B′)⊆(A∩B)′

Proof: (by contradiction)
Assume: (A′∪B′)⊈(A∩B)′
∴∃x such that x∉(A∩B)′, x∈(A′∪B′)
[x∉(A∩B)′]⟹[x∈(A∩B)]
⟹(x∈A∧x∈B)
⟹[x∉A′∧x∉B′)]
⟹[x∉(A′∪B′)]
⊕ (contradiction)
∴(A′∪B′)⊆(A∩B)′ … (2)
By (1) and (2), (A′∪B′)=(A∩B

Answer 6

copied from here
http://www.physicsforums.com/showthread.php?t=31090

Answer 7

...i do not understand a word of that...

Answer 8

mind explaining those notations?

Answer 9

The negation of a conjunction is the disjunction of the negations.
The negation of a disjunction is the conjunction of the negations.

Answer 10

huh?

Answer 11

your taking discrete math? or this is for fun?

Answer 12

this is discrete math

Answer 13

why is (A∩B)′⊈(A′∪B′)

Answer 14

so this is an assignment?

Answer 15

what does that mean?

Answer 16

not exactly. im trying to understand this

Answer 17

http://www.google.com/imgres?imgurl=http://www.math.hawaii.edu/~hile/math100/setsc_files/ven9.jpg&imgrefurl=http://www.math.hawaii.edu/~hile/math100/setsc.htm&h=144&w=207&sz=6&tbnid=OUHN_5F8tFjQmM:&tbnh=90&tbnw=129&zoom=1&usg=__AiYMOi2Z6djecZNwfdu1SPcRuRg=&docid=HiOa6CxjmMcxbM&hl=en&sa=X&ei=YuljUM65MIr89gT-z4G4DA&sqi=2&ved=0CE8Q9QEwDw&dur=535

Answer 18

i tried doing \[\overline {A \cap B} = \{x : x \in \neg (A \cap B)\}\]
then distribute \[\{ x : x \in (\neg A \cup \neg B)\}\]
\[\{x: x \in( \overline A \cup \overline B) \}\]
but it seems wrong

Answer 19

like i said before..it has to be proven via set theory...

Answer 20

hmm \[\{ x : \neg ( x \in (A \cap B) ) \}\]
\[\implies \{ x : \neg(x \in A \text ^ x \in B)\}\]
\[\implies \{x : \neg(x\in A) \text V \neg(x \in B)\}\]
\[\implies \{ x : x \cancel \in A \text V x \cancel \in B\}\]
\[\implies \{x : x \in \overline A \text V x \in \overline B\}\]
\[\implies \overline A \cup \overline B\]
does that look right?

Answer 21

i wonder what are the latex codes for conjuction and disjunction