I understand how to prove the associate law as long as a*b has no division operator in it
I flummoxed when trying to follow division. I can’t figure it out.
I will give you 2 examples and you have go through the steps to show it is or not associative
- a*b=(a+b)/(ab-1)
- ab = ab/(a+b)
Ok I try half of 2.
(a*b)*c=(((a*b)/(a+b))*c)/((ab/(a+b))+c))
=(abc/(ac+bc))/(ab+ca+bc)/(a+b)
= abc(a+b)/((ac+bc)(ab+ca+bc))
I am basing my work of Pinters book A book of Abstract algebra Chapter 3 Solution A3
So I explain what I do . Let a = ab/(a+c).c=b In the numerator it’s says to multiply a*b. In the denominator I add a+b and do that and the rest is simplification.
The first is not associative, take $c=1$, then \begin{align*} (a\ast b)\ast 1&=\frac{a\ast b +1}{a\ast b-1}\nonumber\\ &=\frac{\frac{a+b}{ab-1}+1}{\frac{a+b}{ab-1}-1}\nonumber\\ &=\frac{a+b+ab-1}{a+b-ab+1}\nonumber \end{align*} and \begin{align*} a\ast(b\ast 1)&=\frac{a+(b\ast 1)}{a(b\ast 1)-1}\nonumber\\ &=\frac{a+\frac{b+1}{b-1}}{a\frac{b+1}{b-1}-1}\nonumber\\ &=\frac{ab-a+b+1}{ab+a-b+1}\nonumber\\ \end{align*} The second operation is transitive, indeed \begin{align*} (a\ast b)\ast c&=\frac{(a\ast b)c}{a\ast b +c}\nonumber\\ &=\frac{\frac{ab}{a+b}c}{\frac{ab}{a+b}+c}\nonumber\\ &=\frac{\frac{abc}{a+b}}{\frac{ab+ac+bc}{a+b}}\nonumber\\ &=\frac{abc}{ab+ac+bc}\nonumber \end{align*} and \begin{align*} a\ast(b\ast c)&=\frac{a(b\ast c)}{a+(b\ast c)}\nonumber\\ &=\frac{a(\frac{bc}{b+c})}{a+\frac{bc}{b+c}}\nonumber\\ &=\frac{abc}{ab+ac+bc}\nonumber \end{align*}