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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.

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    $\begingroup$ I think this might be a difficult question to answer in the way you'd like it answered. The reason $0$ is the additive identity in a lot of different structures is because that's just what we call the additive identity. You're right in saying that, in your matrix example, that isn't the number zero. It only involves zeros, though, because something is the additive identity. We might as well involve zero for consistency. Another way to look at it might be "Zero is defined as an element you add to a second element to get back the second element." $\endgroup$ Commented Mar 4, 2014 at 4:16
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    $\begingroup$ But I could simply say that we only write the additive identity as $0$ but if I wanted I could write it as fish$+x=x=x+$fish and this would still be valid, confusing but valid $\endgroup$
    – Wonkman123
    Commented Mar 4, 2014 at 4:19
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    $\begingroup$ Yeah, absolutely. You could write it as whatever you want. $\endgroup$ Commented Mar 4, 2014 at 4:45
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    $\begingroup$ @DanielV: A system with two-sided additive identities has only one; if x + 0_a = x and 0_b + x = x, then 0_a + 0_b = 0_a and 0_b, so 0_a = 0_b. $\endgroup$
    – prosfilaes
    Commented Mar 4, 2014 at 5:46
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    $\begingroup$ Perhaps some more context would help; since you are "trying to explain how" an additive identity is not necessarily $0$, maybe who you are explaining this to or why it needs to be explained would help provide an answer. $\endgroup$
    – abiessu
    Commented Mar 4, 2014 at 22:54

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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: $0\otimes y=y$.

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  • $\begingroup$ I'm tempted to argue this is only because you defined ${\oplus} = \min$ instead of ${\oplus} = \max$. $\endgroup$
    – Zhen Lin
    Commented Mar 4, 2014 at 8:24
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    $\begingroup$ @ZhenLin The same applies to $(\mathbb R\cup\{-\infty\}, \max, +)$. $\endgroup$ Commented Mar 4, 2014 at 9:19
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    $\begingroup$ I think many people would be inclined to call ⊗ "addition" in this system. $\endgroup$
    – Brilliand
    Commented Mar 4, 2014 at 20:27
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    $\begingroup$ @Brilliand We have $a \otimes (b \oplus c) = (a \otimes b) \oplus (a \otimes c)$ and not vice versa; hence the choice of which one we call plus and which times. (But I agree that this point would need to be drawn out if using this example in teaching.) $\endgroup$ Commented Mar 4, 2014 at 20:37
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    $\begingroup$ @Brilland: they might be tempted to call it that, but that would be misleading about the meaning of the operators. The abstract operator '⊗' acts like multiplication. The fact that it is implemented as arithmetic addition is, well not exactly irrelevant, but just not as important is the fact that it acts analogously to arithmetic multiplication. $\endgroup$
    – Mitch
    Commented Mar 5, 2014 at 14:20
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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.

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    $\begingroup$ More usefully, such an operation is useful when converting code from 0-indexing to 1-indexing! $\endgroup$
    – Emily
    Commented Mar 4, 2014 at 4:27
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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.

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  • $\begingroup$ This is really just a boolean table on the inclusive or operation. More often represented with T's, and F's, but often 0 and 1 in many contexts. It's also worth noting that most lights connected by 2 switches, are XORed. $\endgroup$
    – Cruncher
    Commented Mar 4, 2014 at 18:21
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    $\begingroup$ Yes, all of that is true, but the point of this particular answer was to open up the concept of isomorphism, which sits at the core of the question. $\endgroup$
    – Emily
    Commented Mar 4, 2014 at 18:52
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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.

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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?

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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.

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How about string addition? The additive identity is an empty string.

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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.

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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 $\mathbb{Z}$.]

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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 $\mathbb R$].

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