Editing Fixed-point arithmetic (section)

===Decimal fixed point multiplication===
Suppose there is the following multiplication with two fixed-point, 3-decimal-place numbers.
:<math>\begin{align}
(10.500)(1.050)
&=1 \times 10.500 + 0.050 \times 10.500 \\
&= 10.500+0.525000=11.025000
\end{align}</math>
Note how since there are 3 decimal places we show the trailing zeros.  To re-characterize this as an integer multiplication we must first multiply by <math>1000\ (=10^3)</math> moving all the decimal places in to integer places, then we will multiply by <math>1/1000\ (=10^{-3})</math>  to put them back the equation now looks like
:<math>\begin{align}
(10.500)(10^{3}) (1.050)(10^{3})  (10^{-3})(10^{-3})
&= (10500)(1050) (10^{-6}) \\
&= 11\,025\,000  (10^{-6}) \\
&= 11.025000
\end{align}</math>

This works equivalently if we choose a different base, notably base 2 for computing since a bit shift is the same as a multiplication or division by an order of 2. Three decimal digits is equivalent to about 10 binary digits, so we should round 0.05 to 10 bits after the binary point.  The closest approximation is then 0.0000110011.
:<math>\begin{align}
10&= 8+2=2^3+2^1\\
1&=2^0\\
0.5&= 2^{-1}\\
0.05&= 0.0000110011_2
\end{align}</math>
Thus our multiplication becomes
:<math>\begin{align}
(1010.100)(2^3)(1.0000110011)(2^{10}) (2^{-13})
&=(1010100)(10000110011) (2^{-13})\\
&=(10110000010111100) (2^{-13})\\
&=1011.0000010111100
\end{align}</math>

This rounds to 11.023 with three digits after the decimal point.