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Order Laws

There is a relation $<$ (less than) defined on the real numbers such that for each pair $a,b$ of real numbers, the statement ``$a<b$'' is either true or false, and such that the following conditions are satisfied:


Trichotomy law: For each pair $a,b$ of real numbers exactly one of the following statements is true:

\begin{displaymath}
a < b, \hspace{1in}
a = b, \hspace{1in}
b < a.
\end{displaymath} (C.49)

We say that a real number $p$ is positive if and only if $p>0,$ and we say that a real number $n$ is negative if and only if $n < 0.$ Thus as a special case of the trichotomy law we have:

If $a$ is a real number, then exactly one of the following statements is true:

\begin{displaymath}
\mbox{$a$\ is positive,} \hspace{1in} a=0, \hspace{1in} \mbox{$a$\ is negative}.
\end{displaymath} (C.50)

Sign laws: Let $a,b$ be real numbers. Then

$\displaystyle \mbox{ if }\ a > 0 \mbox{ and }\ b > 0$ $\textstyle \mbox{ then }\ $ $\displaystyle ab > 0 \mbox{ and }\ a/b > 0,$ (C.51)
$\displaystyle \mbox{ if }\ a < 0 \mbox{ and }\ b > 0$ $\textstyle \mbox{ then }\ $ $\displaystyle ab < 0 \mbox{ and }\ a/b < 0,$ (C.52)
$\displaystyle \mbox{ if }\ a > 0 \mbox{ and }\ b < 0$ $\textstyle \mbox{ then }\ $ $\displaystyle ab < 0 \mbox{ and }\ a/b < 0,$ (C.53)
$\displaystyle \mbox{ if }\ a < 0 \mbox{ and }\ b < 0$ $\textstyle \mbox{ then }\ $ $\displaystyle ab > 0 \mbox{ and }\ a/b > 0,$ (C.54)
$\displaystyle \mbox{ if }\ a > 0 \mbox{ and }\ b > 0$ $\textstyle \mbox{ then }\ $ $\displaystyle a+b > 0,$ (C.55)
$\displaystyle \mbox{ if }\ a < 0 \mbox{ and }\ b < 0$ $\textstyle \mbox{ then }\ $ $\displaystyle a+b < 0.$ (C.56)

Also,
\begin{displaymath}
\mbox{ $a$\ is positive if and only if $-a$\ is negative},
\end{displaymath} (C.57)

and
\begin{displaymath}
\mbox{ $a$\ is positive if and only if $a^{-1}$\ is positive.}
\end{displaymath} (C.58)


\begin{displaymath}
\mbox{ if }\ ab > 0 \mbox{ then }\ \mbox{either } (a > 0 \mbox{ and }\ b>0)
\mbox{ or }\ (a<0 \mbox{ and }\ b<0).
\end{displaymath} (C.59)


\begin{displaymath}
\mbox{ if }\ ab < 0 \mbox{ then }\ \mbox{either }(a > 0 \mbox{ and }\ b < 0)
\mbox{ or }\ (a < 0 \mbox{ and }\ b > 0).
\end{displaymath} (C.60)


\begin{displaymath}
\mbox{ if }\ a/b > 0 \mbox{ then }\ \mbox{either } (a > 0 \mbox{ and }\ b>0)
\mbox{ or }\ (a<0 \mbox{ and }\ b<0).
\end{displaymath} (C.61)


\begin{displaymath}
\mbox{ if }\ a/b < 0 \mbox{ then }\ \mbox{either }(a > 0 \mbox{ and }\ b < 0)
\mbox{ or }\ (a < 0 \mbox{ and }\ b > 0).
\end{displaymath} (C.62)

It follows immediately from the sign laws that for all real
numbers $a$
\begin{displaymath}
a^2 \geq 0 \mbox{ and }\ \mbox{ if }\ a \not= 0 \mbox{ then }\ a^2 > 0.
\end{displaymath} (C.63)

