Solve for X in this equation of sets.

Taking U to be the universe of discourse, let R be the set of subsets of U and define the operations +, *, for R by

$A + B \equiv A \cup B - A \cap B$

$A \circ B \equiv A \cap B$

for subsets A, B of U. Define $0 \equiv \emptyset$ and $e \equiv U$. Then $(R,0,e,+,\circ)$ is a ring.

Compute the solutions X to the equations below using only $A, B, +, \circ$

i. $A + X = 0$

ii. $A + X = B$

iii. $X = (A-B)(A+B)$

I understand that X in number i. should be A, and I found in ii. that X can be written as $B - A \cup (B' \cap A)$, but I can't write it in terms of $A,B,+,\circ$ like the question asks.

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Hint: This is actually isomorphic to the ring of functions $U\to \mathbb F_2$ under pointwise operations in $\mathbb F_2$. (Every set maps to its indicator function). Since $A+A=0$ for all $A$, it has characteristic 2, and therefore subtraction in the ring is the same as the addition. (It would be notationally easier if you used $\setminus$ instead of $-$ for set difference, such that you could write subtraction in the ring with standard notation). – Henning Makholm Apr 23 '12 at 16:27
Oh man, thank you so much. I just had to realize A + A = A - A – cstack Apr 23 '12 at 16:40

How do we solve an equation of the form $a+x=b$ in the real numbers? We add $-a$ to both sides to get $x=b-a$. What does "$-a$" mean? $-a$ is the unique real number that, when added to $a$, will give you $0$.

So in essence, what we are doing is: \begin{align*} a+x &= b &\text{(assumption)}\\ -a+(a+x) &= -a+b&\\ (-a+a)+x &= -a+b &(+\text{ is associative)}\\ 0+x &= -a+b&\text{(definition of }-a\text{)}\\ x & = -a+b &\text{(property of }0\text{)} \end{align*}

We do the same thing in (i) and (ii). Only... what is "$-A$"? It's the unique subset of $U$ such that $A+A=\varnothing$.

So... what is the element which, when added to $A$, gives $\varnothing$? Once you find it, you can solve both $A+X=0$ and $A+X=B$, in terms of $A$, $B$, and $0$.

For (iii), you need to remember that $A-B$ means $A+(-B)$; so again, the first step is to figure out what is $-B$ in terms of $B$. Once you do, you can compute $A-B$, and compute $A+B$ (using the definition of $+$), and then compute their product (aka their intersection). This will give you $X$ in terms of $A$ and $B$.

Added. Henning raises a good point in comments: that in (iii), the $-$ on the right hand side might refer not to the $-$ operation in the ring, but rather to the set-theoretic difference of $A$ and $B$. This would be bad use of potentially confused and confusing notation; if that's what it means, then you would first compute $A\setminus B = \{a\in A\mid a\notin B\}$, then compute the "product" (intersection) with $A+B$ (aka their symmetric difference) and obtain an expression for $X$ from that.

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In (iii), I suspect that what is meant is actually $X=(A\setminus B)(A+B)$; cf. the use of $-$ in the definition of $+$. – Henning Makholm Apr 23 '12 at 16:44
@HenningMakholm: Humph. That would be bad notation... but you may be right. On the other hand, it uses product, not intersection, so it may not mean that... – Arturo Magidin Apr 23 '12 at 16:46

Hint: Draw Venn diagrams of $A+B$ and $A \circ B$. What happens when you add either of these to $A$?

And these operations are commutative, right? And we've already discovered that $A+A=0$, so what is $A+(A+B)$, I wonder....

Of course, you'll need to show that $+$ is associative--ie $A+(B+C)=(A+B)+C$ for all $A,B,C\in R$--which is doable (with a lot of element chasing).

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