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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1908.0. "Why i^2 = -1 ?" by EVTSG8::ESANU (Au temps pour moi) Fri Nov 18 1994 10:47

Loosely quoting from [1] (a book in the *gem* category):
[ (My comments are inclosed in brackets.)                                 ]


The set RxR of all ordered pairs of real numbers is an Abelian group with
respect to the natural addition defined by

(1)	(x1,y1) + (x2,y2)  =  (x1 + y1, x2 = y2)

The multiplication in RxR is introduced by the definition

(2)	(x1,y1) * (x2,y2)  = (x1*x2 - y1*y2, x1*y2 + y1*x2)

[ If we put  e = (1,0)  and  i = (0,1) , from (2) we obtain               ]
[                                                                         ]
[(3)	i� = -e                                                           ]
[                                                                         ]
[ The multiplicative group  RxR \ {(0,0)}  has  e  as unit, and since  R  ]
[ is naturally imbedded in this group, we identify  e  from this group    ]
[ and  1  from  R , therefore                                             ]
[                                                                         ]
[(3')	i� = -1                                                           ]

Here is Hamilton's motivation for (2): first he finds it suggestive to
define products with real numbers by the rule  r * (x,y) = (r*x, r*y) 
(R-vector space structure). One then already has

	(x,y) = x*e + y*i   with  e = (1,0)  and  i = (0,1)

Now if e is to be the unit element and the distributive laws are to hold,
then one must have

(4)	(x1*e + y1*i) * (x2*e + y2*i) =

		= x1*x2*e + (x1*y2 + y1*x2)*i + y1*y2*i�

The multiplication law is therefore determined as soon as  i� , which must
be of the form  p*e + q*i , is known. There are however infinitely many
ways of choosing  p  and  q  so that the resulting multiplication has an
unique inverse. (The reader may care to find examples.) Hamilton therefore
postulates (as he does later in the case of his quaternions) the product
rule: the length of the product of two factors is equal to the product of
the lengths of the factors, where the length  |z|  of  z = (x,y)  is
defined as  +sqrt(x� +y�) .

[ Hence the condition posed by Hamilton is that  RxR  be a normed         ]
[ R-algebra.                                                              ]

It is then only necessary to apply this product rule to

	i� = p*e + q*i  and  (e + i) * (e - i) = e - i� = (1 - p)*e - q*i

to deduce that  p = -1,  q = 0  (since  |i�| = |i| * |i| = 1  and
|e + i| = |e - i|  = sqrt(2), so that (4) becomes the same as equation (2).


Bibliography:

[1] Ebbinghaus et al., Numbers, Springer-Verlag, GTM 123 (RIM),
New York, 1991
		Full authors list:
	H.-D. Ebbinghaus	H. Hermes
	F. Hirzebruch		M. Koecher
	K. Mainzer		J. Neukirch
	A. Prestel		R. Remmert
The chapter on the complex numbers was written by R. Remmert.
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