# Thread: New to discrete and in need of help!! Set question.

1. ## New to discrete and in need of help!! Set question.

Hello everyone,
I am really new to discrete and totally dont get it!!

I have a question that I have no idea how to get started or really approach it.
It is a Set theory question. I really appreciate it if someone can give me the step by step procedure of how to solve it and explanations so I can use it as a learning guide.

THANK YOU!!!

If the statement is true, give a proof. If it is false then write its negation and prove it.

Assume all sets are subsets of a universal set U.

For all sets A, B,and C, A-(B-C) = (A-B)-C

2. Suppose that you want to prove two sets A and B to be equal. There are two main ways to do that:
1) Take one of them, say A, and performing algebraic transformations try to turn it into B.
2) Prove that $A\subseteq B$ and $B\subseteq A$. To do this, first let $x\epsilon A$ and try to show that $x\epsilon B$ ($A\subseteq B$), and then let $x\epsilon B$ and try to show that $x\epsilon A$ ($B\subseteq A$).

3. Originally Posted by Kitizhi
Hello everyone,
I am really new to discrete and totally dont get it!!

I have a question that I have no idea how to get started or really approach it.
It is a Set theory question. I really appreciate it if someone can give me the step by step procedure of how to solve it and explanations so I can use it as a learning guide.

THANK YOU!!!

If the statement is true, give a proof. If it is false then write its negation and prove it.

Assume all sets are subsets of a universal set U.

For all sets A, B,and C, A-(B-C) = (A-B)-C
Take A={1,2,3,4},B={1,2,3},C={3}.

Now A-(B-C)= {1,2,3,4}-[ {1,2,3}-{3}]= {1,2,3,4}-{1,2}={3,4}

But (A-B)-C = [{1,2,3,4}-{1,2,3}]-{3}= {4}-{3} = {4}

Hence $\displaystyle A-(B-C)\neq (A-B)-C$

4. Awesome guys, you've both been a great help.
However I got another question..

How would I prove this statement is true?
The previous one wasnt too bad cause it was false and I would just provide a counter-example, but this one is true.

For all sets A, B and C, A x (B-C) = (A x B) - (A X C)

Would I assume the negation is true and provide a counter example to prove the negation is false and conclude that the original statement is true?

5. Originally Posted by Kitizhi
Awesome guys, you've both been a great help.
However I got another question..

How would I prove this statement is true?
The previous one wasnt too bad cause it was false and I would just provide a counter-example, but this one is true.

For all sets A, B and C, A x (B-C) = (A x B) - (A X C)

Would I assume the negation is true and provide a counter example to prove the negation is false and conclude that the original statement is true?
No because like that is like proving the above by using only one example .We want to give a general proof

Anyway if the above is true why not try to prove it >

So let xε[Ax(B-C)] THAT implies xεA AND xε(B-C)====> xεA AND xεB AND ~xεC(= x does not belong to C) and that implies:

(xεA AND xεB)AND(~xεC v ~xεA)====> xε(AxB) AND ~xε(ΑxC) ===>..........................BY using de morgan..........

====>xε[(AxB)-(AxC)] ====> $\displaystyle Ax(B-C)\subseteq (AxB)-(AxC)$

You try the converse

6. Originally Posted by benes
No because like that is like proving the above by using only one example .We want to give a general proof

Anyway if the above is true why not try to prove it >

So let xε[Ax(B-C)] THAT implies xεA AND xε(B-C)====> xεA AND xεB AND ~xεC(= x does not belong to C) and that implies:

(xεA AND xεB)AND(~xεC v ~xεA)====> xε(AxB) AND ~xε(ΑxC) ===>..........................BY using de morgan..........

====>xε[(AxB)-(AxC)] ====> $\displaystyle Ax(B-C)\subseteq (AxB)-(AxC)$

You try the converse

Ahh perfect thanks alot. However to complete the proof I have to do the other way around where (AxB)-(AxC) = Ax(B-C) as well right?

7. Originally Posted by Kitizhi
Ahh perfect thanks alot. However to complete the proof I have to do the other way around where (AxB)-(AxC) = Ax(B-C) as well right?
O.K

$\displaystyle x\in[(AxB)-(AxC)]\Longrightarrow[(x\in A\wedge x\in B)\wedge(\neg x\in A\vee\neg x\in C)]\Longrightarrow$$\displaystyle [(x\in A\wedge x\in B)\wedge(x\in A\rightarrow \neg x\in C)]$$\displaystyle \Longrightarrow[x\in A\wedge x\in B\wedge\neg x\in C]\Longrightarrow[x\in A\wedge(x\in B\wedge\neg x\in C)]\Longrightarrow x\in[Ax(B-C)]$

Hence $\displaystyle [(AxB)-(AxC)]\subseteq Ax(B-C)$

8. Thanks alot!!