Question

proof problem - biconditional statement...

Answer 1

latexing in a bit

Answer 2

\[A \subseteq B\] if and only if \[A \backslash B = \emptyset\]

Answer 3

so we have two statements to prove...
one would be that if \[a \subseteq b\] then \[A \backslash B = \emptyset\]

Answer 4

so what that is ... I have an a subset of B, but the complement of B written as \[A \backslash B\] is an empty set

Answer 5

subset def.\[\forall x [ x\in A \rightarrow x \in B]\]

Answer 6

I would use \[
A\setminus B = A\cap \overline{B}
\]

Answer 7

so x belongs in A and x belongs in b. however... I have \[A \backslash B = [ x: x \in A \land x \notin B]\] but as it said in the problem ... \[A \backslash B = \emptyset\]

Answer 8

meaning no elements

Answer 9

At this point in the game, you have enough in your tool box that you don't have to rely just on definitions.

Answer 10

You can start using set algebra.

Answer 11

\(\color{blue}{\text{Originally Posted by}}\) @UsukiDoll
subset def.\[\forall x [ x\in A \rightarrow x \in B]\]
\(\color{blue}{\text{End of Quote}}\)
This can be changed to \[
\[\forall x [ x \notin A \lor x \in B]\]
\]

Answer 12

Since \(p\to q\iff \lnot p \lor q\)

Answer 13

set alg?1

Answer 14

Then finish it off with de Morgan.

Answer 15

argh I'm getting dizzy... after this I'm eating lunch and then finishing the problems from last night. still gotta tackle H.. UGH!

Answer 16

waht the............what if I just want to prove or disprove the statements? there's nothing about alg.

Answer 17

hmm
if \[x \in A then x \in B ,x \notin B^c , A∩ B^c =\phi \]
if
\[ A∩ B^c =\phi , then x \in A .x \notin B^c \rightarrow x \in B , A \subset B\]

Answer 18

?????????????????

Answer 19

I'm dealing with a subset and A\B with an empty set... ak.a... no elements at all

Answer 20

A is a subset of B --> all x belongs to A and B
A\B ---> no elements

Answer 21

huh lol nothing :D
continue plz...

Answer 22

:O! how would I go on with it?

Answer 23

Use set algebra.

Answer 24

\[
A\setminus B = \emptyset
\]This means\[
A\cap \overline{B} = \emptyset
\]Now we \(\cup B\) to both sides: \[
(A\cup B) \cap (\overline{B} \cup B) = \emptyset \cup B
\]Simplify: \[
(A\cup B) \cap \mathcal U = B
\]\[
(A\cup B) = B
\]

Answer 25

dang that looked easy... so what about for the second statement? the a \ b = emptyset then the a is a subset of b... how to tackle that?

Answer 26

You can just use definitions for this uglier part.

Answer 27

well a \ b ... there aren't any elements.... how can a be a subset of b... empty set = nothing

Answer 28

If \(A\subseteq B\), then any element in \(A\) or \(B\) must also be in \(B\), so \(A\cup B= B\).

If \(A\cup B = B\), then there are no elements in \(A\) which are not also in \(B\), therefore \(A\subseteq B\).

Answer 29

If that isn't rigorous enough, then you have to use definitions to show that \[
A \subseteq B \iff A\cup B = B
\]

Answer 30

c rap D:

Answer 31

can't we just disprove this?

Answer 32

like a\b isn't an emptyset oh shi....... then x belongs in a and not b

Answer 33

sounds familiar like the a subset of b but b'

Answer 34

like x belongs to a implies x doesn't belong in b

Answer 35

You want to disprove something that is true?

Answer 36

:S cra..p I shouldn't do that.

Answer 37

Okay about about doing this one direction at a time.

Answer 38

If \(A\subseteq B\) then \(A\setminus B=\emptyset\).

Answer 39

If \(A \subseteq B\) then then \[
\forall x\quad x\in A\to x\in B
\]We use \(p \to q \iff \lnot p \lor q\). \[
\forall x\quad x\notin A\lor x\in B
\]Next we use \(\forall x P(x) \iff \lnot \exists x \lnot P(x)\)\[
\lnot \exists x\quad \lnot ( x\notin A\lor x\in B)
\]We use de Mograns \(\lnot (p\lor q)\iff \lnot p \land \lnot q\)\[
\lnot \exists x\quad x\in A\land x\notin B
\]

Answer 40

O_O lots of demorgans going on here XD

Answer 41

I used it once.

Answer 42

oh .-. whew now that's over....I still have to figure out what was going on with the associative law proof from last night.... apparently I have to use the set property but I don't know where to apply it. >:/