Modulo Operation/Examples/0.11 mod 0.1

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Theorem

$0 \cdotp 11 \bmod 0 \cdotp 1 = 0 \cdotp 01$

where $\bmod$ denotes the modulo operation.


Proof 1

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:

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


Thus:

\(\displaystyle 0 \cdotp 11 \bmod 0 \cdotp 1\) \(=\) \(\displaystyle 0 \cdotp 11 - 0 \cdotp 1 \times \left\lfloor{\dfrac {0 \cdotp 11} {0 \cdotp 1} }\right\rfloor\)
\(\displaystyle \) \(=\) \(\displaystyle 0 \cdotp 11 - 0 \cdotp 1 \times 1\)
\(\displaystyle \) \(=\) \(\displaystyle 0 \cdotp 01\)

$\blacksquare$


Proof 2

From Modulo Operation: $1 \cdotp 1 \bmod 1$:

$1 \cdotp 1 \bmod 1 = 0 \cdotp 1$


From Product Distributes over Modulo Operation:

$z \left({x \bmod y}\right) = \left({z x}\right) \bmod \left({z y}\right)$

and so:

\(\displaystyle 0 \cdotp 11 \bmod 0 \cdotp 1\) \(=\) \(\displaystyle 0 \cdotp 1 \left({1 \cdotp 1 \bmod 1}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle 0 \cdotp 1 \times 0 \cdotp 1\)
\(\displaystyle \) \(=\) \(\displaystyle 0 \cdotp 01\)

$\blacksquare$


Sources