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## Hot answers tagged boolean-ring

6

I believe that the best way to study the properties of this structure is to note that $\mathcal P(X)$ with these operations is isomorphic to the set $\mathcal{F}$ of functions $X \to E$, where $E = \{ 0, 1\}$ is the field with two elements, and $\mathcal{F}$ is endowed with pointwise sum and product, that is, for $f, g \in \mathcal{F}$ we have $$(f + g)(x) =... 4 If I took pairs (x,y) where x and y can be either 0 or 1, I get the set$$ \{ (0,0), (0,1), (1,0), (1,1) \}. $$Multiply pairs by multiplying there coordinates, e.g.:$$ (0,1) * (1,1) = (0 \cdot 1, 1 \cdot 1) = (0,1). $$Then a^2 = a for every element of my four element set. 4 The statement is right, but your proof is not correct. For example, suppose F=\{x_1, x_2\} with x_1< x_2 (that is, x_1\not=x_2 and x_1\vee x_2=x_2). Then the A we get consists of$$a_1=x_1\wedge x_2,\quad a_2=x_1\wedge x_2', \quad a_3=x_1'\wedge x_2,\quad a_4=x_1'\wedge x_2'.$$But distinct elements of the powerset of A do not necessarily ... 4 this, together with the the cofunctor from complete atomic Boolean rings to sets \operatorname{Mor}_{\mathbf{Ring}}({-},\mathbb{Z}/2\mathbb{Z}) yields a coequivalence of categories between the category of sets and the category of complete atomic Boolean rings This is incorrect. To get a coequivalence between the category of sets and the category of ... 3 There is a universal binary (m=2) function for any finite domain. As in Peter Košinár's answer, take to S be the integers modulo |S|. I assume |S|\geq 3. Define$$f(x,y)=\begin{cases}x+1&\text{ if $x=y,$}\\ 0&\text{ otherwise.}\end{cases}$$Some easy gizmos: the successor function s(x)=x+1 can be defined in terms of f by s(x)=f(x,x) ... 3 Let m=4 and S:=\mathbb{Z_k} of residues modulo k\geq 3 (all arithmetical operations will be considered modulo k). We will define f(x,y,a,b) as follows:$$ \begin{equation} f(x,y,a,b)= \begin{cases} (x+1) & \text{if } y=x\\ 0 & \text{if } y=(x+1)\\ a & \text{if } y\not\in\{x,x+1\} \wedge x=0\\ b & \text{if }...

3

$-s$ is the additive inverse of $s$, so $s+(-s)=0$. Then $$0=s0=s(s+(-s))=s^2+s(-s).$$ Also $$0=0(-s)=(s+(-s))(-s)=s(-s)+(-s)^2.$$ Therefore $s^2$ and $(-s)^2$ are both the inverse of $s(-s)$ in the commutative group of $R$ under addition: $s^2=(-s)^2$. But, a better approach to I, is to note that in addition to $s^2=s$ we also have $(s+s)^2=s+s$, that is $... 3 Let symmetric difference be correspond to addition and intersection correspond to multiplication. Also let$\varnothing$be empty set and correspond to$0$, as well as$E$be universe of set and correspond to$1$. Then we can prove that symmetric difference and intersection of set satisfy each ring axiom. Commutativity of addition: $$X\bigtriangleup Y=(X-Y)... 3 No. For instance, if I is any set, then any subring of the product \mathbb{Z}_2^I will also be a Boolean ring, and such subrings usually are not themselves products of copies of \mathbb{Z}_2. For instance, any infinite product of copies of \mathbb{Z}_2 must be uncountable, but you can take the subring of \mathbb{Z}_2^{\mathbb{N}} generated by a ... 2 No, there are many infinite Boolean algebras (equivalent to a Boolean ring), not only the powers of \{0,1\}. E.g the Borel sets in \mathbb{R}. 2 Note that the Boolean algebra corresponding to a ring of the form \prod_{i\in I}\mathbb{Z}/2\mathbb{Z} will have atoms - that is, nonzero elements which don't properly bound any nonzero elements (think about the element corresponding to \delta_i for i\in I ...). So any atomless Boolean algebra will yield a Boolean ring which is not of that form. 2 The free Boolean ring on n generators is nothing more than the polynomial ring \mathbb{F}_2[x_1,x_2,\dots,x_n] modulo the ideal generated by x_i^2-x_i, for i=1,2,\dots n, so it's essentially the set of linear combinations of monomials having degree at most 1 in each variable, with coefficients in \mathbb{F}_2; there are 2^n such monomials, so ... 2 1) yes, any ring with local right identities has this property. That is, if for every x, there exists e such that xe=x. Such an example is \prod _{i=1}^\infty \mathbb Z. Actually if you think about it, this is equivalent to your condition, since \sum_{i=1}^n=y does the trick. 2) rings with local identities are interesting. 3) Let me share a bit ... 2 The answer to Question 2 is yes (but for uninteresting reasons). Fix n. Let N=(n+1)! and consider the ring T=\mathbb Z_N[t]. Let K be the ideal of T generated by p(t)=t(t-1)\ldots(t-n). Since p(t) is monic, T/K is finite. Therefore there exist integers a>b>0 with \bar t^a=\bar t^b, where \bar t=t+K\in T/K. Now suppose R ... 1 \def\peq{\mathrel{\phantom{=}}{}}Step 1: Preliminary analysis. Note that in this system, a + a = 0 and a^2 = a for any a. For any 1 \leqslant k, l \leqslant n, suppose$$ c_n(k) c_n(l) = \sum_{m = 1}^{\min(k + l, n)} b_m c_n(m), $$where b_1, \cdots, b_{\min(k + l, n)} \in \{0, 1\}. Note that for any A, B \subseteq \{1, \cdots, n\}, there is$$ \... 1 The elementary symmetric polynomials are related to your function by $$e_k(a_1,a_2,\dots,a_n)\equiv c_n(k)\pmod{2}$$ where multiplication mod$2$is$\wedge$and addition mod$2$is$\oplus$. For completeness, let$c_n(0):=1$. Vieta's formula is that a polynomial with roots$a_1,\dots,a_n$with multiplicity is given by$f(x)=\sum_{i=0}^n(-1)^ie_i(a_1,\dots,...

