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% Optimal Foraging Theory Revisited: Appendix. Lemma about Ratios
% (note: this material not included in final version of document)
%
% (c) Copyright 2007 by Theodore P. Pavlic
%

\chapter{A Lemma: Relationships Between Ratios}
\label{app:lemma_proof} 

Take an ordered field $(\set{F},{\leq})$ with additive identity $0 \in
\set{F}$. Also take relation ${\vartriangleright} \in
\{{=},{<},{>},{\leq},{\geq}\}$. Let $a,b,c,d,p \in \set{F}$ with $bd>0$
(\ie, $b$ and $d$ have the same ordering with $0$) and $p \geq 0$ and
assume that $a/b \vartriangleright c/d$. This implies that $ad
\vartriangleright bc$. Therefore,
%
\begin{equation*}
        ad + pab \vartriangleright bc + pab 
        \quad \text{ and } \quad
        ad + pdc \vartriangleright bc + pdc 
\end{equation*}
%
and so
%
\begin{equation*}
        a ( d + pb ) \vartriangleright b ( c + pa ) 
        \quad \text{ and } \quad
        d ( c + pa ) \vartriangleright c ( d + pb ) 
\end{equation*}
%
Thus,
%
\begin{equation}
        \frac{a}{b} \vartriangleright \frac{ c + pa }{ d + pb } 
        \quad \text{ and } \quad
        \frac{ c + pa }{ d + pb } \vartriangleright \frac{c}{d} 
        \label{eq:lemma_forward}
\end{equation}
%
By working in reverse, it is easy to show that either of the two
statements in \longref{eq:lemma_forward} imply that $a/b
\vartriangleright c/d$. Therefore,
%
\begin{equation}
        \frac{a}{b} \vartriangleright \frac{c}{d} 
        \iff
        \frac{a}{b} \vartriangleright \frac{ c + pa }{ d + pb } 
        \iff
        \frac{ c + pa }{ d + pb } \vartriangleright \frac{c}{d} 
        \label{eq:lemma_field}
\end{equation}
%
%Defining $\set{F} \triangleq \R$ and ${\vartriangleright} \triangleq
%{\geq}$ makes \longref{eq:lemma_field} equivalent to the statement in
%\longref{eq:lemma}.