Does there exist a system such that the additive identity is non-zero? I am trying to explain how although the additive identity is written as $0$, it is not the same as the number $0$.
For example for a $2\times 2$ matrix the additive identity is $\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$. However this is a bad example since it only involves $0$'s 
So what is a system such that the additve identity is non-zero (preferably not involving a $0$)?
I would use mod p but this (seemingly) contradicts the rule for the uniqueness of the identity
This arose because of a question, find a vector space such that $0=1$, however by "$0,1$" they meant the additive and multiplicative identities. So I was trying to explain that the additve identity is different from $0$ in certain systems.
 A: I am surprised no one mentioned elliptic curves with its "usual" elliptic-curve-addition. You pick two points $a$ and $b$ on an elliptic curve. Then $a+b$ is defined as the negative of the third point of intersection between the elliptic curve and the straight line connecting $a$ and $b$. In this case, the additive identity is denoted $O$ which is the point at infinity. And this example is not just pathological. Elliptic curves are used widely with the most-real-world-example being cryptography.
A: Consider the tropical semiring $(\mathbb R\cup\{\infty\},\oplus,\otimes)$, where $x\oplus y=\min\{x,y\}$ and $x\otimes y=x+y$. The "additive" identity, meaning the identity element for the $\oplus$ operation, is $\infty$. That's pretty far away from $0$!
The number $0$ is present, but it is not the additive identity in this structure. For example, $0\oplus 79 = 0$. Instead, $0$ is the "multiplicative" identity: $0\otimes y=y$.
A: You will probably have to leave number systems if you want to get away from 0 being the additive identity.  For instance, if you take the + operation to be permutations, then you would have a non-number additive identity, which would be the identity permutation (i.e. don't permute anything).  You could also take the operation of symmetries in the plane, where the identity would be not flipping anything.
You may want to make clearer exactly what you want.  Do you actually want a system that has both an addition and multiplication operation, such that multiplication by the additive identity does not necessarily equal the additive identity?
A: Long time since I did math but the easiest example I can think of is modular arithmetic. For a≡b (mod n), n is an additive identity. For instance the modular arithmetic for determining which day of the week has 2 additive identities 0 and 7. Informally, Monday + 0 days = Monday, Monday + 7 days = Monday. 
A: In the integers under the operation $*$ defined by $n * m = n + m - 1$, $1$ acts as the additive identity.
Of course, this is identical to the ordinary additive group structure on the integers except for relabeling. But there are occasional situations where people find themselves using this operation, usually when there are historical off-by-one errors to contend with. For example, $*$ is the operation used to add musical intervals together.
A: How about string addition?  The additive identity is an empty string.
A: A lightswitch has two values, off and on. Assemble two lightswitches in parallel. You now have the following addition table:
$$\begin{array}{c|cc}
+ & \textrm{off} & \textrm{on} \\
\hline\\
\textrm{off} & \textrm{off} & \textrm{on} \\
\textrm{on} & \textrm{on} & \textrm{on}
\end{array}$$
Obviously, $\textrm{off} + \textrm{on} = \textrm{on}$, and $\textrm{off} + \textrm{off} = \textrm{off}$, so the system satisfies the desired properties.
A: If the "additive identity" is the neutral element of some abelian group, ${\Bbb R}^\times={\Bbb R}\setminus\{0\}$ has neutral element/additive identity 1.
A: Just pick any biyection $\psi \colon \mathbb{Z} \to \mathbb{Z}$, and define:
\begin{align}
a \oplus b &= \psi^{-1}(\psi(a) + \psi(b)) \\
a \odot b  &= \psi^{-1}(\psi(a) \cdot \psi(b))
\end{align}
This is a ring, and it's 0 and 1 are $\psi^{-1}(0)$ and $\psi^{-1}(1)$, respectively. So you could pick $\psi(x) = x - 42$, just as a tribute to Douglas Adams.
[Yes, this is blatant cheating, as all this does is to rename the elements of $\mathbb{Z}$.]
A: How about the group with elements $(x\in\mathbb R | x>0)$ and the additive operator defined such that ${x\oplus y}$ is the real-number product of x and y.  The additive identity of that group will then correspond to the multiplicative identity of $\mathbb R$ [which, of course, is non-zero in $\mathbb R$].
