[ 3 / biz / cgl / ck / diy / fa / ic / jp / lit / sci / vr / vt ] [ index / top / reports ] [ become a patron ] [ status ]
2023-11: Warosu is now out of extended maintenance.

/sci/ - Science & Math


View post   

File: 43 KB, 386x395, 397330.jpg [View same] [iqdb] [saucenao] [google]
9279755 No.9279755 [Reply] [Original]

how do i prove

(A ∪ B) \( A ∩ B) =( A \ B) ∪( B \ A)

?

>> No.9279760

>>9279755
Proof: Think

>> No.9279763

say it in words
the union of A and B without where they overlap
equals
the parts of A that don't have B and the parts of B that don't have A

>> No.9279786

>>9279763
I get why its equivalent i do not know how to proof it mathematically

>> No.9279824
File: 142 KB, 948x543, 1375242018674.gif [View same] [iqdb] [saucenao] [google]
9279824

>>9279755
unpack the definition

(A ∪ B)-( A ∩ B)
(x∈A or x∈B)&(not(x∈A & x∈B))
(x∈A or x∈B)&(x∉A or x∉B)
x∈A&x∉A or x∈B&x∉A or x∈A&x∉B or x∈B&x∉B
x∈B&x∉A or x∈A&x∉B
( A - B) ∪( B - A)

>> No.9279831

>>9279755
A\B=A\(AnB)
B\A=B\(AnB)
(A\B)u(B\A)=(AuB)\(AnB)

>> No.9279833

>>9279824
Careful with A-B notation, some use that for {a-b, a in A, b in B}.

>> No.9279846

>>9279833
algebraic difference, amirite?!?
Thanks, Aliprantis and Burkinshaw!

>> No.9279849

prove it with ms paint

>> No.9279884

>>9279824
accurate thx

>> No.9280157

>>9279755
[math](A\cup B)\setminus(A\cap B) = (A\cup B) \cap (A \cap B)^\complement = (A\cup B) \cap (A^\complement \cup B^\complement)[/math]
[math]=[A^\complement\cap(A\cup B)] \cup [B^\complement\cap(A\cup B)] = (A^\complement \cap B) \cup (B^\complement \cap A) = (A\setminus B) \cup (B \setminus A)[/math]

this is elementary set theory OP.
you should not have to ask this

>> No.9280162
File: 85 KB, 680x680, 1495276844960.jpg [View same] [iqdb] [saucenao] [google]
9280162

>>9279755
>Not using the superior triangle notation.
[math]A\triangle B[/math]

>> No.9280170

>>9279755
Write the math statements out in English
Think
Prove
Go to /hm/

>> No.9280180

>>9279755

Forget the responses above; they're ugly.

These types of proofs are usually best done by proving that the left side is a subset of the right, and the right side is a subset of the left.

I don't remember my latex from grad school so let just call the left side L and the right side R.

Proof:

Part 1) Show that L is a subset of R.

Pick x in L. Then x is in either a or b, but not both. Suppose x is in a. Since it's not also in b, its in A\B. Suppose x is in b. Since it's not also in A, it's in B\A. This shows that for all x in L, x is in either A\B or B\A. Therefore x is in R.

Part 2) Show that R is a subset of L.

You can do this part yourself.
You can do the other direction yourself.

>> No.9280217

>>9279755
Just write a truth table