The Road to Reality
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Exercise [13.04]
http://www.roadtoreality.info/viewtopic.php?f=19&t=176
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Author:  Sameed Zahoor [ 12 Jul 2008, 10:38 ]
Post subject:  Exercise [13.04]

We are given,
1.i^{4}=1

2.C^{2}=1

3.Ci=i^{3}C

a)Ci=-iC
We have,
Ci=i^{3}C=-iC(using 3)

b)C(-1)=(-1)C
Consider the identity
i=i
Now,
(Ci)i=(i^{3}C)i(using 3)
orC(ii)=i^{3}(Ci)
orC(-1)=(-i)(i^{3}C)(using 3 again)
orC(-1)=(-1)C

c)C(-i)=iC
We have,
Ci=i^{3}C
Ci(i^{2})=i^{3}C(i^{2})
C(-i)=i^{3}(Ci)i=i^{3}(i^{3}C)i(using 3)
C(-i)=i^{6}Ci=i^{6}i^{3}C(using 3)
C(-i)=i^{9}C=iC

d)CC=1
It is merely a restatement of 2.(C^{2}=1)

Author:  deant [ 18 Jul 2010, 16:11 ]
Post subject:  Re: Exercise [13.04]

Given only 1, 2, and 3, there is no mention of any "-1" element.
There are only 1, i, C, and various multiplications of these.

Now rule (1) allows you to cancel any 4 i's multiplied in a row;
Rule (2) allows you to cancel any 2 C's multiplied in a row;
and rule (3) allows you to take any 'string' of i's and C's, and move all the C's to the right, e.g.

CiiCi

= CiiiiiC (3)
= CiC (1)
= iiiCC (3)
= iii (2)

So, each element of the group can be represented as (0-3 i's)(0 or 1 C).
The 8 distinct elements are thus: 1, i, ii, iii, C, iC, iiC, and iiiC.

..of course there's no reason you can't then denote i^2 as '-1', but it makes no difference to the structure of the group.

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