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Math Help - [SOLVED] Abstract Algebra: Integers Modulo n

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    is up to his old tricks again! Jhevon's Avatar
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    [SOLVED] Abstract Algebra: Integers Modulo n

    GŁten Tag!

    Here's another one. This is one of those questions where the answer seems so obvios to me that I don't know how they expect me to do it.

    Problem:

    Prove that [a] \odot ([b] \oplus [c]) = ([a] \odot [b]) \oplus ([a] \odot [c]) for all [a],[b],[c] \in \mathbb{Z}_n


    Things that may come in handy:

    Nothing really. I assume that anyone who would be able to help me on this will already know what all those symbols I used mean. We really don't need to know anything but their defintions I think.


    What I've Tried:

    Stared at the problem for many minutes wondering..."Uh, isn't this obvios? \odot and \oplus are associative and commutative on \mathbb{Z}_n. Does the result not immediately follow from that? Maybe they are asking me to prove: [a] \odot [b + c] = [ab] \oplus [ac], in which case, the solution is again obvios and trivial! (and when i say that, you know it's true. I'm only a Nascent Mathematician, after all)"

    Can anyone help me to rise above my inclinations and be rigorous with this problem?


    Thanks guys and gals (of course, for future reference, unless otherwise stated, gals refers to JaneBennet -- are there anymore female mathematicians otu there?)
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  2. #2
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    Quote Originally Posted by Jhevon View Post
    Prove that [a] \odot ([b] \oplus [c]) = ([a] \odot [b]) \oplus ([a] \odot [c]) for all [a],[b],[c] \in \mathbb{Z}_n
    I assume by \odot you mean multiplication by congruences classes. And \oplus you mean addition of congruences classes. Then use the fact that [a] + [b] = [a+b] is well-defined operations. Thus, [a]([b]+[c]) = [a]([b+c]) = [ab+ac] = [ab]+[ac]=[a][b]+[a][c].
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I assume by \odot you mean multiplication by congruences classes. And \oplus you mean addition of congruences classes. Then use the fact that [a] + [b] = [a+b] is well-defined operations. Thus, [a]([b]+[c]) = [a]([b+c]) = [ab+ac] = [ab]+[ac]=[a][b]+[a][c].
    That is exactly what i was thinking of doing. but it seemed too easy. here is how i would write it out (because my professor is the type that wants you to give a reason for anything you do--which is not necessarily bad, i guess).


    we wish to show that [a] \odot ([b] \oplus [c]) = ([a] \odot [b]) \oplus ([a] \odot [c])<br />

    Note that: [a] \odot ([b] \oplus [c]) = [a] \odot [b + c] ..........................by the definition of \oplus

    ............................................ = [a(b + c)] ..............................by the definition of \odot

    ............................................ = [a \cdot b + a \cdot c] ........................ by the the distributive property of real numbers

    ............................................ = [a \cdot b] \oplus [a \cdot c] .......................by the definition of \oplus

    ............................................ = ([a] \odot [b]) \oplus ([a] \odot [c]) ........by the definition of \odot

    as was to be shown
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