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Unformatted text preview: 89 2.1. FIRST RESULTS but by the associative law, the ﬁrst two and the last two are equal. Thus
there are at most three different product of four elements:
a.bcd /; .ab/.cd /; .abc/d: Using the associative law, we see that all three are equal:
a.bcd / D a.b.cd // D .ab/.cd / D ..ab/c/d D .abc/d:
Thus there is a well-deﬁned product of four elements, which is independent
of the way the elements are grouped for multiplication.
There are 14 ways to group ﬁve elements for multiplication; we won’t
bother to list them. Because there is a well-deﬁned product of four or less
elements, independent of the way the elements are grouped for multiplication, there are at most four distinct products of ﬁve elements:
a.bcde/; .ab/.cde/ .abc/.de/; .abcd /e: Using the associative law, we can show that all four products are equal,
a.bcde/ D a.b.cde// D .ab/.cde/;
etc. Thus the product of ﬁve elements at a time is well-deﬁned, and independent of the way that the elements are grouped for multiplication.
Continuing in this way, we obtain the following general associative
Proposition 2.1.19. (General associative law) Let M be a set with an
associative operation, M M ! M , denoted by juxtaposition. For
every n 1, there is a unique product M n ! M ,
.a1 ; a2 ; : : : ; an / 7! a1 a2 an ; such that
(a) The product of one element is that element .a/ D a.
(b) The product of two elements agrees with the given operation
.ab/ D ab .
(c) For all n 2, for all a1 ; : : : an 2 M , and for all 1 Ä k Ä n 1,
a1 a2 an D .a1 ak /.ak C1 an /: Proof. For n Ä 2 the product is uniquely deﬁned by (a) and (b). For n D 3
a unique product with property (c) exists by the associative law. Now let
n > 3 and suppose that for 1 Ä r < n, a unique product of r elements
exists satisfying properties (a)-(c). Fix elements a1 ; : : : an 2 M . By the
induction hypothesis, the n 1 products
pk D .a1 ak /.ak C1 an /; ...
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This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.
- Fall '08