Understanding how $mathcal{P}(A)$ is a Boolean algebra











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I'm currently doing work on discrete mathematics in my free time and am having some difficulties with understanding Boolean algebra. To be specific, I'm stuck on the following practice question:




Let $A = {a, b}$ and list the four elements of the power set $mathcal{P}(A)$. We consider the operation $+$ to be $cup$, $cdot$ to be $cap$, and complement to be set complement. Consider $1$ to be $A$ and $0$ to be $∅$.




  1. Explain why the description above defines a Boolean algebra.

  2. Find two elements $x$, $y$ in $mathcal{P}(A)$ such that $xy = 0$, $x neq 0$ and $y neq 0$.




Starting with the power set: $$mathcal{P}(A) = {∅, {a},{b},{a,b}}.$$



How would I go about finding the elements of $x$ and $y$ to satisfy part two of the question using algebraic axioms? Also, for explaining how the above defines a Boolean algebra, do you think it would suffice to simply mention how there are two binary operations and a set associated with the Boolean algebra?










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  • Can you describe the "multiplication" in words?
    – pjs36
    Sep 24 '15 at 2:20










  • I would assume so. Would that make the solution that much easier to find/prove?
    – Johnathan Scott
    Sep 24 '15 at 2:21










  • I would start by trying the various options.
    – JonMark Perry
    Sep 24 '15 at 2:24










  • I think it would make finding $x neq varnothing neq y$ with $x cdot y = 0$ easier, yes.
    – pjs36
    Sep 24 '15 at 2:25






  • 1




    @CameronBuie; its okay - i figured it out
    – JonMark Perry
    Sep 24 '15 at 3:42















up vote
1
down vote

favorite
3












I'm currently doing work on discrete mathematics in my free time and am having some difficulties with understanding Boolean algebra. To be specific, I'm stuck on the following practice question:




Let $A = {a, b}$ and list the four elements of the power set $mathcal{P}(A)$. We consider the operation $+$ to be $cup$, $cdot$ to be $cap$, and complement to be set complement. Consider $1$ to be $A$ and $0$ to be $∅$.




  1. Explain why the description above defines a Boolean algebra.

  2. Find two elements $x$, $y$ in $mathcal{P}(A)$ such that $xy = 0$, $x neq 0$ and $y neq 0$.




Starting with the power set: $$mathcal{P}(A) = {∅, {a},{b},{a,b}}.$$



How would I go about finding the elements of $x$ and $y$ to satisfy part two of the question using algebraic axioms? Also, for explaining how the above defines a Boolean algebra, do you think it would suffice to simply mention how there are two binary operations and a set associated with the Boolean algebra?










share|cite|improve this question
























  • Can you describe the "multiplication" in words?
    – pjs36
    Sep 24 '15 at 2:20










  • I would assume so. Would that make the solution that much easier to find/prove?
    – Johnathan Scott
    Sep 24 '15 at 2:21










  • I would start by trying the various options.
    – JonMark Perry
    Sep 24 '15 at 2:24










  • I think it would make finding $x neq varnothing neq y$ with $x cdot y = 0$ easier, yes.
    – pjs36
    Sep 24 '15 at 2:25






  • 1




    @CameronBuie; its okay - i figured it out
    – JonMark Perry
    Sep 24 '15 at 3:42













up vote
1
down vote

favorite
3









up vote
1
down vote

favorite
3






3





I'm currently doing work on discrete mathematics in my free time and am having some difficulties with understanding Boolean algebra. To be specific, I'm stuck on the following practice question:




Let $A = {a, b}$ and list the four elements of the power set $mathcal{P}(A)$. We consider the operation $+$ to be $cup$, $cdot$ to be $cap$, and complement to be set complement. Consider $1$ to be $A$ and $0$ to be $∅$.




  1. Explain why the description above defines a Boolean algebra.

  2. Find two elements $x$, $y$ in $mathcal{P}(A)$ such that $xy = 0$, $x neq 0$ and $y neq 0$.




Starting with the power set: $$mathcal{P}(A) = {∅, {a},{b},{a,b}}.$$



How would I go about finding the elements of $x$ and $y$ to satisfy part two of the question using algebraic axioms? Also, for explaining how the above defines a Boolean algebra, do you think it would suffice to simply mention how there are two binary operations and a set associated with the Boolean algebra?










share|cite|improve this question















I'm currently doing work on discrete mathematics in my free time and am having some difficulties with understanding Boolean algebra. To be specific, I'm stuck on the following practice question:




Let $A = {a, b}$ and list the four elements of the power set $mathcal{P}(A)$. We consider the operation $+$ to be $cup$, $cdot$ to be $cap$, and complement to be set complement. Consider $1$ to be $A$ and $0$ to be $∅$.




