Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to prove the following 2 equations:
A,B are subsets of a group U

A ⊕ ∅ = A
now as I understood this equation means {a,∅},{b,∅}... if a,b -> A
Right?

Second equation is

(A ⊕ B) ⊕ B = A

I think if I would understand the method on one of them,
I'll understand it on the other

share|improve this question
1  
Do you have a definition of the $\oplus$ symbol? –  Henning Makholm Nov 5 '11 at 18:00
    
The question is unclear: your first equation is missing $B$ and you haven't defined the circled plus, which you should do for a homework question. If you're having trouble with math formatting and don't know LaTeX, I suggest writing your question out in words. This post How do I type math in my question/answer/comment? was helpful for me--it's from the FAQ. –  sasha Nov 5 '11 at 18:01
1  
Also, are you working with groups (an algebraic structure) or simply with sets? The two questions make sense if $A\oplus B$ means the symmetric difference $(A\cup B)\setminus(A\cap B)$, but your "$\{a,\varnothing\},\{b,\varnothing\},\ldots$" could suggest that you're confusing it with direct sums. –  Henning Makholm Nov 5 '11 at 18:06
    
Asaf, if you want to write the name of $\oplus$ in Hebrew and I could translate it to the proper term. Do you mean הפרש סימטרי, also known as symmetric difference and "xor"? –  Asaf Karagila Nov 5 '11 at 18:10
    
I also note that you used "group" which can be the naive translation of קבוצה, however it is not the correct term. Group is a distinct concept translated as חבורה. –  Asaf Karagila Nov 5 '11 at 21:24
show 2 more comments

2 Answers

up vote 1 down vote accepted

If I remember correctly, $A⊕B=(A\setminus B) \cup ( B \setminus A).$

Well, let's use it on your examples :)

$$A⊕\varnothing=(A\setminus \varnothing) \cup (\varnothing \setminus A)=A\cup\varnothing=A$$

$$(A⊕B)⊕B=((A\setminus B) \cup ( B \setminus A))⊕B=$$ $$(((A\setminus B) \cup ( B \setminus A))\setminus B)\cup(B \setminus((A\setminus B) \cup ( B \setminus A)))=$$ $$(A \setminus B)\cup(A\cap B)=A$$

share|improve this answer
add comment

The simplest definition I know of symmetric set difference is $$ x \in A \oplus B \;\equiv\; x \in A \not\equiv x \in B $$ Using this, we can calculate the elements of the left hand side of first equality, as follows: \begin{align} & x \in A \oplus \varnothing \\ \equiv & \;\;\;\;\;\text{"definition of $\;\oplus\;$"} \\ & x \in A \not\equiv x \in \varnothing \\ \equiv & \;\;\;\;\;\text{"definition of $\;\varnothing\;$"} \\ & x \in A \not\equiv \text{false} \\ \equiv & \;\;\;\;\;\text{"logic: simplify"} \\ & x \in A \\ \end{align} By set extensionality this proves the first equality.

To prove the second, we calculate similarly \begin{align} & x \in (A \oplus B) \oplus B \\ \equiv & \;\;\;\;\;\text{"definition of $\;\oplus\;$, twice; drop parentheses since $\;\not\equiv\;$ is associative"} \\ & x \in A \not\equiv x \in B \not\equiv x \in B \\ \equiv & \;\;\;\;\;\text{"logic: simplify"} \\ & x \in A \not\equiv \text{false} \\ \equiv & \;\;\;\;\;\text{"logic: simplify"} \\ & x \in A \\ \end{align}

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.