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 \section{Week 2}
 
 \exercise*[1.1.1]
-Exercise 1.1.1 comes here.
+\begin{tcolorbox}
+    Prove:
+
+    Let $F$ be an ordered field and $x,y,z \in F$. If $x < 0$ and $y < z$, then $xy > xz$.
+\end{tcolorbox}
+
+So let's assume the premise. $F$ is an ordered field and $x,y,z \in F$,
+and we choose $x,y$ and $z$ such that $x < 0$ and $y < z$.
+
+From $x < 0$ it follows that $(-x) > 0$. From $y < z$ it follows that $0 < z - y$.
+From both of these, we can conclude that $0 < (-x)(z-y)$.
+Working out the right side with the distributive law, gives $0 < (-x*z)-(-x*y)$.
+Using $-1*-1 = 1$, gives $0 < (-xz)-(-xy)$, thus $0 < xy - xz$. The right part can be split again: $xz < xy$.
+Then, the < can be flipped, which gives $xy > xz$. \qed
 
 \end{document}