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Thread: Boolean Algebra

  1. #1
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    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.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dwsmith View Post
    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 $\displaystyle B_n=2^{[n]}$.
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    So if $\displaystyle \displaystyle |L|=8$, what am I showing it is isomorphic too?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dwsmith View Post
    So if $\displaystyle \displaystyle |L|=8$, what am I showing it is isomorphic too?
    $\displaystyle 2^{[3]}$.
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    Quote Originally Posted by Drexel28 View Post
    $\displaystyle 2^{[3]}$.
    Unfortunately, that didn't help. I understand $\displaystyle 8=2^3$ but how do I show it is isomorphic to $\displaystyle 2^3$?
    Last edited by dwsmith; Dec 7th 2010 at 03:50 PM.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dwsmith View Post
    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?
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    Haven't a clue unless you are talking about the science sense or as in bomb.
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dwsmith View Post
    Haven't a clue unless you are talking about the science sense or as in bomb.
    Hahaha. Nice. Then what do you have?
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    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

    Boolean Algebra-hasse-diagram.jpg
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dwsmith View Post
    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

    Click image for larger version. 

Name:	Hasse Diagram.jpg 
Views:	4 
Size:	10.9 KB 
ID:	20015
    Here is the Hasse diagram for $\displaystyle 2^{[3]}$
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  11. #11
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    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?
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  12. #12
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dwsmith View Post
    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 $\displaystyle \{1,2\}$.
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    So 4 is the P(s)={1,2,3,4} for instance?
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  14. #14
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dwsmith View Post
    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.
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