The numbers in notations like $|n\rangle$ are the analogues of indices in matrix notation. That is, $|0\rangle=e_0$, $|1\rangle=e_1$, etc., where $e_n$ is the vector which has a $1$ in the $n$th position and $0$ in the other entries. Unfortunately, this notation is unspecific about the dimension of the base space. For qubits in quantum computers, the dimension is $2$, so we only have $|0\rangle=e_0=(1,0)$ and $|1\rangle=e_1=(0,1)$. It is also common to have a countable infinity of basis vectors, so we get $|n\rangle$ for each $n\in\Bbb N$. In quantum mechanics one also deals with this notation for larger dimensional spaces; for example we may have $|x\rangle$ for each $x\in\Bbb R^3$ (the position basis), which is a vector space of uncountable dimension $|\Bbb R^3|=2^{\aleph_0}$.
In any case, these vectors are usually enumerating a basis of some kind, and the details beyond that depend on the context.
The notation $\langle 0|0\rangle$ is written in linear algebra notation as $e_0^Te_0$, which is a $1\times 1$ matrix whose value can be identified with the dot product $e_0\cdot e_0$. Provided that the vector is normalized, this will always be $1$. So a general answer is $\langle m|n\rangle=0$ if $m\ne n$, and $\langle n|n\rangle=1$, which expresses that the vectors $(|n\rangle)_{n\in\Bbb N}$ are an orthonormal basis for the space.
For some general rules, then, we have $|n\rangle=e_n$ and $\langle n|=e_n^T$ (or $e_n^\dagger$ in complex vector spaces), where we understand the first as a $d\times 1$ matrix so that the second is $1\times d$, where $d$ is the dimension of the space. Then the inner product is $\langle m|n\rangle=e_m^Te_n=e_m\cdot e_n$, and the outer product is $|n\rangle\langle m|=e_ne_m^T$, which is a $d\times d$ matrix with a single $1$ at the index $(n,m)$. Note that these notations are also used for arbitrary vectors; for example we might write $|\psi\rangle=v$ for some vector $v$, and then $\langle\psi|=v^T$, $\langle\psi|\psi\rangle=\|v\|^2$, and $|\psi\rangle\langle\psi|$ is the projection matrix in the direction of $v$.
For you background, there is a discrepancy between notations in the U.S and some countries in Europe (mainly France and Italy, but not only). This is of course very general, there may be exceptions, but it explains why at some point the need for an unambiguous notation appeared.
Rem: I use U.S and E.U below, but again it may vary from people to people, don't be offended if you do differently...
- $\mathbb N$ in U.S does not contains $0$ while contains $0$ in E.U
Thus the unambiguous notation is $\mathbb N_0$ for $\{0,1,2,\cdots\}$.
In E.U we use $\mathbb N^*$ for $\{1,2,\cdots\}$
- Positive or negative is taken in strict sense in U.S and loose sense in E.U (this means $0$ is both a negative and a positive number).
Thus $\mathbb Z^+$ or $\mathbb R^+$ in E.U contains $0$ and not in U.S, as for the case of $\mathbb N$ we can use $\mathbb Z_0^+$ and $\mathbb R_0^+$ for expliciting the set contains $0$.
In E.U we would use $\mathbb Z^{+*}$ or $\mathbb R^{+*}$ to exclude $0$.
As you can see U.S add zero via the notation $\mathbb X_0$ while E.U exclude it via $\mathbb X^*$.
- Even with the interval notation there are differences (but only in the symbols used), U.S would generally note open bounds with parenthesis while E.U would use outward facing square brackets.
i.e. $(a,b) \text{ vs } ]a,b[$ or $[0,+\infty)\text{ vs }[0,+\infty[$
Best Answer
$z \in \mathbb{C} \setminus \mathbb{R}$ means that $z$ is a complex number that is not a real number. I.e., any number of the form $a+bi$ where $b \not= 0$.
The "backslash" $\setminus$ is the set-difference or set-minus operation. In general $A \setminus B$ is the set of all $x \in A$ such that $x \not\in B$.
The "forward slash" $/$ is a quotient operator. $\mathbb{C}/\mathbb{R}$ would be the set of all cosets of $\mathbb{R}$ in $\mathbb{C}$; these cosets each contain complex numbers whose imaginary components are equal.