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diff --git a/assignments/week1/1.pdf b/assignments/week1/1.pdf
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diff --git a/assignments/week1/1.tex b/assignments/week1/1.tex
index 7811909..05de978 100644
--- a/assignments/week1/1.tex
+++ b/assignments/week1/1.tex
@@ -156,7 +156,7 @@ 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:] We want to show that $f$ is 1-1,
+            \item[Injectivity:] We want to show that $f$ is 1-1,
                 i.e. $f(x_1)=f(x_2) \implies x_1 = x_2$.
 
                 So, let's assume that for any $x_1, x_2 \in \{ q > 0 : q \in \Q \}$, $f(x_1) = f(x_2)$.
@@ -180,7 +180,7 @@ See the assignment PDF for the full assignment specification and theorem.
                 \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.
+                $x_2$ are uniquely derived. This is why the implications in the equation above hold.
                 Because both $x_1$ and $x_2 > 0$, $x_1 = x_2$.
 
                 Now for the case where $x \in \Q\backslash\N$. 
@@ -204,8 +204,34 @@ See the assignment PDF for the full assignment specification and theorem.
                     \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. $f(\{ q > 0 : q \in \Q \}) = \N$.
+                I'm kinda stuck at this point. I see that this is definitely injective, since the way the exponents
+                are defined, you will always know which prime factors belong to the numerator or to the denominator.
+                But I fail to prove this using the direct definition of $f$ like we could do for the natural numbers.
+                This is because the products of the denominators in the last equation are not unique.
+                So maybe I simplified them too much and shouldn't try and write them in terms of $x_1$ and $x_2$ like
+                we did earlier, and try and focus more on just the exponents, but I feel it becomes really hard
+                to show that $x_1 = x_2$ that way.
+
+            \item[Surjectivity:] We want to show that $f$ is onto, i.e. $f(\{ q > 0 : q \in \Q \}) = \N$.
+
+                In order to prove this, we will take an arbitrary $y \in \N$, and show that $\exists x : f(x)=y$.
+                We know from the \textbf{Theorem} that $y$ can be written as a product of unique prime factors,
+                $p_1^{r_1}***p_N^{r_N}$. From the definition of $f$ we know that if the exponents of the prime factors
+                $r$ are even, they belong to the numerator of $x$ and if the exponents are
+                odd, they belong to the denominator of $x$. If there are no prime factors with odd exponents,
+                $x$ will be a natural number. If $y=1$, $x=1$.
+
+                We will now only consider the case that $y$ is a prime factorization with factors with odd exponents
+                \footnote{The case for a prime factorization with solely even exponents can be backtracked
+                in a similar fashion, just without the case for odd exponents and making $x$ a fraction.}.
+                Then, we can find $x$ in the following way: we multiply each prime factor $p_i^{2r_i-1}$ with $p_i$
+                and take the square root. We know that the square root of $p_i^{2r_i}$ is defined, since the exponent
+                is multiplied by a factor 2, which the root negates. This will yield a prime factorization that we will
+                put in the denominator of a fraction. We do the same for the prime factors with even exponent, but
+                without multiplying with $p_i$. The prime factors we gain like that we put as a product in the
+                numerator of the fraction. So, we gain a fraction with both the numerator and the denominator
+                consisting of products of prime numbers, which are natural numbers, and so the fraction is positive
+                and in fact a fraction. \qed
         \end{description}
 \end{enumerate}