0
$\begingroup$

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

$\endgroup$
7
  • 1
    $\begingroup$ Do you have a definition of the $\oplus$ symbol? $\endgroup$ – hmakholm left over Monica Nov 5 '11 at 18:00
  • $\begingroup$ 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. $\endgroup$ – sasha Nov 5 '11 at 18:01
  • 1
    $\begingroup$ 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. $\endgroup$ – hmakholm left over Monica Nov 5 '11 at 18:06
  • $\begingroup$ 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"? $\endgroup$ – Asaf Karagila Nov 5 '11 at 18:10
  • $\begingroup$ 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. $\endgroup$ – Asaf Karagila Nov 5 '11 at 21:24
2
$\begingroup$

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

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

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

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

$\endgroup$
1
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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