[SPCA] Fix more typos

semester3/spca/parts/01_c/06_floating-point/02_representation.tex

@@ -25,7 +25,8 @@ Single precision and Double precision floating point numbers store the $3$ param
     Bias: $1023$, Exponent range: $[-1022, 1023]$
 \end{center}
 
-Most of the extra precision in $64$b floating point numbers is associated to the mantissa. Note how double precision is necessary to represent all $32$b signed Integers, and not all $64$b signed Integers can be represented in either format.
+Most of the extra precision in $64$b floating point numbers is associated to the mantissa. Note how double precision is necessary to represent all $32$b signed Integers,
+and not all $64$b signed Integers can be represented in either format.
 
 \newpage
 
@@ -33,7 +34,8 @@ The way these bitfields are interpretd \textit{differs} based on the exponent fi
 
 \begin{enumerate}
     \item \textbf{Normalized Values}: Exponent bit field $e$ is neither all $1$s nor all $0$s.\\
-          In this case, $E$ is read in \textit{biased} form: $E = e - b$. The bias is $b=2^{k-1}-1$, where $k$ is the amount of bits reserved for $e$. This produces the exponent ranges $E \in [-(b-1), b]$.\\
+          In this case, $E$ is read in \textit{biased} form: $E = e - b$. The bias is $b=2^{k-1}-1$, where $k$ is the number of bits reserved for $e$.
+          This produces the exponent ranges $E \in [-(b-1), b]$.\\
           The mantissa field $m$ is interpreted as $M = 0.m_{n-1}\ldots m_1 m_0 + 1$, where $n$ is the amount of bits reserved for $m$
     \item \textbf{Denormalized Values}: Exponent bit field $e$ is all $0$s.\\
           In this case, $E$ is read in \textit{biased} form $E = 1 - b$. (Instead of $E = e - b$)\\