Re: [代數] subring

[nicewine1 (不知該不該考研究所)]

※ 引述《wsx02 ()》之銘言：
: for the commutative ring R=(Z,⊕,⊙)
: a⊕b = a+b-1
: a⊙b = a+b-ab
: 要證(Z_odd,⊕,⊙)為(Z,⊕,⊙)的subring
: a in Z_odd, 請問為什麼 -a = 2-a ?
: 我是知道zero=1, unity=0
: 謝謝
  ~~~~~~~~~~~~~~~~~~
  If a is odd and b is odd ,too

  then  a(+)b = a+b-1 =  odd + odd -1 = even -1 = odd
        a(*)b = a+b -a*b = odd + odd - odd*odd = even - odd = odd

        so  (Z-odd,+,*) is the subring of (Z,+,*)

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