-In the graph of the difference between representations, a single isolated point should be visible; this is not an error, but due to the greater discontinuity between the denormalised and normalised values ($e = 0$ and $1$ respectively).
+Figure \ref{float.pdf} shows the positive real numbers which can be represented exactly by an 8 bit floating point number encoded in the IEEE-754 format\footnote{Not quite; we are ignoring the IEEE-754 definitions of NaN and Infinity for simplicity}, and the distance between successive floating point numbers. We show two encodings using (1,2,5) and (1,3,4) bits to encode (sign, exponent, mantissa) respectively. We have also plotted a fixed point representation. For each distinct value of the exponent, the successive floating point representations lie on a straight line with constant slope. As the exponent increases, larger values are represented, but the distance between successive values increases; this can be seen on the right. The marked single point discontinuity at \verb/0x10/ and \verb/0x20/ occur when $e$ leaves the denormalised region and the encoding of $m$ changes.
+
+The earlier example $7.25$ would be converted to a (1,3,4) floating point representation as follows:
+\begin{enumerate}
+ \item Determine the fixed point representation $7.25 = 111.01_2$
+ \item Determine the sign bit; in this case $s = 0$
+ \item Calculate the exponent by shifting the point $111.01_2 = 1.1101_2 \times 2^2 \implies e = 2 = 10_2$
+ \item Determine the exponent encoding; in IEEE-754 equal to the number of exponent bits is added so $e_{enc} = e+3 = 5 = 101_2$
+ \item Remove the implicit bit if the encoded exponent $\neq 0$; $1.1101_2 \to .1101_2$
+ \item Combine the three bit strings$0,101,1101$
+ \item The final encoding is $01011101 \equiv \text{0x5D}$
+\end{enumerate}
+This particular example can be encoded exactly; however as there are an infinite number of real values and only a finite number of floats, in general a value must be $7.26$ must be rounded or truncated at Step 3.