inequalities for real numbersInequalityForRealNumbers2003-09-26 01:14:442006-03-04 14:36:42Definitionstrict inequalityinequality% this is the default PlanetMath preamble. as your knowledge
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\newtheorem{thm}{Theorem}
\newtheorem{defn}{Definition}
\newtheorem{prop}{Proposition}
\newtheorem{lemma}{Lemma}
\newtheorem{cor}{Corollary}Suppose $a$ is a real number.
\begin{enumerate}
\item If $a<0$ then $a$ is a \emph{negativ{e} number}.
\item If $a>0$ then $a$ is a \emph{positiv{e} number}.
\item If $a\le 0$ then $a$ is a \emph{non-positiv{e} number}.
\item If $a> 0$ then $a$ is a \emph{non-negativ{e} number}.
\end{enumerate}
The first two inequalities are also called {\bf strict inequalities}.
\subsubsection*{Properties}
Suppose $a$ and $b$ are real numbers.
\begin{enumerate}
\item If $a>b$, then $-a<-b$. If $a<b$, then $-a>-b$.
\item If $a\ge b$, then $-a\le -b$. If $a\le b$, then $-a\ge -b$.
\end{enumerate}
\begin{lemma}\, $0<a$\, iff\, $-a<0$.
\end{lemma}
\begin{proof} If $0<a$, then adding $-a$ on both sides of the inequality gives $-a=-a+0<-a+a=0$.\, This process can also be reversed.
\end{proof}
\begin{lemma} For any $a\in \R$, either $a=0$ or $0<a^2$.
\end{lemma}
\begin{proof}
Suppose\, $a\ne 0$, then by trichotomy, we have either\, $0<a$\, or\, $a<0$, but not both.\, If\, $0<a$,\, then\, $0=0\cdot a<a\cdot a=a^2$.\, On the other hand, if\, $-(-a)=a<0$, then\, $0<-a$\, by the previous lemma.\, Then repeating the previous \PMlinkescapetext{argument},\, $0 = 0\cdot(-a) < (-a)(-a)=a^2$.
\end{proof}
Three direct consequences follow:
\begin{cor}\, $0<1$ \end{cor}
\begin{cor} For any $a\in \R$, $0<1+a^2$. \end{cor}
\begin{cor} There is no real solution for $x$ in the equation $1+x^2=0$.
\end{cor}
\subsubsection*{Inequality for a converging sequence}
Suppose $a_0,a_1,\ldots$ is a sequence of real numbers converging to a real
number $a$.
\begin{enumerate}
\item If $a_i < b$ or $a_i \le b$
for some real number $b$ for each $i$, then $a\le b$.
\item If $a_i > b$ or $a_i \ge b$
for some real number $b$ for each $i$, then $a\ge b$.
\end{enumerate}