-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).
+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.