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:

$x \bmod y := x - y \floor {\dfrac x y}$

for $y \ne 0$.


We have:

\(\ds \dfrac {0 \cdotp 11} {-0 \cdotp 1}\) \(=\) \(\ds \dfrac {1 \cdotp 1} {-1}\)
\(\ds \) \(=\) \(\ds -1 \cdotp 1\)

and so:

$\floor {\dfrac {0 \cdotp 11} {-0 \cdotp 1} } = -2$


Thus:

\(\ds 0 \cdotp 11 \bmod -0 \cdotp 1\) \(=\) \(\ds 0 \cdotp 11 - \paren {-0 \cdotp 1} \times \floor {\dfrac {0 \cdotp 11} {-0 \cdotp 1} }\)
\(\ds \) \(=\) \(\ds 0 \cdotp 11 - \paren {-0 \cdotp 1} \times \paren {-2}\)
\(\ds \) \(=\) \(\ds 0 \cdotp 11 - 0 \cdotp 2\)
\(\ds \) \(=\) \(\ds -0 \cdotp 09\)

$\blacksquare$


Sources