  1. Explain why the description above defines a Boolean algebra.

  2. Find two elements $x$, $y$ in $mathcal{P}(A)$ such that $xy = 0$, $x neq 0$ and $y neq 0$.




Starting with the power set: $$mathcal{P}(A) = {∅, {a},{b},{a,b}}.$$



How would I go about finding the elements of $x$ and $y$ to satisfy part two of the question using algebraic axioms? Also, for explaining how the above defines a Boolean algebra, do you think it would suffice to simply mention how there are two binary operations and a set associated with the Boolean algebra?







boolean-algebra






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edited Sep 24 '15 at 3:56









user642796

44.4k558115




44.4k558115










asked Sep 24 '15 at 2:17









Johnathan Scott

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  • Can you describe the "multiplication" in words?
    – pjs36
    Sep 24 '15 at 2:20










  • I would assume so. Would that make the solution that much easier to find/prove?
    – Johnathan Scott
    Sep 24 '15 at 2:21










  • I would start by trying the various options.
    – JonMark Perry
    Sep 24 '15 at 2:24










  • I think it would make finding $x neq varnothing neq y$ with $x cdot y = 0$ easier, yes.
    – pjs36
    Sep 24 '15 at 2:25






  • 1




    @CameronBuie; its okay - i figured it out
    – JonMark Perry
    Sep 24 '15 at 3:42


















  • Can you describe the "multiplication" in words?
    – pjs36
    Sep 24 '15 at 2:20










  • I would assume so. Would that make the solution that much easier to find/prove?
    – Johnathan Scott
    Sep 24 '15 at 2:21










  • I would start by trying the various options.
    – JonMark Perry
    Sep 24 '15 at 2:24










  • I think it would make finding $x neq varnothing neq y$ with $x cdot y = 0$ easier, yes.
    – pjs36
    Sep 24 '15 at 2:25






  • 1




    @CameronBuie; its okay - i figured it out
    – JonMark Perry
    Sep 24 '15 at 3:42
















Can you describe the "multiplication" in words?
– pjs36
Sep 24 '15 at 2:20




Can you describe the "multiplication" in words?
– pjs36
Sep 24 '15 at 2:20












I would assume so. Would that make the solution that much easier to find/prove?
– Johnathan Scott
Sep 24 '15 at 2:21




I would assume so. Would that make the solution that much easier to find/prove?
– Johnathan Scott
Sep 24 '15 at 2:21












I would start by trying the various options.
– JonMark Perry
Sep 24 '15 at 2:24




I would start by trying the various options.
– JonMark Perry
Sep 24 '15 at 2:24












I think it would make finding $x neq varnothing neq y$ with $x cdot y = 0$ easier, yes.
– pjs36
Sep 24 '15 at 2:25




I think it would make finding $x neq varnothing neq y$ with $x cdot y = 0$ easier, yes.
– pjs36
Sep 24 '15 at 2:25




1




1




@CameronBuie; its okay - i figured it out
– JonMark Perry
Sep 24 '15 at 3:42




@CameronBuie; its okay - i figured it out
– JonMark Perry
Sep 24 '15 at 3:42










2 Answers
2






active

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up vote
3
down vote













Let $x={a}$ and $y={b}$. We have $xneemptyset$ and $yneemptyset$. We also have $xy=xcap y={a}cap{b}=emptyset=0$, so that seems to work.



A boolean algebra has variables that can only be true or false (or $1$ or $0$). Normal algebraic operations can continue as usual, unless of course they don't work.






share|cite|improve this answer





















  • +1: Excellent! Now you need only prove that/explain why the Boolean algebra axioms hold on the given structure.
    – Cameron Buie
    Sep 24 '15 at 3:44










  • As a bonus problem: you should be able to prove that we can replace the given set $A$ with any set, define the operations the same way, and still get a Boolean algebra. This is because Boolean algebras were pretty much invented as generalizations of power sets under intersection and union.
    – Cameron Buie
    Sep 24 '15 at 3:47










  • en.wikipedia.org/wiki/Boolean_algebra#Concrete_Boolean_algebras
    – JonMark Perry
    Sep 24 '15 at 3:48










  • /sigh/ Yes, of course you could just look it up, but that won't necessarily help you get a feel for anything.
    – Cameron Buie
    Sep 24 '15 at 3:51










  • i should do, but before i got into this question, i was going to look at en.wikipedia.org/wiki/Quadratrix
    – JonMark Perry
    Sep 24 '15 at 3:55


















up vote
1
down vote













A boolean algebra is any set together with two binary operation $land$ and $lor$, one unary operation $lnot$ and two nullary operations $0$ and $1$ (that is, two constants) satisfying a suitable set of axioms - see, for instance, the definition section from the Wikipedia article.



