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Linear Algebra (Proof help)

  • #1

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!


Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
This is not linear algebra; it is set theory.

Anyway, where are you having problems and what have you tried ?
 
  • #3
Sorry, I wrote linear algebra because we are having to learn this in my linear algebra class. Thanks for the correction.

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
Can you post what you have so that we can help you out ? Perhaps there is a logical error somewhere, or perhaps just a simple mistake.
 
  • #5
Well I've tried many different variations but the one I'm looking at now is this:
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
Well I've tried many different variations but the one I'm looking at now is this:
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
Sorry I am confused. What exactly do you mean by x is automatically in A-B?
 
  • #8
Sorry I am confused. What exactly do you mean by x is automatically in A-B?
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.

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
Ok i was thinking about these operations the wrong way. I've only had one lecture on set theory so I its expected. Thank you very much. You've helped my understanding on the matter.
 
  • #10
Please, post more if you need more help.
 
  • #11
Will do. Thank you very much.
 

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