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So in $R$ I have $$aTb=a+b-ab$$ I have to study the structure $R'=R-2$ which is part of $R$. So I have $a ≠ 2,b≠2$ so $a+b-ab≠2$ let's suppose for a moment that $a+b-ab=2$, $a+b-ab-2=0$, now how do I solve this and find $a$ and $b$

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Spend more time formatting your question. It is really problematic to read. What is R? If R were the real numbers, then $a=3$ and $b=1/2$ would make $a+b-ab=2$, so the conclusion that it cannot be 2 doesn't make sense at the moment. Please work on the formatting. – rschwieb Jan 11 '13 at 19:53
I love you :) :) :) This was all I needed. – rere Jan 11 '13 at 19:56
Hrm, I think you ran with what I said rather quickly. The solution I gave isn't the only solution... I was just trying to give you an example that your fourth line does not follow from your third line. – rschwieb Jan 11 '13 at 20:39
Please use descriptive titles. – leonbloy Jan 11 '13 at 22:56

Let $(R,+,\cdot)$ be any (associative, commutative, unital) ring.

Define the translation $\iota : R\to R : x\mapsto x+1$. Then we may define a 'translated' ring structure by defining a new addition $(R,\oplus,\odot)$ where $$x\oplus y := \iota^{-1}(\iota x + \iota y) = (x+1)+(y+1) - 1 = x+y+1$$ and $$x\odot y := \iota^{-1} (\iota x\cdot \iota y) = (x+1)(y+1)-1 = xy+x+y.$$

Then clearly this new structure is a ring and in particular the map $\iota : (R,+,\cdot) \to (R,\oplus,\odot)$ is a morphism of rings. This tells us that $(R - \{2\},\odot)$ has the same structure as $(R-\{1\},\cdot)$. (In particular it is not closed, unless there are no units (except 1) in $(R,+,\cdot)$.)

So if you wish to solve the equation $a\odot x=b$, applying $\iota$ yields $\iota(a)\cdot\iota(x) = \iota(b)$.

Hopefully these comments are of some help to your question, that I find hard to understand anyway.

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