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I am given a "multiplication" acting on real numbers such that \begin{equation} x * y = 2xy - 6x - 6y + 21.\end{equation}

So far I have proven that this is commutative, associative, closed on the set of real numbers, the identity element with respect to it is $\frac{7}{2}$.

I have to construct an "addition" operation so that the multiplication will be distributive with respect to the addition, ie. \begin{equation}x * (y \oplus z) = x * y \oplus x*z.\end{equation}

I've tried $x \oplus y = \frac{x+y}{2}$, but it is not associative. Please help me find an "addition" that is associative.

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    $\begingroup$ Hint: $x * y = g^{-1}(g(x)g(y))$ where $g(x) = 2x-6$. $\endgroup$ – achille hui Jan 5 '15 at 19:21
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    $\begingroup$ Without further restrictions on $\oplus$, something like $a \oplus b = 3$ for all $a,b$ should work. $\endgroup$ – user133281 Jan 5 '15 at 19:25
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    $\begingroup$ $x\oplus y=x+y-3$ also works. $\endgroup$ – ryagami Jan 5 '15 at 19:27
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    $\begingroup$ @user13281 by the title it seems it has to be the operation for an abelian group since we're trying to make a field. $\endgroup$ – Matt Samuel Jan 5 '15 at 19:27
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    $\begingroup$ think I would start by finding a zero element, some $a$ such that we always get $a \ast x = a.$ $\endgroup$ – Will Jagy Jan 5 '15 at 19:28
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Instead of $*$, let us use $\otimes$ to denote the "multiplication" at hand. Let $+, - $ and $\times$ be the ordinary addition, subtraction and multiplication of $\mathbb{R}$.

Let $g(x) = 2 \times x - 6$, we have

$$x \otimes y = g^{-1}( g(x) \times g(y) ) = 2 \times x \times y - 6 \times x - 6 \times y + 21$$

If we define
$$x \oplus y = g^{-1}( g(x) + g(y) ) = x + y -3 $$
We will find $\otimes$ and $\oplus$ satisfy following distributive law:

$$\begin{align} x \otimes ( y \oplus z ) &= g^{-1}( g(x) \times g(g^{-1}( g(y) + g(z) )) )\\ &= g^{-1}( g(x) \times ( g(y) + g(z) ) )\\ &= g^{-1}( (g(x) \times g(y) ) + (g(x) \times g(z) ) )\\ &= g^{-1}( g( x \otimes y) + g( x \otimes z) )\\ &= ( x \otimes y) \oplus ( x \otimes z ) \end{align} $$ The main point is the function $g$ provides an isomorphism between the algebra $( R, +, \times )$ and $( R, \oplus, \otimes )$. Other properties of the algebraic structure $( R, \oplus, \otimes )$ can be derived from those of $(R, +, \times )$ by repeat application of $g$ and unwinding with $g^{-1}$. The distributive law above is an example of this general procedure. Another example is the associativity of our "addition" $\oplus$:

$$\begin{align} x \oplus ( y \oplus z ) &= g^{-1}(g(x) + g(g^{-1}(g(y) + g(z)))\\ &= g^{-1}(g(x) + (g(y) + g(z)))\\ &= g^{-1}((g(x) + g(y)) + g(z))\\ &= g^{-1}(g( x \oplus y ) + g(z) )\\ &= (x \oplus y ) \oplus z \end{align}$$

Update

About the question how to discover $g(x)$. Given a complicated expression, one strategy to simplify it is repeatly get rid of pieces that look most ugly. In an expression like $$z \stackrel{def}{=} x \otimes y = 2xy - 6x - 6y + 21$$ the most ugly pieces are the $-6x$ and $-6y$, so I start by getting rid of them first. $$\begin{array}{rrl} & z &= 2xy - 6x - 6y + 21\\ \iff & z &= 2(x-3)(y-3)+3\\ \end{array}\tag{*1}$$ Since I suspect the underlying algebra will be ordinary addition/multiplication, I try to transform the expression to the form $g(z) = g(x)\times g(y)$. With this as guideline, what need to do is sort of obvious. $$\begin{array}{rrl} (*1) \iff & z - 3 &= 2(x-3)(y-3)\\ \iff & 2(z-3) &= (2(x-3))(2(y-3))\\ \iff & 2z - 6 &= (2x - 6)(2y - 6) \end{array}$$ and the last equality leads to the choice $g(x) = 2x - 6$.

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  • $\begingroup$ up to you, i think this is a little deep for this OP; it all can be done at a very concrete level. $\endgroup$ – Will Jagy Jan 5 '15 at 19:56
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    $\begingroup$ A short observation that saves us from checking umpteen axioms is worth leaving the concrete level. $\endgroup$ – Hagen von Eitzen Jan 5 '15 at 20:03
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    $\begingroup$ @HagenvonEitzen, a hand in the bush is worth two birds. $\endgroup$ – Will Jagy Jan 5 '15 at 20:07
  • $\begingroup$ Yeah, sure. I think the specific manipulations of the bijection are what would teach this OP something about bijections, rather than the reverse. Meanwhile, there is just one letter 'g' in my last name, maybe your software is doubling it... Of course, I was hoping the OP would go through the concrete calculations himself. I also hope for world peace. $\endgroup$ – Will Jagy Jan 5 '15 at 20:15
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    $\begingroup$ @user3237992 I updated the answer to describe some meta-principle used to find $g(x)$. $\endgroup$ – achille hui Jan 5 '15 at 22:10

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