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I am trying to prove that axioms A1, A2, A3, A4, M2, D hold.

so A1 :

$$ a+b=b+a$$ $$x+x+x+x=x+x+x+x$$ $$4x=4x$$ Am I on the right way?

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    I don't think you are. I understand that your Axiom 1 says that if you take two elements $a$ and $b$ then $a+b=b+a.$ You have only one element in this set, that is $x$. So you have to take $a=x$ and $b=x$. We have $a+b=x+x=x=x+x=b+a.$2012-03-09
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    What does $x+x+x+x=x+x+x+x$ have to do with anything?2012-03-09
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    @Chris: Best guess: the OP is reckoning that the only element is $x+x$ and substituting that for both $a$ and $b$ in the statement of commutativity.2012-03-09
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    Name your element $x$ by $0$ and everything will make more sense.2012-03-09
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    You need to tell us what you mean by A1, etc. These are labels in your texbook, not universally accepted labels.2012-03-09

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No, you are not. In order to show that $$a+b=b+a\tag{1}$$ for all elements $a,b\in S$, you merely have to observe that since $x$ is the only element of $S$, the only thing that you can substitute for $a$ and $b$ in $(1)$ is $x$. When you do that, you get $x+x=x+x$; is that true? Yes, because both sides are equal to $x$ by the definition of $+$.

I’ll do one other example as an illustration. I don’t know for sure, but I suspect that your D is $$a\cdot(b+c)=(a\cdot b)+(a\cdot c)\tag{2}$$ for all $a,b,c\in S$. Again, the only thing in $S$ is $x$, so the only possible instance of $(2)$ to be checked is $x\cdot(x+x)=(x\cdot x)+(x\cdot x)$; is it true? On the lefthand side we have $$x\cdot(x+x)=x\cdot x=x\;,$$ and on the righthand side we have $$(x\cdot x)+(x\cdot x)=x+x=x\;;$$ these are indeed equal, so $(2)$ is true for all possible choices of $a,b$, and $c$.

One last comment: $4x$ makes no sense here. We have not defined any operation that combines natural numbers and this object $x$.

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    $4x$ is always defined for a ring, $4x = x + x + x + x$ by definition.2012-03-09
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    @David: As a notational convention, yes. But given the overall confusion, I think it important to be clear about what is really going on, which is *not* the multiplication that it might naïvely be taken as.2012-03-09
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Hint $\ $ The definitions simply transport the ring structure from the subring $\:\{0\}\subset \mathbb Z\:$ to $\{x\}$.

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Most of the required properties of a ring are immediate for the trivial ring, since any expressions on the left- or right-hand side evaluate to $x$, the unique element.

That leaves the existence of a additive identity and inverse, met by $x+x = x$ and a multiplicative identity, met by $x \cdot x = x$.