Working on the injectivity of the case where x is a fraction of the last exercise

assignments/week1/1.tex

@@ -156,8 +156,56 @@ See the assignment PDF for the full assignment specification and theorem.
         $\frac{2^1 * 3^1}{5^1} = \frac{6}{5}$.
     \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[Surjectivity:] We want to show that $f$ is 1-1,
-            \item[Injectivity:]
+                i.e. $f(x_1)=f(x_2) \implies x_1 = x_2$.
+
+                So, let's assume that for any $x_1, x_2 \in \{ q > 0 : q \in \Q \}$, $f(x_1) = f(x_2)$.
+                Since the function $f$ has 3 parts, based on the input, we have to prove this statement for those 3
+                parts separately as well. First, the easiest case, where the input set is $\{1\}$.
+                Then, $f(x) = 1 \forall x$, so $f$ is injective.
+
+                For the case where $x \in \N\backslash\{1\}$, $f(x) := p_1^{2r_1}***p_N^{2r_N}$.
+                We know from the \textbf{Theorem} that any fraction can be uniquely written as a product of prime
+                factors with exponents, so when we assume $f(x_1) = f(x_2)$, we can also assume that $x_1$ and $x_2$
+                have a unique prime factorization associated with them. So let's assume that $f(x_1) = f(x_2)$.
+                This means that $p_1^{2r_1}***p_N^{2r_N} = q_1^{2s_1}***q_M^{2s_M}$, where $p_i^{r_i}$ and $q_j^{s_j}$
+                denote the prime factors for both sides. We can further expand this expression into:
+
+                \begin{align}
+                    p_1^{r_1} * p_1^{r_1} *** p_N^{r_N} * p_N^{r_N} &=
+                    q_1^{s_1} * q_1^{s_1} *** q_M^{s_M} * q_M^{s_M} \implies \\
+                    p_1^{r_1}***p_N^{r_N} * p_1^{r_1}***p_N^{r_N} &=
+                    q_1^{s_1}***q_M^{s_1} * q_1^{s_1}***q_N^{s_M} \implies \\
+                    x_1 * x_1 &= x_2 * x_2
+                \end{align}
+
+                Because we know that each fraction constitutes a unique prime factorization, we also know that $x_1$ and
+                $x_2$ are uniquely derived. This is why the implications in the equation above holds.
+                Because both $x_1$ and $x_2 > 0$, $x_1 = x_2$.
+
+                Now for the case where $x \in \Q\backslash\N$. 
+                Then $f(x) := p_1^{2r_1}***p_N^{2r_N}q_1^{2s_1-1}***q_M^{2s_M-1}$, using the unique factorization
+                derived from the \textbf{Theorem}. So again, we assume that for any $x_1, x_2 \in \Q\backslash\N$,
+                $f(x_1) = f(x_2)$.
+                Using the definition of $f$, we get:
+                $p_1^{2r_1}***p_N^{2r_N}q_1^{2s_1-1}***q_M^{2s_M-1} =
+                v_1^{2t_1}***v_n^{2t_n}w_1^{2u_1-1}***w_m^{2u_m-1}$.\footnote{The super- and subscripts become a bit
+                abracadabra, but I think everything is unique and readable this way.}
+                Expanding this expression further, we get:
+
+                \begin{align}
+                    \frac{p_1^{2r_1}}{p_1}***\frac{p_N^{2r_N}}{p_N}\frac{q_1^{2s_1}}{q_1}***\frac{q_M^{2s_M}}{q_M} &=
+                    \frac{v_1^{2t_1}}{v_1}***\frac{v_n^{2t_n}}{v_n}\frac{w_1^{2u_1}}{w_1}***\frac{w_m^{2u_m}}{w_m} 
+                    \implies \\
+                    \frac{p_1^{r_1}*p_1^{r_1}}{p_1}***\frac{p_N^{r_N}*p_N^{r_N}}{p_N}
+                    \frac{q_1^{s_1}*q_1^{s_1}}{q_1}***\frac{q_M^{s_M}*q_M^{s_M}}{q_M} &=
+                    \frac{v_1^{t_1}*v_1^{t_1}}{v_1}***\frac{v_n^{t_n}*v_n^{t_n}}{v_n}
+                    \frac{w_1^{u_1}*w_1^{u_1}}{w_1}***\frac{w_m^{u_m}*w_m^{u_m}}{w_m} \implies \\
+                    \frac{x_1*x_1}{p_1***p_N*q_1***q_M} &= \frac{x_2*x_2}{v_1***v_n*w_1***w_m}
+                \end{align}
+
+            \item[Injectivity:] We want to show that $f$ is onto,
+                i.e. $f(\{ q > 0 : q \in \Q \}) = \N$.
         \end{description}
 \end{enumerate}