Modulo Operation/Examples/0.11 mod 0.1

Jump to navigation Jump to search

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

$1 \cdotp 1 \bmod 1 = 0 \cdotp 1$
$z \left({x \bmod y}\right) = \left({z x}\right) \bmod \left({z y}\right)$

and so:

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

$\blacksquare$