Are there in pure mathematics ensembles of number's which not divided by them self except $0$? In pure mathematics we know well that each number divided by him self except $0$ , the question that let me confused is: Is there a proof  in pure mathematics show to us that there are others ensembles of number's that's don't divide them self .
-in other way, how we could proof that only the number : "0" that's doesn't divide him self ?  
I would be interest for any replies or any comment's .Thank you 
 A: Well, what is a 'number'?
Though among complex numbers indeed only $0$ is the only one that cannot be a divisor, if instead of an imaginary number $i$ whose square is $-1$, we extend $\Bbb R$ the same way by an 'imaginary number' $E$ whose square is $1$, then $(1+E)(1-E)=0$,  hence if we had $(1-E)/(1-E)$, then 
$$1+E \ =\ \frac{(1+E)\cdot(1-E)}{1-E}\ = 0\,.$$
This shows that you cannot find a very general proof.
Anyway, the problem is rather with dividing by $0$ than only $0/0$.
A set equipped with operations (satisfying similar rules as) $+\ -\ \cdot$ is called a ring, and a nonzero element $x$ of a ring is called zero divisor if $xy=0$ for some nonzero $y$.  
If $x$ is a zero divisor in a ring (with $xy=0$), then we cannot have $x/x$, as $y\ne 0$ but
$$y=\frac{x\cdot y}x=\frac0x=0\,.$$
A: By definition, $a/b$ is the unique solution of the equation $x\cdot b=a$. That is, if there is no solution or if the solution is not unique, we do not have a quotient. That being said, how can anything not be divisible by itself? How can $x\cdot a=a$ not have $1$ as the unique solution? 
Firstly, $1$ could fail to be a solution if there is no $1$ in our number system; for example if we for some reason should decide to work only with even integers then divisions like $24/6=4$ are fine, but neither $24/8$ nor $6/6$ are defined because there is no suitable even integer that could work as quotient. Well, this might suggest tha we should revise our weird decision to ignore odd numbers ...
More importantly, the solution could fail to be unique: $x\cdot a=a$ is equivalent to $(x-1)\cdot a=0$. In rationl, real, complex numbers, a product is zero if and only if at least one of the factors is zero. Thus if $a\ne0$, then necessarily $x-1=0$, i.e. the solution $x=1$ is unique. But if $a=0$, then $x-1$ (and $x$) an be arbitrary.
There are other "number systems" (or better: unitary rings) where such problems may occur even if $a\ne 0$. They have so-callled divisors of zero. As an example take the set of $2\times 2$ matrices with real number entries. Under addition and matrix multiplication they form a ring with the identity matrix $E$ as "one". Whenever $A$ has determinant $0$, there are several $C$ with $CA=0$, and each such $C$ gives rise to a solution $X=C+E$ of $XA=A$ - no unique quotient $A/A$ can be defined.
A: No. All numbers, real or complex, are divisible by themselves... the sole exception being zero.
Someone might suggest that infinity is another, but infinity is not considered a number... at least not a real number.
There are of course types of mathematics that dont involve numbers at all, and operations on them can be defined rather trickily. Vectors, matrices, etc., and other non-numeric entities. Even some treatments of infinity.  The resolution to your question might rest with these higher, more abstract notions.
