# Boolean Algebra

• Dec 7th 2010, 02:44 PM
dwsmith
Boolean Algebra
In order to show a Lattice is a Boolean Algebra, the diagram needs to be bounded, distributive, complemented, and $\displaystyle |L|=2^n \ n\geq 1 \ n\in\mathbb{Z}$.

However, my book says, "A finite lattice is called a Boolean Algebra if it is isomorphic to $\displaystyle B_n$ for some nonnegative integer n."

What is $\displaystyle B_n$?

I know $\displaystyle D_n$ are the numbers that divide n but have no clue about this $\displaystyle B_n$.

Thanks.
• Dec 7th 2010, 02:58 PM
Drexel28
Quote:

Originally Posted by dwsmith
In order to show a Lattice is a Boolean Algebra, the diagram needs to be bounded, distributive, complemented, and $\displaystyle |L|=2^n \ n\geq 1 \ n\in\mathbb{Z}$.

However, my book says, "A finite lattice is called a Boolean Algebra if it is isomorphic to $\displaystyle B_n$ for some nonnegative integer n."

What is $\displaystyle B_n$?

I know $\displaystyle D_n$ are the numbers that divide n but have no clue about this $\displaystyle B_n$.

Thanks.

I'm not quite sure what they mean, but a fundamental result in the study of Boolean algebras is that every finite Boolean algebra $\displaystyle B$ is isomorphic to $\displaystyle 2^{[n]}$ (here $\displaystyle [n]=\{1,\cdots,n\}$) for some $\displaystyle n\in\mathbb{N}$. This fact would make me guess that $\displaystyle B_n=2^{[n]}$.
• Dec 7th 2010, 03:01 PM
dwsmith
So if $\displaystyle \displaystyle |L|=8$, what am I showing it is isomorphic too?
• Dec 7th 2010, 03:26 PM
Drexel28
Quote:

Originally Posted by dwsmith
So if $\displaystyle \displaystyle |L|=8$, what am I showing it is isomorphic too?

$\displaystyle 2^{[3]}$.
• Dec 7th 2010, 03:30 PM
dwsmith
Quote:

Originally Posted by Drexel28
$\displaystyle 2^{[3]}$.

Unfortunately, that didn't help. I understand $\displaystyle 8=2^3$ but how do I show it is isomorphic to $\displaystyle 2^3$?
• Dec 7th 2010, 03:43 PM
Drexel28
Quote:

Originally Posted by dwsmith
Unfortunately, that didn't help. I understand $\displaystyle 8=2^3$ but how do I show is isomorphic to $\displaystyle 2^3$?

This isn't in general easy. The result takes a fair amount of background. Do you know what 'atomic' means?
• Dec 7th 2010, 03:50 PM
dwsmith
Haven't a clue unless you are talking about the science sense or as in bomb.
• Dec 7th 2010, 03:53 PM
Drexel28
Quote:

Originally Posted by dwsmith
Haven't a clue unless you are talking about the science sense or as in bomb.

Hahaha. Nice. Then what do you have?
• Dec 7th 2010, 04:13 PM
dwsmith
The book mentions making a Hasse diagram of $\displaystyle B_n$. What would that look like?

Here is the Hasse diagram I need to show it is isomorphic to B_n.

Labeled a to h. a=O and h=I

Attachment 20015
• Dec 7th 2010, 04:21 PM
Drexel28
Quote:

Originally Posted by dwsmith
The book mentions making a Hasse diagram of $\displaystyle B_n$. What would that look like?

Here is the Hasse diagram I need to show it is isomorphic to B_n.

Labeled a to h. a=O and h=I

Attachment 20015

Here is the Hasse diagram for $\displaystyle 2^{[3]}$
• Dec 7th 2010, 04:23 PM
dwsmith
How is that the Hasse Diagram for $\displaystyle 2^3\mbox{?}$ How do I come up with the Hasse diagram when I have $\displaystyle 2^n$ is the better question?
• Dec 7th 2010, 04:26 PM
Drexel28
Quote:

Originally Posted by dwsmith
How is that the Hasse Diagram for $\displaystyle 2^3\mbox{?}$ How do I come up with the Hasse diagram when I have $\displaystyle 2^n$ is the better question?

Because it's the set of all subsets of a set with three elements. You do the same thing. Just draw out the diagram of the lattice for the power set of $\displaystyle \{1,2\}$.
• Dec 7th 2010, 04:30 PM
dwsmith
So 4 is the P(s)={1,2,3,4} for instance?
• Dec 7th 2010, 06:12 PM
Drexel28
Quote:

Originally Posted by dwsmith
So 4 is the P(s)={1,2,3,4} for instance?

What I mean, is that if you have a Boolean algebra $\displaystyle B$ with $\displaystyle \#(B)=2^n$ then $\displaystyle B\cong\mathcal{P}\left(\{1,\cdots,n\}\right)$ if that notation is more comfortable.