R3 – Multiplication – Simple

This post looks at some simple R3 multiplication of Opposite Values together and using counters. The examples show the distributivity of R3 multiplication. However, they also show that R3 multiplication is not associative or commutative.

R3 Unitary Multiplication Table

*	1^	1v	i^	iv	j^	jv
1^	1^	1v	j^	jv	iv	i^
1v	1^	1v	jv	j^	i^	iv
i^	jv	j^	i^	iv	1^	1v
iv	j^	jv	i^	iv	1v	1^
j^	i^	iv	1v	1^	j^	jv
jv	iv	i^	1^	1v	j^	jv

Commutativity

• 3i^v*4^{^} = 12j^{^}
• 4^{^}*3i^v = 12j^v   
  • So, multiplication is not commutative in R3      
• 4i^v*3j^v = 12^{^}
• 3j^v*4i^v = 12^v 
  • So, multiplication is not commutative in R3           
• 1^{^}*1^v = 1^v
• 1^v*1^{^} = 1^{^}
  • So, multiplication is not commutative in R3   
• j^{^}*j^{^} = j^{^}
• i^v*i^v = i^v

• (5i^v + 2j^{^})(6i^{^} + 3^v) = (5i^v*6i^{^} + 5i^v*3^v + 2j^{^}*6i^{^} + 2j^{^}*3^v)
  • =  30i^{^} + 15j^v + 12^v + 6i^v
  • = 12^v + 24i^{^} + 15j^v
• (6i^{^} + 3^v) (5i^v + 2j^{^}) = (6i^{^}*5i^v + 6i^{^}*2j^{^} + 3^v*5i^v + 3^v*2j^{^})  
  • =  30i^v  + 12^{^}  + 15j^{^}  + 6i^{^}
  • = 12^{^}+ 24i^v + 15j^{^}
    • Note that the signs of the results are different
    • So, multiplication is not commutative

Associativity

• (4i^v*3^v)*2i^{^}  = 12j^v*2i^{^} = 24^{^}
• 4i^v*(3^v*2i^{^})  = 4i^v*6j^v =  24^{^}               
  • It looks like multiplication might be associative
• (j^{^}*j^{^})*1^v  = j^{^}*1^v = i^v
• j^{^}*(j^{^}*1^v) = j^{^}*i^v = 1^{^}                          
  • But it is not, as shown in this case
• (2i^v*4j^{^})*3^{^} = 8^v*3^{^} = 24^{^}
• 2i^v*(4j^{^}*3^{^} ) = 2i^v*12i^{^} = 24i^{^}            
  • And not in this case
  • So, multiplication is not associative.

Distributivity

• 5j^{^}(6^{^} + 3i^v) = 5j^{^}(6^{^}) + 5j^{^}(3i^v)
  • = 30i^{^} + 15^{^}
• (5j^{^}+ 4i^v)(6^{^} + 3i^v) = 5j^{^}*6^{^} + 5j^{^}*3i^v + 4i^v*6^{^} + 4i^v*3i^v
  • = 30i^{^} + 15^{^} + 24j^{^} + 12i^v
  • = 15^{^} + 18i^{^} + 24j^{^}
• 5i^v(6^{^} + 3j^v)  = 5i^v*6^{^} + 5i^v*3j^v
  • = 30j^{^} + 15^{^}
• (2^v + 10^{^}) (9^v+ 44^{^}) = 8^v*35^{^} = 280^{^}  
  • but
    • 2^v*9^v + 2^v*44^{^} + 10^{^}*9^v + 10^{^}*44^{^}= 18^v + 88^{^} +90^v + 440^{^}
    • = 420^{^} which is different to the previous answer
    • So, multiplication is distributive as long as all Opposite Values of the same Opposite Types are added together before distribution.

R3 Multiplication with Counters

• 5(6^{^} + 3i^v + 4j^{^}) = 30^{^} + 15i^v + 20j^{^}

A flip sign in front of a Counter indicates that the Opposite Sign should be changed to the other Opposite Sign after multiplication.

• ^–5(6^{^} + 3i^v + 4j^{^}) = 30^v + 15i^{^} + 20j^v
• ^–2*^–2*2j^{^} = 8j^{^}                                                     
  • Note that two flips cancel each other out

Counters cannot be multiplied with only themselves, as the result of any calculation must be an Opposite Value. Therefore, 3*5*2j^{^} = 30j^{^} is valid, but 3*5 = 15 is not permitted as 15 is not an Opposite Value.

Other Examples

• (1^{^} + i^{^} + j^{^})*1^{^} = (1^{^} + j^v + i^{^} )
• (1^{^} + i^{^} + j^{^})(1^{^} + j^v + i^{^} )
  • = (1^{^} + j^{^} + i^{^} ) + (j^v +1^v + i^{^}) + (i^{^} + j^v + 1^v )
  • = (1^v + 3i^{^} + j^v )
• (6^{^} + ^–4^{^} )(2^v + 4i^v + 3j^v) = 2^{^}(2^v +4i^v + 3j^v)
  • = 4^v + 6i^{^}+ 8j^v
  • = 6^{^}*2^v + 6^{^}*4i^v  + 6^{^}*3j^v + 4^v*2^v + 4^v*4i^v  + 4^v*3j^v
    • = 12^v +24j^v + 18i^{^} +8^v + 16j^{^} + 12i^v
    • = 20^v + 6i^{^} + 8j^v
    • which is different to the previous answer
    • Again, multiplication is distributive as long all possible additions are performed before distribution

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