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Multiplication modulo algebra

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  1. Feb 20, 2014 #1
    Question: Let n be a positive integer and consider x,y,z to be elements of Zn. Prove that if x . y = 1 and x . z = 1, then y = z. (Since working in Zn the sign '.' means "multiplication modulo n".)

    Conclude that if x has an inverse element in Zn, then the inverse element is unique.

    Attempt: Well if x . y = 1 then this means that xy = n + 1 since were working in Zn. This means the same can be said about y . z = 1 that yz = n + 1. This suggests that z = x by substitution. I'm not sure of this is correct or if I'm going down right path here.

    Other thing I thought about was multiplying x . y = 1 by z on both sides to prove y=z but I'm a bit stuck.

    I'm also a bit confused about the last bit of the question where I have to conclude if x has inverse element (not entirely sure what this is) in Zn then the inverse element is unique.

    Can anyone help please?


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  3. Feb 20, 2014 #2

    HallsofIvy

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    This is NOT true. for example, 5*5= 1 (mod 12) because 5*5= 2(12+ 1). That is, 2n+ 1, NOT n+ 1.

    Why "a bit stuck"? Where did you get stuck? Since multiplication is commutative, yes, if xy= 1 then zxy= (xz)y= z so y= z.

    An "inverse element" of x is another element, y, such that xy= yx= 1. For any "other inverse element", z, xz= 1.

    I'm a bit puzzled why, on recognizing that you did not know the definitions well ("not entirely sure what this is"), you did not immediately look up and review the definitions.
     
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