[Discrete Mathematics] Symmetric Difference Example

In this video we look at the symmetric difference operator and do some proofs.

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We introduce the symmetric difference operator and do some questions.

Hello, welcome to TheTrevTutor. I'm here to help you learn your college courses in an easy, efficient manner. If you like what you see, feel free to subscribe and follow me for updates. If you have any questions, leave them below. I try to answer as many questions as possible. If something isn't quite clear or needs more explanation, I can easily make additional videos to satisfy your need for knowledge and understanding.

[Trevtutor]

It comes with video lectures, text lectures, practice problems, solutions, and a practice final exam!

[mattholsten]

1:02 that is the cleanest crispest looking A I've ever seen written

[SewonKimMusic]

you are the greatest teacher for discrete mathematics. Thank you so much for this godly work.

[iandrsaurri]

For that last proof, I found it more intuitive to just expand certain parts using unions and simplifying it. Would that still be valid as a proof?

[lewisconnolly]

Here's a more verbose proof for the last question

XOR = exclusive OR = in one or the other, not both
=> = then

Proof:

Let x ∈ (A ⨁ B) ⨁ B ⑴
=> x ∈ (A ⨁ B) XOR x ∈ B
=> (x ∈ A XOR x ∈ B ) XOR x ∈ B

If x ∈ B
=> x ∉ A (from above: x ∈ A XOR x ∈ B)
=> x ∈ (A ⨁ B)
=> x ∉ (A ⨁ B) ⨁ B ❌ this contradicts statement 1, so x can't be in B

If x ∉ B
=> x ∈ A
=> x ∈ (A ⨁ B)
=> x ∈ (A ⨁ B) ⨁ B ✔ no contradiction with statement 1

So if x is in A then it's not in B (a fact that produces no contradiction with our initial premise). That results in anything else from B (also not in A) in the set (A ⨁ B) being removed by ⨁ B. All you're left with is A—the same as the right side of the equation i.e. (A ⨁ B) ⨁ B = A

[myonlynick]

is it possible the following incorrect solution to become correct?
(A xor B) xor B => (A xor B) = [ (A-B)U(B-A) ] = (AUB) - (A^B) (intersection symbol is ^), as you showed in the previous video. Take all that and XOR it with B:
[(AUB) - (A^B)] xor B =
( [(AUB) - (A^B)] - B ) U ( B - [(AUB) - (A^B)] ). Equation 1.

Now, having my venn diagram in front of me i deduct that:
( [(AUB) - (A^B)] - B ) = A. partial result 1.
explanation: [(AUB) - (A^B)] it means A without any traces of B.That is, A without their intersection. Also, i deduct that
( B - [(AUB) - (A^B)] ) = B !! partial result 2. In order to get a correct solution It should give A here.
Putting together partial result 1 and partial result 2 in equation 1: AUB. Thus,
(A xor B) xor B = AUB !! which is wrong.

[j_blue]

for the last proof ((A xor B) xor B), if you made a venn diagram the result would be A\B, right?

[spicy_wizard]

for A ^ B ^ B = A, it is so much easier to explain in graphs

[cherrynaresh]

Does A symmetric difference B <==> to (A n B)'. Does (A n B)' includes elements from universal set too

[simransngh]

can we write symmetric difference as (AuB)-(AnB) instead of (A-B)u(B-A)?

[JohnCena-mulv]

If A = {1, 2, 3, 4, 5, 6, 7} and B = {4, 5, 6, 7, 8, 9}, then the resulting set from these operations would be {1, 2, 3}. So if A and B overlap, then this would be false, correct?

[rockerguy]

your handwriting reminds me of the disney font

[ROCLife]

I guess the symmetric difference between A and B would be the same as the compliment of the intersection of A and B, correct?

[Jojo-mcho]

This subject is crazy, I have plenty of WHY'S since there are a lot of things that haven't explained further

[Ryan_Martin]

I got it till you used 1's and 0's. Can you help me understand, please?

[LoveMakesMeBrave]

Doesnt equivalent to need to be proved in both directions? In the video we just proved that from left to right it holds true, but what about the right to left part? or its to say that when it comes to statement like this with only part of the involved sets on one side, we only need to prove one way true.

[tryingtolearn]

isn't the final conclusion supposed to be that x is in A and x is not in B? (A-B)?

[rivenort]

The solution for last problem is not perfect. You actually NEED to solve the logical equation for EACH case (EACH possible "trivial" subset). Lets assume that u want to solve A∩B using your method. For case: x in A we've got: 1 and 0 = 0 ; for case x in B we've got: 0 and 1 = 0 ; Sooo 'x' is neither in A nor B ? No because it is in (A∩B) but u didn't check that case. For (A∩B) we've got 3 trivial sets A \ (A∩B), B \ (A∩B), A∩B.

[XXgamemaster]

Wouldn't this definition of the symmetric difference between two sets A and B also work?

A (sym) B := the set of all x such that x is not in A intersect B.

[woodenstick]

I started losing it here, i can't image what happens in the middle of the course.