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Topic: Cube Roots of 1 and Multiplicative Identity
Replies: 7   Last Post: May 31, 2000 5:07 PM

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 Chuck Cadman Posts: 16 Registered: 12/12/04
Re: Cube Roots of 1 and Multiplicative Identity
Posted: May 21, 2000 11:40 PM

Pagadala wrote in message <8ga2sm\$26l\$1@nnrp1.deja.com>...
>Again, from the book "Abstarct Algebra" by Beachy/Blair, 2nd ed.
>Chapter 4, Polynomials.
>
><quote>
>v) Identity elements:
>The set F contains an additive identity element, denoted by 0, such
>that for all a in set F,
>a + 0 = a and 0 + a = a.
>The set F also contains a multiplicative identity element, denoted by 1
>(and assumed to be different from 0) such that for all a in set F,
>a*1 = a and 1*a = a.
><unquote>
>
>Lets consider this -
>
>x^3 = N
>x = N^(1/3)
>x = (1*N)^(1/3)
>x = 1^(1/3) * N^(1/3)
>x = 1^(1/3) * x
>
>According to the definition of multiplicative identity,
>if x = a*x, then "a" is the multiplicative identity, that is 1.
>And since mulitplicative identity is not defined to be equal to any
>other number except 1, it follows that -
>
>All cube roots of 1 are equal to 1.
>Let the complex cube roots be x + iy and x - iy, then
>x + iy = x - iy = 1
>iy = 0 and x = 1
>y = 0 or i = 0

I'm not sure how what you've done is related to the algebra book. Suppose
we were dealing with the field of complex numbers. Then, I hope you know,
the equation x=y*x where x is nonzero will certainly imply that y=1. So
have you just found some contradiction in the definition of the complex
numbers, or is there something wrong with your proof?

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