Modulo Operation/Examples/0.11 mod -0.1

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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:

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

$\blacksquare$


Sources