Modulo Operation/Examples/0.11 mod 0.1/Proof 1
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Theorem
- $0 \cdotp 11 \bmod 0 \cdotp 1 = 0 \cdotp 01$
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:
- $\floor {\dfrac {0 \cdotp 11} {0 \cdotp 1} } = 1$
Thus:
\(\ds 0 \cdotp 11 \bmod 0 \cdotp 1\) | \(=\) | \(\ds 0 \cdotp 11 - 0 \cdotp 1 \times \floor {\dfrac {0 \cdotp 11} {0 \cdotp 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0 \cdotp 11 - 0 \cdotp 1 \times 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0 \cdotp 01\) |
$\blacksquare$