1

Here's another approach. Let $\mathbf B$ be an atomic Boolean Algebra (not all infinite Boolean Algebras are atomic) and $A$ be its set of atoms. Prove that, for $a \in A$, $$\bigvee (A \setminus \{a\}) = a'.$$ It follows that if $A_0$ is a proper subset of $A$, then $\bigvee A_0 \leq a'<1$, for each $a \in A \setminus A_0$.

1

The idea of the proof is good, but should be probably called 'proof by contradiction' instead of 'induction': If we have $a_1+\dots+a_n=1$ for atoms $a_i$, then every element $x$ of the Boolean algebra is a sum of certain $a_i$'s (specifically, of those that are smaller than $x$).

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Rephrasing what zero divisor means in this context may help you: Since the zero is the empty set, and intersection is the product, finding a zero divisor is equivalent to finding two non-empty sets whose intersection is empty.

1

In a Boolean ring $(-1)^2=-1$, so $1=-1$. In particular $$xy-yx=xy+yx$$

1

The "function" you use is $a_{n+1}= a_n- 1$. Given any positive initial number that gets every number less that the initial number so certainly 1.

1

The equation is true if we agree that a sum with only finitely many non-zero summands is the um of those finitely many summands. (And of course, $B$ must at least be an abelian group )

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As an added answer, George Boole originally had sets in mind. If $X$ and $Y$ are two sets we can put $$X + Y = X \cup Y, \quad X \cdot Y = X \cap Y.$$ Then $$X^2 = X \cdot X = X \cap X = X.$$ Take the set $\{0,1\}$, then the set of all subsets is $$\{ \emptyset, \{0\}, \{1\}, \{1,2\} \}.$$ This is closed under union and intersection and has four ...

1

You have indeed two options. If you equip the set $\{0, 1\}$ with $OR$ and $AND$ you have a semiring structure. If you make use of $XOR$ and $AND$, you have a Boolean ring. If $S$ is one of these two structures, the product $S^\mathbb{N}$ has the same structure, defined componentwise as you did.

1

As the commenter said: use the inclusion-exclusion property, which for two events says: $P(A \lor B) = P(A) + P(B) - P(A \land B)$ and (in your case more relevant!) for three events says: $P(A \lor B \lor C) = P(A) + P(B) + P(C) - P(A \land B) - P(A \land C)- P(B \land C) + P(A \land B \land C)$ So, use this formula, where: $A = a_1(a_4+a_8a_5+a_8a_7a_6)... 1 There are only finitely many sets that are intersections of sets in$V$and their complements. Every set in$W$must be a union of those. (This straightforward argument may be just a rephrasing of the idea behind the BIG HAMMER you want to avoid.) Edit in response to comment from the OP Not doubting of course, but not understanding the reasoning. Here'... 1 Ring of Sets by corresponding union to addition and intersection to multiplication ($\varnothing$to$0$and$E$(universe of set) to$1$as well) satisfy most ring axioms. However the fundamental difference is that it doesn't have an additive inverse because$X\cup Y=\varnothing$if and only if$X=\varnothing$and$Y=\varnothing\$. Thus your definition of ...

1

A "ring of sets" should really be called a "distributive lattice of sets." Does that answer your question?

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