The simplest boolean algebra is the already mentioned two element set ${0,1}$ together with the traditional truth table operations, but - as you seem to know already - any set $S$ gives rise to a boolean algebra defined on its power set $mathcal{P}(S)$: just define $land, lor, lnot, 0$ and $1$ to be $cap, cup, Ssetminus, emptyset$ and $S$ respectively ($Ssetminus$ means complement with relation to $S$).



As for proving that something is indeed a boolean algebra, there's no free lunch. You must check that the definitions you give for the boolean operations satisfy the axioms. So besides mentioning that there are two binary operations, you must be sure to have them properly defined so that anyone can understand what you take them to be. Also you must define the other operations as well and verify that the axioms hold for them.



For instance, one axiom says that:



$a land lnot a = 0$



To show that your particular instance of boolean algebra satisfies this axiom, you must show that:



$a cap (Ssetminus a) = emptyset$.






share|cite|improve this answer























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    2 Answers
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    2 Answers
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    up vote
    3
    down vote













    Let $x={a}$ and $y={b}$. We have $xneemptyset$ and $yneemptyset$. We also have $xy=xcap y={a}cap{b}=emptyset=0$, so that seems to work.



    A boolean algebra has variables that can only be true or false (or $1$ or $0$). Normal algebraic operations can continue as usual, unless of course they don't work.






    share|cite|improve this answer





















    • +1: Excellent! Now you need only prove that/explain why the Boolean algebra axioms hold on the given structure.
      – Cameron Buie
      Sep 24 '15 at 3:44










    • As a bonus problem: you should be able to prove that we can replace the given set $A$ with any set, define the operations the same way, and still get a Boolean algebra. This is because Boolean algebras were pretty much invented as generalizations of power sets under intersection and union.
      – Cameron Buie
      Sep 24 '15 at 3:47










    • en.wikipedia.org/wiki/Boolean_algebra#Concrete_Boolean_algebras
      – JonMark Perry
      Sep 24 '15 at 3:48










    • /sigh/ Yes, of course you could just look it up, but that won't necessarily help you get a feel for anything.
      – Cameron Buie
      Sep 24 '15 at 3:51










    • i should do, but before i got into this question, i was going to look at en.wikipedia.org/wiki/Quadratrix
      – JonMark Perry
      Sep 24 '15 at 3:55















    up vote
    3
    down vote













    Let $x={a}$ and $y={b}$. We have $xneemptyset$ and $yneemptyset$. We also have $xy=xcap y={a}cap{b}=emptyset=0$, so that seems to work.



    A boolean algebra has variables that can only be true or false (or $1$ or $0$). Normal algebraic operations can continue as usual, unless of course they don't work.






    share|cite|improve this answer





















    • +1: Excellent! Now you need only prove that/explain why the Boolean algebra axioms hold on the given structure.
      – Cameron Buie
      Sep 24 '15 at 3:44










    • As a bonus problem: you should be able to prove that we can replace the given set $A$ with any set, define the operations the same way, and still get a Boolean algebra. This is because Boolean algebras were pretty much invented as generalizations of power sets under intersection and union.
      – Cameron Buie
      Sep 24 '15 at 3:47










    • en.wikipedia.org/wiki/Boolean_algebra#Concrete_Boolean_algebras
      – JonMark Perry
      Sep 24 '15 at 3:48










    • /sigh/ Yes, of course you could just look it up, but that won't necessarily help you get a feel for anything.
      – Cameron Buie
      Sep 24 '15 at 3:51










    • i should do, but before i got into this question, i was going to look at en.wikipedia.org/wiki/Quadratrix
      – JonMark Perry
      Sep 24 '15 at 3:55













    up vote
    3
    down vote










    up vote
    3
    down vote









    Let $x={a}$ and $y={b}$. We have $xneemptyset$ and $yneemptyset$. We also have $xy=xcap y={a}cap{b}=emptyset=0$, so that seems to work.



    A boolean algebra has variables that can only be true or false (or $1$ or $0$). Normal algebraic operations can continue as usual, unless of course they don't work.






    share|cite|improve this answer












    Let $x={a}$ and $y={b}$. We have $xneemptyset$ and $yneemptyset$. We also have $xy=xcap y={a}cap{b}=emptyset=0$, so that seems to work.



