Algebraic Models For Accounting Systems by Salvador Cruz Rambaud, José García Pérez, Robert A Nehmer,

By Salvador Cruz Rambaud, José García Pérez, Robert A Nehmer, Derek J S Robinson

This booklet describes the development of algebraic versions which signify the operations of the double access accounting process. It provides a singular, complete, facts established remedy of the subject, utilizing such suggestions from summary algebra as automata, digraphs, monoids and quotient constructions.

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For any a, b ∈ R, exactly one of the statements a < b, a = b, b < a holds. There is another important and easily deduced consequence of (9) and (10): 12. if ab = 0 with a, b ∈ R, then a = 0 or b = 0. A commutative ring with identity which satisfies (11) is called an integral domain, or simply a domain. Thus the ordered commutative rings with identity are exactly the ordered domains. 2. The State of an Accounting System 33 the sets of integers, rational numbers, real numbers respectively, where the standard arithmetic operations of addition and multiplication, and the usual meaning of “positive”, are used.

For example, when n = 4,     0 −1  1     and e(3, 1) =  0  . e(2, 3) =   −1   1  0 0 The e(i, j) are called elementary balance vectors: evidently every simple balance vector is a scalar multiple of an elementary balance vector. The number of elementary balance vectors in Rn is equal to n(n − 1), for this is the number of ways of choosing two objects from a set of n in a definite order. In the following section it will be seen that the elementary balance vectors play a special role in the module Baln (R).

This is done in the next result. 1). Let R be an ordered domain and let n > 1 be an integer. Then the elementary balance vectors e(1, 2), e(2, 3), . . , e(n − 1, n) constitute an R-basis of Baln (R). Thus Baln (R) is a free R-module of rank n − 1. Proof In the first place e(1, 2), . . , e(n − 1, n) are linearly independent. For if r1 , . . , rn−1 ∈ R, then  r1  r2 − r 1     r3 − r 2  , r1 e(1, 2) + r2 e(2, 3) + · · · + rn−1 e(n − 1, n) =  ..   .   r  − r n−1 n−2 −rn−1  and the only way this can equal 0 is if r1 = r2 = · · · = rn−1 = 0.

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