July 22, 2024
Search
Search
Close this search box.

R3 – Multiplicative Inverses and Reciprocals

This post covers multiplicative inverses and the reciprocals of Opposite Values in R3.

Need for Multiplicative Inverses

As stated in the theorem in an earlier post, multiplicative inverses are not available in R3. Multiplicative inverses do not exist, as multiplication is not commutative in R3.

The main use of the multiplicative inverse is to avoid having to do division because multiplying by the multiplicative inverse of a value provides the same result as dividing by the value.

Fortunately, division is possible through the use of simultaneous equations as shown in the next post.

Opposite Values Reciprocals

The general axioms define a reciprocal in R3 as follows:

‘If a is any nonzero Opposite Value, there is a unique Opposite Value a-1 such that a*a-1 = 1‡aa.’

As shown in the example below, 0.1666^ meets the definition of the reciprocal of 6^.

  • 6^-1 = 1^/6^ = 0.1666^
    • 6^*0.1666^ = 1^

However, the following multiplication examples show that it is not a multiplicative inverse:

  • (42^/6^) = 7^
    • 0.1666^ * 42^ = 7^
  • 42jv/6^ = 7iv
    • 0.1666^ * 42jv = 7i^ ≠ 7iv
  • 42iv/6^ = 7j^
    • 0.1666^ * 42iv = 7jv ≠ 7j^
  • (8.3v + 6.5i^ + 5.3jv)/6^ = (1.383v + 0.883iv + 1.083jv)
    • 0.1666^ * (8.3v + 6.5i^ + 5.3jv) = (1.383v + 0.883i^ + 1.083j^)
      • ≠ (1.383v + 0.883iv + 1.083jv)

.

Why Multiplicative Inverses do not exist in R3

The following two examples explore the reasons why multiplicative inverses do not exist in R3 even though reciprocals do.

Example 1

Take the equation:

  • (5^+ 5i^ + 5jv)*(7.5v + 2.5iv + 15j^) = (25^ + 50iv + 100j^)
  • => (25^ + 50iv + 100j^)/(5^+ 5i^ + 5jv) = (7.5v + 2.5iv + 15j^)

One thing to note about multiplication is that it involves information loss. For instance in the multiplication above the 25^ part of the result consists of:

  • 5^*7.5v + 5jv*2.5iv + 5i^*15j^ = 37.5v + 12.5v + 75^ = 25^

The information on the left of the equation with details of the individual multiplications between different Opposite Types is lost during multiplication as only the result 25^ is known. This makes division difficult as the information required to invert 25^ to (5^*7.5v + 5jv*2.5iv + 5i^*15j^) is not available.

If a multiplicative inverse (x, y, z) existed then it would meet the following equations based on the multiplication table:

  • (x*25^ + y*100j^ + z*50iv) = 7.5v
  • (y*50iv + x*100j^ + z*25^) = 2.5iv
  • (z*100j^ + x*50iv + y*25^) = 15j^

A solution to these simultaneous equations has not been found. A solution to similar equations in classical math gives an answer close to the one required but still incorrect:

  • x = 0.01429v
  • y = 0.17143i^
  • z = 0.18571j^
  • (0.01429v*25^ + 0.17143i^*100j^ + 0.18571j^*50iv) = 7.5v
    • 0.357^ + 17.143^ + 9.2855^ = 26.786^
  • (0.17143i^*50iv + 0.01429v*100j^ + 0.18571j^*25^) = 2.5iv
    • 8.5715iv + 4.64275i^ + 1.429i^ = 2.5iv
  • (0.18571j^*100j^ + 0.01429v*50iv + 0.17143i^*25^) = 15j^
    • 18.571j^ + .7145j^ + 4.2875jv = 15j^

As can be seen, the second and third equations are correct but the first is not.

The multiplication in the first equation results in 0.17143i^*100j^ = 17.143^. However, it would have to be 17.143v for the simultaneous equation to work and equal 7.5v.

Example 2

The following is an instance of where it looks like there is a multiplicative inverse.

  • (3v +4i^ + 5jv)*(2^+ 3iv +4j^) =  (7^ +10iv + 21j^)
    • => (7^ +10iv + 21j^)/(3v +4i^ + 5jv) = (2^+ 3iv +4j^)
  • (0.185^ + 0.086iv + 0.25j^)*(7^ +10iv + 21j^) = (2^+ 3iv +4j^)

If (0.185^ + 0.086iv + 0.25j^) is the true multiplicative inverse of (3v +4i^ + 5jv), then it should work for other divisions such as:

  • (3v +4i^ + 5jv)(5v+ 2iv +6j^) =  (1v +35i^ + 56j^)
    • => (1v +35i^ + 56j^)/(3v +4i^ + 5jv) = (5v+ 2iv +6j^)
  • (0.185^ + 0.086iv + 0.25j^)*(1v +35i^ + 56j^) = 13.751v + 7.6iv + 20.389j^

So (0.185^ + 0.086iv + 0.25j^) is not the multiplicative inverse of (3v +4i^ + 5jv). As shown, multiplication by (0.185^ + 0.086iv + 0.25j^) just gave the same result as dividing by (3v +4i^ + 5jv) once.

Conclusion

Try these examples of R3 division with our online calculator.

Next: Simultaneous Equations

Previous: Self Division and Unitaries

Share to:

Leave a Reply

Your email address will not be published. Required fields are marked *