\textbf{From now on the passages are going to include only the calculations to avoid verbosity. Nonetheless, the same reasoning will be applied for each conversion.}
\item[b)]\textbf{\large 5/64};
Sign bit = 0
$\frac{5}{64}=0.078125$
Conversion to binary:\\
\begin{itemize}
\item$0.078125\times2=\textbf{0}+0.15625$;
\item$0.15625\times2=\textbf{0}+0.3125$;
\item$0.3125\times2=\textbf{0}+0.625$;
\item$0.625\times2=\textbf{1}+0.25$;
\item$0.25\times2=\textbf{0}+0.5$;
\item$0.5\times2=\textbf{1}+0$;
\end{itemize}
Result of conversion: $0.000101$.
Normalization to scientific notation: $1.\textbf{01}\times2^{-4}$.
Exponent bits in decimal = $127-4=123$
Conversion of exponent:\\
\begin{tabular}[h]{l|c}
\textbf{Fraction}&\textbf{Rest}\\
\hline\\
$123/2=61$& 1 \\\\
$61/2=30$& 1 \\
$30/2=15$& 0 \\
$15/2=7$& 1 \\
$7/2=3$& 1 \\
$3/2=1$& 1 \\
$1/2=0$& 1 \\
\end{tabular}
Exponent bits in binary = $01111011$
Final binary number = $0\ 01111011\ 01000000000000000000000=$
$00111101101000000000000000000000$
Conversion to hexadecimal:
$0011\ 1101\ 1010\ 0000\ 0000\ 0000\ 0000\ 0000=3$ D A $0\ 0\ 0\ 0\ 0=3$DA$00000$.
\item[c)]\textbf{\large -5/64};
With this being the negative counterpart of the previous number we just need to flip
the sign bit of $\frac{5}{64}$.
With that we get: $10111101101000000000000000000000$.
Which, converted to hexadecimal, gives:
$1011\ 1101\ 1010\ 0000\ 0000\ 0000\ 0000\ 0000=$ B D A $0\ 0\ 0\ 0\ 0=$ BDA$00000$.
From this we can observe that the sign bit is 0 (i.e. positive).
Conversion of exponent from binary to decimal:
$10000101=2^7+2^2+2^0=128+4+1=133$
To find how many bits have been moved beyond the floating point we now find the
exponent for our notation:
$133-127=6$
And we then take the mantissa to multiply it by $2^6$ (where 6 is the number we found through the last passage) to get the final binary number to convert to decimal:
and y $=\ 1.0\times2^{26}$ such that the sequence of operations $r =(x+y)- y $ results in r = 0.0 which is incorrect as the actual value should be 1.5.
This is due to catastrophic cancellation which occurred during the subtraction.
This kind of occurrences can't be identified unless data is further examined leaving no