T tegra97 New member Joined Sep 2, 2006 Messages 38 Apr 30, 2007 #1 Hi I have a quick question. If I=<2+2i> how would you determine how many elements are in Z/I? Thanks
D daon Senior Member Joined Jan 27, 2006 Messages 1,284 Apr 30, 2007 #2 tegra97 said: Hi I have a quick question. If I=<2+2i> how would you determine how many elements are in Z/I? Thanks Click to expand... Z/I = (a+bi) + <2+2i> Also note that (2+2i) + <2+2i> = 0 + <2+2i>, so 2+2i=0. Or... 2=-2i, 1=-i. So in this particular factor ring, -1 = i. Then squaring both sides we see 1 = -1, i.e. that 2=0. Therefore when we have something like 9+8i + <2+2i> we actually get 9 + <2+2i> = (8+1)+<2+2i> = 1+<2+2i>. Counting the number of distinct ones shouldn't be hard now.
tegra97 said: Hi I have a quick question. If I=<2+2i> how would you determine how many elements are in Z/I? Thanks Click to expand... Z/I = (a+bi) + <2+2i> Also note that (2+2i) + <2+2i> = 0 + <2+2i>, so 2+2i=0. Or... 2=-2i, 1=-i. So in this particular factor ring, -1 = i. Then squaring both sides we see 1 = -1, i.e. that 2=0. Therefore when we have something like 9+8i + <2+2i> we actually get 9 + <2+2i> = (8+1)+<2+2i> = 1+<2+2i>. Counting the number of distinct ones shouldn't be hard now.