Here, as usual $a^2$ means $a \cdot a.$


Transitivity of $<$: Let $a,b,c$ be real numbers. Then

\begin{displaymath}
\mbox{ if }\ a < b \mbox{ and }\ b < c \mbox{ then }\ a < c.
\end{displaymath} (C.64)

We write $ a \leq b$ as an abbreviation for ``either $a<b$ or $a = b$'', and we write $b>a$ to mean $a<b$. We also nest inequalities in the following way:

\begin{displaymath}a < b \leq c = d < e \end{displaymath}

means

\begin{displaymath}a < b \mbox{ and }\ b \leq c \mbox{ and }\ c = d \mbox{ and }\ d < e. \end{displaymath}

Addition of Inequalities: Let $a,b,c,d$ be real numbers. Then

$\displaystyle \mbox{ if }\ a < b \mbox{ and }\ c < d$ $\textstyle \mbox{ then }\ $ $\displaystyle a+c < b+d,$ (C.65)
$\displaystyle \mbox{ if }\ a \leq b \mbox{ and }\ c \leq d$ $\textstyle \mbox{ then }\ $ $\displaystyle a+c \leq b+d,$ (C.66)
$\displaystyle \mbox{ if }\ a < b \mbox{ and }\ c \leq d$ $\textstyle \mbox{ then }\ $ $\displaystyle a+c < b+d,$ (C.67)
$\displaystyle \mbox{ if }\ a < b$ $\textstyle \mbox{ then }\ $ $\displaystyle a-c < b-c,$ (C.68)
$\displaystyle \mbox{ if }\ c < d$ $\textstyle \mbox{ then }\ $ $\displaystyle -c > -d,$ (C.69)
$\displaystyle \mbox{ if }\ c < d$ $\textstyle \mbox{ then }\ $ $\displaystyle a-c > a-d.$ (C.70)

Multiplication of Inequalities: Let $a,b,c,d$ be real numbers.

$\displaystyle \mbox{ if }\ a < b \mbox{ and }\ c > 0$ $\textstyle \mbox{ then }\ $ $\displaystyle ac < bc,$ (C.71)
$\displaystyle \mbox{ if }\ a < b \mbox{ and }\ c > 0$ $\textstyle \mbox{ then }\ $ $\displaystyle a/c < b/c ,$ (C.72)
$\displaystyle \mbox{ if }\ 0 < a < b \mbox{ and }\ 0 < c < d$ $\textstyle \mbox{ then }\ $ $\displaystyle 0 < ac < bd,$ (C.73)
$\displaystyle \mbox{ if }\ a < b \mbox{ and }\ c < 0$ $\textstyle \mbox{ then }\ $ $\displaystyle ac > bc,$ (C.74)
$\displaystyle \mbox{ if }\ a < b \mbox{ and }\ c < 0$ $\textstyle \mbox{ then }\ $ $\displaystyle a/c > b/c ,$ (C.75)
$\displaystyle \mbox{ if }\ 0 < a \mbox{ and }\ a < b$ $\textstyle \mbox{ then }\ $ $\displaystyle a^{-1} > b^{-1}.$ (C.76)

Discreteness of Integers: If $n$ is an integer, then there are no integers between $n$ and $n+1$, i.e. there are no integers $k$ satisfying $n < k < n+1.$ A consequence of this is that

\begin{displaymath}
\mbox{If } k,n \mbox{ are integers, and } k < n+1, \mbox{ then } k \leq n.
\end{displaymath} (C.77)

If $x$ and $y$ are real numbers such that $y - x > 1$ then there is an integer $n$ such that
\begin{displaymath}
x < n < y.
\end{displaymath} (C.78)

Archimedean Property: Let $x$ be an arbitrary real number. Then

\begin{displaymath}
\mbox{there exists an integer $n$\ such that $n > x$.}
\end{displaymath} (C.79)


next up previous
Next: Miscellaneous Properties Up: C. prerequisites Previous: Algebraic Laws
Ray Mayer 2007-08-31