Lemma 1.21

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\title{Lemma 1.21} %Change this to the appropriate number.
\author{Yanan Zhang} %Change this to your name.
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If $x > 0$, then $x^{-1} >0$
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\begin{proof}
Proof by contradiction. Case 1: Let $x>0$ and suppose $x^{-1}<0$. Then $x\cdot x^{-1}<x\cdot0$ by axiom 1.16. Since $x\cdot$$x^{-1}=1$ by axiom 1.11 and $x\cdot0=0$ by axiom 1.17, then $1<0$. which is a contradiction. Case 2: Let $x>0$ and suppose $x^{-1}=0$. By Lemma 1.11, $x^{-1}\in \mathbb{R}$ and $x\cdot$$x^{-1}$=1. If $x^{-1}=0$, then $x\cdot$$x^{-1}$=$x\cdot0$=0 by Lemma 1.17. That is, 1=0. which is a contradiction. Thus $x^{-1}\neq0$. Therefore, if $x>0$, then $x^{-1}>0$ .
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