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It seems to me that operators usually come in pairs: Plus and minus, Multiplication and Division, Exponentiation and Logarith, Derivative and Integral.

Is there such inverse operation of modulus?

Carlos Eduardo
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    As with any other many-to-one function, there are inverses, but you then need to choose one element from an equivalence class of inverses; e.g.: $-\frac\pi2\le\sin^{-1}(x)\le\frac\pi2$. In the case of modulus, you can just use the identity to be the inverse. – robjohn Feb 26 '21 at 03:03
  • Now I know where to start. I also understood a little better others functions. Thanks! – Carlos Eduardo Feb 26 '21 at 03:11

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I don't think there's an inverse operation to modulus, however, there are additive inverses. Consider that $x = a$ mod $m$. Then the additive inverse is $(m - a)$ mod $m$. In particular, $x + m - a = a + m - a = m = 0$ mod $m$. Notice though additive inverses are not unique. For instance, an additive inverse to $x$ is $k(m-a)$ for $k \in \mathbb{Z}$

ENV
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