36 lines
1.6 KiB
TeX
36 lines
1.6 KiB
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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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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\exercise[1.1.2]
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\begin{tcolorbox}
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Let $S$ be an ordered set. Let $A \subset S$ be a non-empty finite subset. Then A is bounded. Furthermore,
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$\text{inf} A$ exists and in in $A$ and $\text{sup} A$ exists and is in $A$.
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\end{tcolorbox}
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In order to prove that $A$ is bounded, we have to prove that it has an upper and a lower bound. Let us prove
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that $A$ is bounded above first.
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In particular, we have to prove that $\exists a \in A$ such that $x \leq b$ for all $x \in E$.
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Since $A$ is non-empty and finite, we can use induction on the cardinality of $A$, since that will always
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be some natural number $n$. So, we have to prove two cases: the base case, where $|A|=1$, and the inductive step,
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where we will assume that when $A$ has an upper bound when it has cardinality $m$,
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then it also has an upper bound when its carindality is equal to $m+1$.
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\end{document}
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