Definition of $0$? In real number axioms, it is defined that there is $0$ such that $x+0=x$ for all $x.$ I was wondering an example is there any other algebraic structure than real numbers which satisfy the real number axioms other than $x+0=x$ but where $x\cdot 0=0\cdot x =0$ for all elements $x$.
This question popped on my mind when I was wondering why $0x=0$ is not an axiom. I know how to prove it from the other axioms but I'm not sure if $0x=0$ implies that $x+0=x$.
 A: You may be interested in the notion of a real closed field. Among the many properties is that all real closed fields are elementarily equivalent -- any statement you can make using only first-order logic and $+,-,\cdot,/,$ and $<$ is true in one real closed field if and only if it is true in all real closed fields.
In particular, $\mathbb{R}$ is a real closed field, and there are many other examples of such things as well, such as $\overline{\mathbb{Q}} \cap \mathbb{R}$.
A: If you view the real numbers as an ordered field, then $\Bbb Q$ and any intermediate field between $\Bbb Q$ and $\Bbb R$ as well. There are other ordered fields, much much larger than $\Bbb R$, too.
But you can also think of the real numbers as a field, without concern of the order, in which case every field has this property. That is to say, in every field $0+x=x=x+0$ and $0\cdot x=0\cdot x=0$. Examples for fields are $\Bbb C$, and even finite fields like $\Bbb Z/3\Bbb Z$.
You can also remove some of the properties and view the real numbers as a ring, or a commutative ring, or a domain, or many other stronger or weaker types of rings. And in rings it holds that $0$ has these properties as well. However in an arbitrary ring not all elements have a multiplicative inverse, and sometimes the multiplication is not commutative, and sometimes other bad things happen as well.
A: This does not really have anything to do with real numbers. This is a question about fields, or rather, rings.
A ring $(R, +, \cdot)$ is an abelian group $(R, +)$ with a multiplication $\cdot$ that is associative and compatible with the group structure, i.e. that it fulfills the distributive properties $x\cdot(y+z) = x\cdot y + x \cdot z$ and $(x+y)\cdot z = x\cdot z + y \cdot z$ for all $x, y, z \in R$. 
From this automatically follows $0 \cdot x = 0$ as 
$$0 \cdot x = (0 + 0)\cdot x = 0 \cdot x + 0 \cdot x \Longrightarrow 0 = 0 \cdot x$$
This is why you do not need $0 \cdot x = 0$ as an axiom. It follows from the other axioms. 
On the other hand, it does not make sense to not require $0 + x = x$, as you want that to have a well-formed addition.
As for the other axioms for the reals: You want $(R \setminus \{0\}, \cdot)$ to be an abelian group as well (i.e. to be a field), and you require that it is 1) complete and 2) archimedically ordered. Then you can show that there is only one such set (up to isomorphy).
A: How are you going to establish an axion that  $0\cdot x=0$ where $x=\infty$? $0\cdot \infty$ is indeterminate.
Hence $0\cdot x=0\implies0+x=0$ isn't true $\forall x \in \Bbb R$
