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 $x+0=x$.
On
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.
On
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 $\forall x \in \Bbb R$
On
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).
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 $\overline{\mathbb{Q}} \cap \mathbb{R}$.