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@@ -133,5 +133,32 @@ Each set is definitely finite, because they all contain just one element.
 Finally, the union of the collection of sets is equal to $\N$, which is not a finite set.
 
 \exercise*
+\begin{tcolorbox}
+    \begin{enumerate}[label=\emph{\alph*)}, wide]
+        \item Compute $f(4/15)$. Find $q$ such that $f(q) = 108$.
+        \item Use the \textbf{Theorem} to prove that $f$ is a bijection.
+    \end{enumerate}
+\end{tcolorbox}
+
+See the assignment PDF for the full assignment specification and theorem.
+
+\begin{enumerate}[label=\emph{\alph*)}, wide]
+    \item $\frac{4}{15}$, if written as a product of prime factors, is equal to $\frac{2^2}{3^1*5^1}$.
+        Since this fraction is not a natural number, we have to use the second part of the definition of $f$.
+        So, $f(q) = 2^{2*2} * 3^{2 * 1 - 1} * 5^{2 * 1 - 1} = 240$.
+
+        For the inverse of $f$, it is still necessary to compute the factorization in prime numbers.
+        Using the powers of the primes we can deduce whether the prime present is, if applicable,
+        part of either the numerator or the denominator.
+        $180 = 2^2 * 3^2 * 5^1$. Because of the way $f$ is defined, we know that all the prime factors with an even
+        power are part of the numerator and all prime factors with an odd power are part of the denominator (except 1,
+        which just maps to itself). When we backtrack using this information, we then get the following fraction:
+        $\frac{2^1 * 3^1}{5^1} = \frac{6}{5}$.
+    \item In order to prove that $f$ is a bijection, we have to prove that $f$ is injective and surjective.
+        \begin{description}
+            \item[Surjectivity:] 
+            \item[Injectivity:]
+        \end{description}
+\end{enumerate}
 
 \end{document}