R3 – Division – Simultaneous Equations

Why Simultaneous Equations are Needed

According to the theorem of multiplicative inverses, R3 lacks multiplicative inverses. Therefore, the Wave Number system employs an alternative method to compute the result of division in R3, using simultaneous equations.

Consider the following equation:

$(3^v + 4i\hat{ } +5j^v )(2\hat{ } + 3i\hat{ } + 4j\hat{ }) = (7\hat{ } + 10i^v + 21j\hat{ })$
$\implies (7\hat{ } + 10i^v + 21j\hat{ })/(3^v + 4i\hat{ } +5j^v) = (2\hat{ } + 3i^v + 4j\hat{ })$

To obtain the result of the division, you need to solve the following equation:

$(7\hat{ } + 10i^v + 21j\hat{ })/(3^v + 4i\hat{ } +5j^v) = (x? + y? + z?)$

The following method involves using simultaneous equations to determine the result of dividing R3.

Method

Step 1: Identifying the R3 Simultaneous Equations Needed for Division

Rewrite the equation above as:

$(3^v + 4i\hat{ } +5j^v)*(x + y + z) = (7\hat{ } +10i^v + 21j\hat{ })$

The expression $(3^v + 4i\hat{ } +5j^v)$ serves as the Operator of the multiplication and is represented by $(x_a, y_a, z_a)$ . Each element corresponds to an axis of one of the three rotations of the multiplication. The Operand of the division is $(7\hat{ } +10i^v + 21j\hat{ })$ and is represented by $(x_0, y_0, z_0)$ .

The formulae for calculating multiplication are:

$\hat{ }/^v$ value: $(x_a*x_o + y_a*z_o + z_a*y_o)$
$i\hat{ }/i^v$ value: $(y_a*y_o + x_a*z_o + z_a*x_o)$
$j\hat{ }/j^v$ value: $(z_a*z_o + x_a*y_o + y_a*x_o)$

Accordingly, this gives three equations:

A: $3^v*x + 4i\hat{ }*z + 5j^v*y = 7\hat{ }$
B: $3^v*z + 4i\hat{ }*y + 5j^v*x = 10i^v$
C: $3^v*y + 4i\hat{ }*x + 5j^v*z = 21j\hat{ }$

Step 2: Transforming into Simultaneous Equations

Equation A

Transform Equation A into a simultaneous equation as follows:

$3^v*x + 4i\hat{ }*z + 5j^v*y = 7\hat{ }$
$\implies 3x + 4z + 5y = 7$
$\implies x = (7 + ^-4z + ^-5y)/3$

The multiplication table indicates that the product of Opposite Values of a $^v$ with a $\hat{ }$ results in a $\hat{ }/^v$ value that retains the sign of the operand. Thus, $3^v*x$ becomes $3x.$

Similarly $4i\hat{ }*z$ produces a $\hat{ }/^v$ value with the same sign as in $z$ . Thus, $4i$ becomes $4z$ .

Likewise, $5j^v*y$ results in a $\hat{ }/^v$ value with the same sign as in $y$ . Thus, $5j^v*y$ becomes $5y$ .
The Opposite Sign of the result is $\hat{ }$ and so the result becomes $7$ .

The Opposite Types and Signs have been eliminated, yielding equation A-modified:

$3x + 4z + 5y = 7$
$x = (7\hat{ } + ^-4z + ^-5y)/3$

Equation B

Transform Equation B into a simultaneous equation as follows:

$3^v*z + 4i\hat{ }*y + 5j^v*x = 10i^v$
$\implies 3z + 4y + ^-5x = ^-10$
$\implies z = (^-10 + ^-4y + 5x)/3$

The product of a $^v$ and a $j^v$ or $j\hat{ }$ result in an $i\hat{ }/i^v$ that retains the sign of the operand. So, $3^v*z$ becomes $3z$ .

Similarly $4i\hat{ }*y$ becomes $4y$ with the same sign as in $y$ .

However, the product of a $j^v$ and an $x$ results in a sign reversal, thus $5j^v*x$ becomes $^-5x$ .

The Opposite Sign of the result is $^v$ and so the result becomes $^-10$ .

Equation C

Finally, equation C generates the following simultaneous equation C-modified:

$^-3y + ^-4x + 5z = 21$
$\implies y = (^-21 + ^-4x + 5z)/3$

Step3: Solving the R3 Simultaneous Equations for Division

Substituting for $y = (^-21 + ^-4x + 5z)/3$ from Equation C-modified into Equation A-modified above gives:

$3x + 4z + (5/3)(^-21 + ^-4x + 5z) = 7$
$\implies 9x + 12z + ^-105 + ^-20x + 25z = 21$
$\implies ^-11x + 37z = 21 +105 = 126$ giving Equation D

Secondly, a similar substitution for $y$ in Equation B-modified above gives:

$3z + (4/3)(^-21 + ^-4x + 5z) + ^-5x = ^-10$
$\implies 9z + ^-84 + ^-16x + 20z + ^-15x = ^-30$
$\implies 29z + ^-31x = 54 => z = (54 + 31x)/29$ giving Equation E

Thirdly, substituting for $z = (54 + 31x)/29$ from Equation E into Equation D gives:

$^-11x + 37(54+ 31x)/29 = 126$
$\implies ^-319x + 1998 +1147x = 3654$
$\implies 828x = 1656 => x = 2$

Next, substituting for $x$ with $2$ in Equation D gives:

$^-11*2 + 37z = 126$
$\implies 37z = 126 + 22$
$\implies z = 148/37 = 4$

Finally, substituting for $x$ with $2$ and $z$ with $4$ in Equation A-modifed gives:

$3*2 + 4*4 + 5y = 7$
$\implies 5y = 7 + ^-22 = ^-15$
$\implies y = ^-3$

The variable $x$ represents the $\hat{ }/^v$ Counter and is $2$ and so gives a solution for $x$ of $2\hat{ }$ . The variable $y$ represents the $i\hat{ }/i^v$ Counter and is $^-3$ and so gives a solution for $y$ of $3i^v$ . The variable $z$ represents the $j\hat{ }/j^v$ Counter and is $4$ and so gives a solution for $z$ of $4j\hat{ }$ . So, the overall solution is $(2\hat{ } + 3i^v + 4j\hat{ })$ .