# Modulo Operation/Examples/0.11 mod -0.1

## Theorem

$0 \cdotp 11 \bmod -0 \cdotp 1 = -0 \cdotp 09$

where $\bmod$ denotes the modulo operation.

## Proof

By definition of modulo operation:

$x \bmod y := x - y \left \lfloor {\dfrac x y}\right \rfloor$

for $y \ne 0$.

We have:

 $\displaystyle \dfrac {0 \cdotp 11} {-0 \cdotp 1}$ $=$ $\displaystyle \dfrac {1 \cdotp 1} {-1}$ $\displaystyle$ $=$ $\displaystyle -1 \cdotp 1$

and so:

$\left\lfloor{\dfrac {0 \cdotp 11} {-0 \cdotp 1} }\right\rfloor = -2$

Thus:

 $\displaystyle 0 \cdotp 11 \bmod -0 \cdotp 1$ $=$ $\displaystyle 0 \cdotp 11 - \left({-0 \cdotp 1}\right) \times \left\lfloor{\dfrac {0 \cdotp 11} {-0 \cdotp 1} }\right\rfloor$ $\displaystyle$ $=$ $\displaystyle 0 \cdotp 11 - \left({-0 \cdotp 1}\right) \times \left({-2}\right)$ $\displaystyle$ $=$ $\displaystyle 0 \cdotp 11 - 0 \cdotp 2$ $\displaystyle$ $=$ $\displaystyle -0 \cdotp 09$

$\blacksquare$