1.9: Inexact Numbers

Example:

$$\frac{1}{3}=(-1)^{0} \times \frac{1}{4} \times\left(1+\frac{1}{3}\right) \text {, } \nonumber$$

so that $p=e-127=-2$ and $e=125=128-3$, or in binary, $e=01111101$. How is $f=1 / 3$ represented in binary? To compute binary number, multiply successively by 2 as follows:

$\begin{array}{lr}0.333 \ldots & 0 . \\ 0.666 \ldots & 0.0 \\ 1.333 \ldots & 0.01 \\ 0.666 \ldots & 0.010 \\ 1.333 \ldots & 0.0101 \\ \text { etc. } & \end{array}$

so that $1 / 3$ exactly in binary is $0.010101 \ldots$ With only 23 bits to represent $f$, the number is inexact and we have

$$f=01010101010101010101011 \nonumber$$

where we have rounded to the nearest binary number (here, rounded up). The machine number $1 / 3$ is then represented as

$$00111110101010101010101010101011 \nonumber$$

or in hex

$$\text { Зeaaaaab. } \nonumber$$

Find smallest positive integer that is not exact in single pre- cision

Let $N$ be the smallest positive integer that is not exact. Now, I claim that

$$N-2=2^{23} \times 1.11 \ldots 1 \nonumber$$

and

$$N-1=2^{24} \times 1.00 \ldots 0 \nonumber$$

The integer $N$ would then require a one-bit in the $2^{-24}$ position, which is not available. Therefore, the smallest positive integer that is not exact is $2^{24}+1=16777217$. In MATLAB, single $\left(2^{24}\right)$ has the same value as single $\left(2^{24}+1\right) .$ Since single $\left(2^{24}+1\right)$ is exactly halfway between the two consecutive machine numbers $2^{24}$ and $2^{24}+2$, MATLAB rounds to the number with a final zero-bit in $f$, which is $2^{24} .$