- #1

- 16

- 0

## Homework Statement

Hi, I am having trouble trying to prove the following operation: (A union B)-(A intersect B)=(A-B)union(B-A) given that: A-B = {x:x belong to A and x does not belong to B}. Thank you!

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter TheIconoclast
- Start date

- #1

- 16

- 0

Hi, I am having trouble trying to prove the following operation: (A union B)-(A intersect B)=(A-B)union(B-A) given that: A-B = {x:x belong to A and x does not belong to B}. Thank you!

- #2

- 830

- 1

Anyway, where are you having problems and what have you tried ?

- #3

- 16

- 0

It seems like I have tried everything i can think of. I have began by letting x belong to one side of the equation and logically taking that side apart to turn it into the other side of the equation. I have tried this for both sides and have come up short. I have also tried entering x belong to A and x does not belong to B for (A-B) on multiple occasions and have come up short there too. I don't know if there is some special trick to the proof.

- #4

- 830

- 1

- #5

- 16

- 0

Let x belong to (A-B)U(B-A)

(x belongs to A and x does not belong to B) or ((x belongs to B)-(x belongs to A))

I chose this side of the equation because on the other side there are two operations (union, and intersect) and if I substitute:x belongs to A and x does not belong to B, for A-B I can introduce another operation.

- #6

- 830

- 1

Let x belong to (A-B)U(B-A)

(x belongs to A and x does not belong to B) or ((x belongs to B)-(x belongs to A))

I chose this side of the equation because on the other side there are two operations (union, and intersect) and if I substitute:x belongs to A and x does not belong to B, for A-B I can introduce another operation.

I would start the following way.

[tex] x \in (A \cup B)-(A \cap B) [/tex]

What this means is that either [tex] x \in A \quad or \quad x \in B[/tex] but not both.

If [tex] x \in (A \cup B)-(A \cap B) [/tex] and x is in A. Then x is automatically in A-B.

If [tex] x \in (A \cup B)-(A \cap B) [/tex] but not in A then it is in B. Hence, it is automatically in B-A.

Then continue...

- #7

- 16

- 0

Sorry I am confused. What exactly do you mean by x is automatically in A-B?

- #8

- 830

- 1

If [tex] x \in (A \cup B)-(A \cap B) [/tex]then x is either in A or B but x cannot be in both A and B.Sorry I am confused. What exactly do you mean by x is automatically in A-B?

So if [tex] x \in (A \cup B)-(A \cap B) [/tex] then [tex] x \in A-B[/tex] or [tex] B -A [/tex].

x in A-B means x is only in A and not B ! Which is exactly was the left handside of the orginal equation is saying.

or

x in B-A means the same thing as the above but with A and B exchanged.

- #9

- 16

- 0

- #10

- 830

- 1

Please, post more if you need more help.

- #11

- 16

- 0

Will do. Thank you very much.

Share: