Algebraic Models for Accounting Systems

• Balance vectors
  • A balance vector is a column vector or column matrix the sum of whose entries equals zero.
  • Balance vectors are able to represent the state of an accounting system at any instant.
  • They are also capable of encoding the transactions that are applied to the system.
• Digraph
  • The vertices represent accounts and the edges indicate where there are flows of value within the system. Thus a digraph gives a picture of how value can flow around an accounting system.
• Automaton
  • The applicability to accounting is clear: the states of the accounting system are the balance vectors, the inputs are the transactions and the outputs are the new balance vectors.

Let there be given a set R together with two binary operations on R called addition and multiplication, denoted in the usual way, such that the following rules hold for all elements \(a, b, c \in R\):

\begin{align*} (a + b) + c &= a + (b + c) \tag{1}\\ a + b &= b + a \tag{2}\\ \exists 0 \in R, \forall a \in R, a + 0_r &= a \tag{3}\\ \forall a \in R, \exists (-a) \in R, a + (-a) &= 0 \tag{4}\\ a \times (b \times c) &= (a \times b) \times c \tag{5}\\ a \times b &= b \times a \tag{6}\\ a \times (b + c) &= a \times b + a \times c \tag{7}\\ \exists 1 \in R, \forall a \in R, a \times 1 &= a \tag{8} \end{align*}

The first four of these requirements assert that R is an algebraic structure called an . With the additional properties (5) through (8) R becomes a with identity.

A commutative ring with identity R is said to be linearly ordered if there is a non-empty subset P of R not containing 0, called the set of positive elements, such that the following conditions are satisfied:

\begin{align*} a,b \in P \implies a + b \in P &\wedge a \times b \in P \tag{9}\\ \forall a \in R, (a \in P) \lor (a = 0) &\lor (-a \in P) \tag{10}\\ \end{align*}

The actual concept of a linear order arises when one defines a < b to mean that b − a ∈ P. The negative elements of R are the elements of the set R \ (P ∪ {0}). On the basis of (9) and (10) it can be shown that the following holds:

\begin{align*} \forall a,b \in R, (a < b) \lor (a = b) &\lor (a > b) \tag{11}\\ a,b \in R \wedge a \times b = 0 \implies a = 0 &\lor b = 0 \tag{12}\\ \end{align*}

Let R be an ordered domain, which will be the universal set for all account values, and let n be a positive integer, which will be the number of accounts in the accounting system. The state of the system at any instant can be described by listing the values of the accounts, which are assumed to be in some agreed order, in the form of an n-column vector over R.

\begin{equation*} V = \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \\ \end{bmatrix} \end{equation*}

Thus v_i ∈ R is the value of the ith account. The set of all n-column vectors over R is denoted by Rⁿ.

Of particular importance is the zero vector:

\begin{equation*} 0 = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ \end{bmatrix} \end{equation*}

Addition can also be defined:

\begin{equation*} u + v = \begin{bmatrix} u_1 + v_1 \\ u_2 + v_2 \\ \vdots \\ u_n + v_n \\ \end{bmatrix} \end{equation*}

as well as scalar products:

\begin{equation*} r \times v = \begin{bmatrix} r \times v_1 \\ r \times v_2 \\ \vdots \\ r \times v_n \\ \end{bmatrix} \end{equation*}

Let \(u, v, w \in R^n\) and \(r, s \in R\):

\begin{align*} (u + v) + w &= u + (v + w) \tag{1}\\ u + v &= v + u \tag{2}\\ v + 0 &= v \tag{3}\\ v + (-v) &= 0 \tag{4}\\ r \times (u + v) &= r \times u + r \times v \tag{5}\\ (r + s) \times u &= r \times u + s \times u \tag{6}\\ (r \times s) \times v &= r \times (s \times v) \tag{7}\\ 1_R \times v &= v \tag{8} \end{align*}

These properties demonstrate that the set Rⁿ has a recognizable algebraic structure. Indeed properties (1) – (4) assert that Rⁿ is an , while the additional properties (5) through (8) make Rⁿ into a .