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For university, it was my excercise to proof the associativity of the monoid $$H=(\mathbb{Q},\circ)\text{ with } a \circ b := a+b-ab\quad(a,b \in \mathbb{Q})$$

The excercise instructor gave us the following solution: $$ a \circ(b \circ c) = (a \circ b) \circ c $$ Hence $$ (a+b-ab) \circ c = a+c-ac \quad+\quad b+c-bc \quad +\quad-ab-c+abc$$ He also did it for $a\circ(b+c-bc)$ and pointed out that both equations have the same summands. That was the proof for the assoiciativity and I do understand this so far.

What I do not understand is how he determined the third part with $ -ab-c+abc$.

If I rewrite the operation to $\alpha \circ \beta = \alpha + \beta - \alpha\beta$ for not mixing up the variables later, I get:

$\alpha = -ab \\ \beta = c$

And hence in my mind it is not $-ab-c+abc$ but $-ab+c +abc$. But when I keep on calculating with $+c$ instead of $-c$, my proof fails. So why has it to be $-c$?

Michael Hardy
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Drudge
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  • It seems you have some typos on line 6. Since $(a \circ b) \circ c = (a+b-ab)\circ c = a+b-ab + c - (a+b-ab)c$. – user35603 Mar 21 '14 at 17:04
  • Well, I checked it again and in line 6 is character by character the solution he gave us. Is line 6 wrong? I thought this was correct since I could follow his argumentation (without why I had to use $-c$ instead of $+c$) – Drudge Mar 21 '14 at 17:09
  • As i understood the definition of $\circ$, in order to find $(a+b-ab) \circ c$ we need: the first component ($a+b-ab$) plus the second ($c$) and minus the product of the first ($a+b-ab$) and the second ($c$). – user35603 Mar 21 '14 at 17:13
  • Well I think you are right. This makes a lot more sense. I wonder what my excercise instructor did. I simply think it's completely wrong? – Drudge Mar 21 '14 at 17:16
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    I hope that there was some noise introduced between what your instructor wrote and what is written above. A first and major error is "$a\circ (b\circ c)=(a\circ b)\circ c)$. Hence $\dots$". We want to prove that for all $a,b,c$ we have $a\circ (b\circ c)=(a\circ b)\circ c)$. So we cannot start by asserting that the equality holds, and derive conclusions from that. For the proof, just calculate, substituting correctly. – André Nicolas Mar 21 '14 at 17:21
  • Note that if f(x) = 1-x, then your x * y is nothing but f ^{-1}(f(x)f(y)), where f(x)f(y) is usual multiplication. – Eric Mar 21 '14 at 17:38
  • See also this question. – Bill Dubuque Mar 21 '14 at 18:04
  • @AndréNicolas Sorry, this was my fault. He stated that we have to proof $ a \circ (b \circ c) = (a \circ b) \circ c$ for associativity. He didn't mentioned it as a fact. – Drudge Mar 21 '14 at 21:30

2 Answers2

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For every $a,b,c\in \mathbb{Q}$ we have \begin{eqnarray} a\circ(b\circ c)&=&a\circ\underbrace{(b+c-bc)}_{d}=a+d-ad=a+(b+c-bc)-a(b+c-bc)\\ &=&a+b+c-bc-ab-ac+abc=a+b-ab+c-ac-bc+abc\\ &=&\underbrace{(a+b-ab)}_{e}+c-\underbrace{(a+b-ab)}_{e}c=e+c-ec=e\circ c\\ &=&(a\circ b)\circ c. \end{eqnarray}

HorizonsMaths
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  • It's a great idea to substitute it this way. I wonder why we didn't learned it that way. That's a solution I understand, thank you for that! – Drudge Mar 21 '14 at 17:45
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It may be easier if you look at the bijective map $$ f\colon \mathbb{Q}\to \mathbb{Q},\quad f(a)=1-a. $$ You can notice that $$f(a\circ b)=1-a\circ b=1-a-b+ab=(1-a)(1-b)=f(a)\cdot f(b).$$ Let's call “property H” the equality $$ f(a\circ b)=f(a)\cdot f(b) $$ that holds for all $a,b\in \mathbb{Q}$.

Thus $f$ is an isomorphism between $(\mathbb{Q},\circ)$ and $(\mathbb{Q},\cdot)$. Since one operation is associative, also the other one is.

To see why, consider \begin{align} f((a\circ b)\circ c)&=f(a\circ b))\cdot f(c) &&\text{by property H} \\[2ex] &=(f(a)\cdot f(b))\cdot f(c) &&\text{by property H} \\[2ex] &=f(a)\cdot (f(b)\cdot f(c)) &&\text{associativity of multiplication} \\[2ex] &=f(a)\cdot f(b\circ c) &&\text{by property H} \\[2ex] &=f(a\circ(b\circ c)) &&\text{by property H} \end{align} Since $f$ is injective, we conclude that $(a\circ b)\circ c=a\circ(b\circ c)$.

This is a general fact; if $X$ and $Y$ are endowed with two operations, say $(X,\circ)$ and $(Y,*)$ and $f\colon X\to Y$ is a bijective map with the property that $$ f(a\circ b)=f(a)*f(b) $$ for all $a,b\in X$, then one operation is associative if and only if the other one is and the proof is just the same as before. We are basically transferring one operation to the other set, so any “algebraic property” holding for one operation holds true also for the other one. This is what's called an isomorphism between algebraic structures.

egreg
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