    A boolean algebra has variables that can only be true or false (or $1$ or $0$). Normal algebraic operations can continue as usual, unless of course they don't work.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Sep 24 '15 at 3:41









    JonMark Perry

    11.2k92237




    11.2k92237












    • +1: Excellent! Now you need only prove that/explain why the Boolean algebra axioms hold on the given structure.
      – Cameron Buie
      Sep 24 '15 at 3:44










    • As a bonus problem: you should be able to prove that we can replace the given set $A$ with any set, define the operations the same way, and still get a Boolean algebra. This is because Boolean algebras were pretty much invented as generalizations of power sets under intersection and union.
      – Cameron Buie
      Sep 24 '15 at 3:47










    • en.wikipedia.org/wiki/Boolean_algebra#Concrete_Boolean_algebras
      – JonMark Perry
      Sep 24 '15 at 3:48










    • /sigh/ Yes, of course you could just look it up, but that won't necessarily help you get a feel for anything.
      – Cameron Buie
      Sep 24 '15 at 3:51










    • i should do, but before i got into this question, i was going to look at en.wikipedia.org/wiki/Quadratrix
      – JonMark Perry
      Sep 24 '15 at 3:55


















    • +1: Excellent! Now you need only prove that/explain why the Boolean algebra axioms hold on the given structure.
      – Cameron Buie
      Sep 24 '15 at 3:44










    • As a bonus problem: you should be able to prove that we can replace the given set $A$ with any set, define the operations the same way, and still get a Boolean algebra. This is because Boolean algebras were pretty much invented as generalizations of power sets under intersection and union.
      – Cameron Buie
      Sep 24 '15 at 3:47










    • en.wikipedia.org/wiki/Boolean_algebra#Concrete_Boolean_algebras
      – JonMark Perry
      Sep 24 '15 at 3:48










    • /sigh/ Yes, of course you could just look it up, but that won't necessarily help you get a feel for anything.
      – Cameron Buie
      Sep 24 '15 at 3:51










    • i should do, but before i got into this question, i was going to look at en.wikipedia.org/wiki/Quadratrix
      – JonMark Perry
      Sep 24 '15 at 3:55
















    +1: Excellent! Now you need only prove that/explain why the Boolean algebra axioms hold on the given structure.
    – Cameron Buie
    Sep 24 '15 at 3:44




    +1: Excellent! Now you need only prove that/explain why the Boolean algebra axioms hold on the given structure.
    – Cameron Buie
    Sep 24 '15 at 3:44












    As a bonus problem: you should be able to prove that we can replace the given set $A$ with any set, define the operations the same way, and still get a Boolean algebra. This is because Boolean algebras were pretty much invented as generalizations of power sets under intersection and union.
    – Cameron Buie
    Sep 24 '15 at 3:47




    As a bonus problem: you should be able to prove that we can replace the given set $A$ with any set, define the operations the same way, and still get a Boolean algebra. This is because Boolean algebras were pretty much invented as generalizations of power sets under intersection and union.
    – Cameron Buie
    Sep 24 '15 at 3:47












    en.wikipedia.org/wiki/Boolean_algebra#Concrete_Boolean_algebras
    – JonMark Perry
    Sep 24 '15 at 3:48




    en.wikipedia.org/wiki/Boolean_algebra#Concrete_Boolean_algebras
    – JonMark Perry
    Sep 24 '15 at 3:48












    /sigh/ Yes, of course you could just look it up, but that won't necessarily help you get a feel for anything.
    – Cameron Buie
    Sep 24 '15 at 3:51




    /sigh/ Yes, of course you could just look it up, but that won't necessarily help you get a feel for anything.
    – Cameron Buie
    Sep 24 '15 at 3:51












    i should do, but before i got into this question, i was going to look at en.wikipedia.org/wiki/Quadratrix
    – JonMark Perry
    Sep 24 '15 at 3:55




    i should do, but before i got into this question, i was going to look at en.wikipedia.org/wiki/Quadratrix
    – JonMark Perry
    Sep 24 '15 at 3:55










    up vote
    1
    down vote













    A boolean algebra is any set together with two binary operation $land$ and $lor$, one unary operation $lnot$ and two nullary operations $0$ and $1$ (that is, two constants) satisfying a suitable set of axioms - see, for instance, the definition section from the Wikipedia article.



