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




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?



