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