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Math Help - Boolean algebra,isomorphism

  1. #1
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    Boolean algebra,isomorphism

    Let (B1,+,*,c) is Boolean algebra where B1 is set of every positive dividers of 2310, in which are defined:x+y = lcm(x,y), x*y = gcd(x,y) and xc = 2310/x.
    Let (B2,,,c) is Boolean algebra of every subsets of {a,b,c,d,e} with standard operations.

    Let f:B1 → B2 is isomorphism where f(2)={a} , f(3)={b} , f(5)={c} , f(7)={d} , f(11)={e}.
    - How f(35), f(110), f(210), f(330) are defined?
    - Determine f((30 + 5*7)c).
    - How many elements B1 have?
    - Write all atoms of B1.
    - How many different isomorphism can be defined between B1 and B2?
    thanks for your time
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  2. #2
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    Re: Boolean algebra,isomorphism

    Quote Originally Posted by kljoki View Post
    How f(35), f(110), f(210), f(330) are defined?
    Since * corresponds to ∪, factor each of these numbers and use the fact that f(x * y) = f(x) ∪ f(y).
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  3. #3
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    Re: Boolean algebra,isomorphism

    hint: 2310 = 2*3*5*7*11

    make a conjecture about "primes" in B1, and "singleton sets" in B2.

    suppose σ is a bijection on (the underlying set of) B2.

    show that σof is also an isomorphism from B1 to B2. how many such permutations (bijections) exist?

    prove that every bijection from {2,3,5,7,11} to {a,b,c,d,e} can be written as σof, for some permutation σ of {a,b,c,d,e}.

    conclude that the number of such isomorphisms is 120 <---where does this number come from?

    one specific example worked out:

    f(35) = f(lcm(5,7)) = f(5+7) = f(5) + f(7) = f(5) U f(7) = {c} U {d} = {c,d}.
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  4. #4
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    Re: Boolean algebra,isomorphism

    lets see
    f(35), f(110), f(210), f(330)

    f(110) = f(lcm(11,2,5)) = f(11+2+5) = f(11) + f(2) + f(5) = f(11) U f(2) U f(5) = {e} U {a} U {c} = {a,c,e}

    f(210) = f(lcm(7,5,3,2)) = f(7+5+3+2) = f(7) + f(5) + f(3) +f(2) = f(7) U f(5) U f(3) U f(2) = {d} U {c} U {b} U {a} = {a,b,c,d}

    f(330) = f(lcm(11,5,3,2)) = f(11+5+3+2) = f(11) + f(5) + f(3) +f(2) = f(11) U f(5) U f(3) U f(2) = {e} U {c} U {b} U {a} = {a,b,c,e}
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  5. #5
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    Re: Boolean algebra,isomorphism

    f((30 + 5*7)c) = 2310 / f( f(lmc(2,3,5)) + f(gcd(5,7))) = {a,b,c,d,e} / {{a,b,c} U } = {d,e}
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  6. #6
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    Re: Boolean algebra,isomorphism

    B1 have 5 elements and every element is an atom
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  7. #7
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    Re: Boolean algebra,isomorphism

    there is 5! = 120 (permutation) different isomorphism between B1 and B2 because we have 2 sets with 5 elements.
    For the first element we have 5 combinations (f(2)={a}, f(2)={b}, f(2)={c}, f(2)={d}, f(2)={e}), second element have 5-1 combinations (because one element applies to f(2)), third 5-2 etc.
    so came to the result of 120 different isomorphisms
    Last edited by kljoki; August 19th 2012 at 08:52 AM.
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