# Set Theory (help again)

• Jul 12th 2012, 07:56 AM
PennCow
Set Theory (help again)
Thanks for the help from the previous problem! Anyways, I just got stuck on another "All that apply" problem... These are really killing me :(
I listed my logic below (Own = I just interpreted it for myself [could be wrong for all I know :(], Text = I double checked with the text and found them there, and Unsure = no idea....)

Which of the following are true? Select all that apply.

http://i45.tinypic.com/35900o4.png

A. Unsure
B. True (Text)
C. True (Text)
D. False (Own)
E. False (Own)
F. Unsure
G. True (Own)
H. True (Own)
I. False (Text)
J. True (Text)
K. True (Text)

Thanks again!
• Jul 12th 2012, 08:28 AM
emakarov
Re: Set Theory (help again)
For any propositions A and B, I'll write ~A to mean the negation of A and A => B to mean "A implies B." Also, If P is a property (or predicate, such as "even" or "mortal"), I'll write P(x) to mean that x satisfies P. Therefore, ~P(x) means that x does not satisfy P.

If A and B are propositions, then ~A => ~B does not in general imply A => B. However, ~A => ~B is equivalent to B => A. The formula ~A => ~B is called the contrapositive of B => A.

Question A says "For every x, ~A(x) => ~B(x)." Using the information above, this is the same as "For every x, B(x) => A(x)," but not the same as "For every x, A(x) => B(x)."

B is in fact false. If X, Y and Z are propositions, then (X or Y) => Z is equivalent ((X => Z) and (Y => Z)), but x ∈ A ∩ B does not imply that x ∈ A ∩ C.

E is true: consider two cases, when x ∈ A and x ∈ B.

G is false: consider x ∈ B, but x ∉ A and x ∉ C.

I'll let someone else to do the rest, or I'll return to this later.
• Jul 12th 2012, 09:07 AM
PennCow
Re: Set Theory (help again)
Thanks again!

Wouldn't G be True?
If x is in A∪B or x is in C then x is in A∪C and x is in B∪C.

Since there is an 'or', only one of the condition needs to be met correct? That being in the case, we only use "x is in C" and not "x is in A∪B":
When x is in C, then x is in A∪C as well as in B∪C right? I'm not sure if I'm misunderstanding anything though :(.

Also are A and F False?
• Jul 12th 2012, 09:26 AM
emakarov
Re: Set Theory (help again)
Quote:

Originally Posted by PennCow
Wouldn't G be True?
If x is in A∪B or x is in C then x is in A∪C and x is in B∪C.

Since there is an 'or', only one of the condition needs to be met correct? That being in the case, we only use "x is in C" and not "x is in A∪B"

First, I stand by my suggestion to consider the situation where x ∈ B, but x ∉ A and x ∉ C. Second, as I said with respect to question B, (X or Y) => Z is equivalent ((X => Z) and (Y => Z)). When you are proving (X or Y) => Z, you assume X or Y and need to show Z. You don't know which of X and Y is true, only that one of them is. Therefore, you need to show Z assuming just X, and also show Z assuming just Y. In question G, you don't have the right to assume that the premise (x ∈ A ∪ B or x ∈ C) means x ∈ C. Instead, you need to show that x ∈ A ∪ C assuming just x ∈ A ∪ B and also show x ∈ A ∪ C assuming just x ∈ C. The first of these two implications is false in general.

Quote:

Originally Posted by PennCow
Also are A and F False?

As I said,
Quote:

Originally Posted by emakarov
Question A says "For every x, ~A(x) => ~B(x)." Using the information above, this is the same as "For every x, B(x) => A(x)," but not the same as "For every x, A(x) => B(x)."

• Jul 12th 2012, 10:08 AM
PennCow
Re: Set Theory (help again)
Ahh I see, thanks. Been up all night so I think I am misreading your stuff as we go... gotcha now >_<! Sorry!