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Math Help - Show that N is multiplicative

  1. #1
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    Show that N is multiplicative

    Define  N:Z[w] \rightarrow Z by N(a+bw) = a^2 - ab + b^2 with  w = \frac {-1}{2} + \frac {3^{1/2}}{2}i .

    Show that N is multiplicative.

    I'm a bit rusty on this, showing it is multiplicative means closure, associative, inverse, identity, right? Or N(ab) = N(a)N(b)?
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    Quote Originally Posted by tttcomrader View Post
    Define  N:Z[w] \rightarrow Z by N(a+bw) = a^2 - ab + b^2 with  w = \frac {-1}{2} + \frac {3^{1/2}}{2}i .

    Show that N is multiplicative.

    I'm a bit rusty on this, showing it is multiplicative means closure, associative, inverse, identity, right? Or N(ab) = N(a)N(b)?
    You need to show that N((a+bw)(c+dw)) = N(a+bw)N(c+dw).
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    Alright, that would be easy, thank you!
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    I was trying to work this out, but here is what I got.

    Let x=a+bw and y=c+dw.

    N(xy) = N(ac+adw+cbw+bdw), I can't get it into a the form of u+vw, how would I do it? Thanks.
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  5. #5
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    Quote Originally Posted by tttcomrader View Post
    I was trying to work this out, but here is what I got.

    Let x=a+bw and y=c+dw.

    N(xy) = N(ac+adw+cbw+bdw), I can't get it into a the form of u+vw, how would I do it? Thanks.
    If x=a+bw and y=c+dw then xy = ac +adw +bcw +bdw^2. But w^2 = -1-w, so xy = (ac-bd) + (ad+bc-bd)w.

    Thus N(xy) = (ac-bd)^2 -(ac-bd)(ad+bc-bd) + (ad+bc-bd)^2. You need to verify that this is equal to N(x)N(y) = (a^2-ab+b^2)(c^2-cd+d^2). Good luck!
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