    The simplest boolean algebra is the already mentioned two element set ${0,1}$ together with the traditional truth table operations, but - as you seem to know already - any set $S$ gives rise to a boolean algebra defined on its power set $mathcal{P}(S)$: just define $land, lor, lnot, 0$ and $1$ to be $cap, cup, Ssetminus, emptyset$ and $S$ respectively ($Ssetminus$ means complement with relation to $S$).



    As for proving that something is indeed a boolean algebra, there's no free lunch. You must check that the definitions you give for the boolean operations satisfy the axioms. So besides mentioning that there are two binary operations, you must be sure to have them properly defined so that anyone can understand what you take them to be. Also you must define the other operations as well and verify that the axioms hold for them.



    For instance, one axiom says that:



    $a land lnot a = 0$



    To show that your particular instance of boolean algebra satisfies this axiom, you must show that:



    $a cap (Ssetminus a) = emptyset$.






    share|cite|improve this answer



























      up vote
      1
      down vote













      A boolean algebra is any set together with two binary operation $land$ and $lor$, one unary operation $lnot$ and two nullary operations $0$ and $1$ (that is, two constants) satisfying a suitable set of axioms - see, for instance, the definition section from the Wikipedia article.



      The simplest boolean algebra is the already mentioned two element set ${0,1}$ together with the traditional truth table operations, but - as you seem to know already - any set $S$ gives rise to a boolean algebra defined on its power set $mathcal{P}(S)$: just define $land, lor, lnot, 0$ and $1$ to be $cap, cup, Ssetminus, emptyset$ and $S$ respectively ($Ssetminus$ means complement with relation to $S$).



      As for proving that something is indeed a boolean algebra, there's no free lunch. You must check that the definitions you give for the boolean operations satisfy the axioms. So besides mentioning that there are two binary operations, you must be sure to have them properly defined so that anyone can understand what you take them to be. Also you must define the other operations as well and verify that the axioms hold for them.



      For instance, one axiom says that:



      $a land lnot a = 0$



      To show that your particular instance of boolean algebra satisfies this axiom, you must show that:



      $a cap (Ssetminus a) = emptyset$.






      share|cite|improve this answer

























        up vote
        1
        down vote










        up vote
        1
        down vote









        A boolean algebra is any set together with two binary operation $land$ and $lor$, one unary operation $lnot$ and two nullary operations $0$ and $1$ (that is, two constants) satisfying a suitable set of axioms - see, for instance, the definition section from the Wikipedia article.



        The simplest boolean algebra is the already mentioned two element set ${0,1}$ together with the traditional truth table operations, but - as you seem to know already - any set $S$ gives rise to a boolean algebra defined on its power set $mathcal{P}(S)$: just define $land, lor, lnot, 0$ and $1$ to be $cap, cup, Ssetminus, emptyset$ and $S$ respectively ($Ssetminus$ means complement with relation to $S$).



        As for proving that something is indeed a boolean algebra, there's no free lunch. You must check that the definitions you give for the boolean operations satisfy the axioms. So besides mentioning that there are two binary operations, you must be sure to have them properly defined so that anyone can understand what you take them to be. Also you must define the other operations as well and verify that the axioms hold for them.



        For instance, one axiom says that:



        $a land lnot a = 0$



        To show that your particular instance of boolean algebra satisfies this axiom, you must show that:



        $a cap (Ssetminus a) = emptyset$.






        share|cite|improve this answer














        A boolean algebra is any set together with two binary operation $land$ and $lor$, one unary operation $lnot$ and two nullary operations $0$ and $1$ (that is, two constants) satisfying a suitable set of axioms - see, for instance, the definition section from the Wikipedia article.



        The simplest boolean algebra is the already mentioned two element set ${0,1}$ together with the traditional truth table operations, but - as you seem to know already - any set $S$ gives rise to a boolean algebra defined on its power set $mathcal{P}(S)$: just define $land, lor, lnot, 0$ and $1$ to be $cap, cup, Ssetminus, emptyset$ and $S$ respectively ($Ssetminus$ means complement with relation to $S$).



        As for proving that something is indeed a boolean algebra, there's no free lunch. You must check that the definitions you give for the boolean operations satisfy the axioms. So besides mentioning that there are two binary operations, you must be sure to have them properly defined so that anyone can understand what you take them to be. Also you must define the other operations as well and verify that the axioms hold for them.



        For instance, one axiom says that:



        $a land lnot a = 0$



        To show that your particular instance of boolean algebra satisfies this axiom, you must show that:



        $a cap (Ssetminus a) = emptyset$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 25 at 6:35

























        answered Sep 24 '15 at 12:09









        Tarc

        339411




        339411






























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