Take $a \leftrightarrow[s, m], b \leftrightarrow[t, n]$, and $c \leftrightarrow[u, p] .$ Suppose first that
$$
a+c>b+c \quad \text { or } \quad([s, m]+[u, p])>([t, n]+[u, p])
$$
Now this implies
which in tum implies $(s+u)+((n+p)>(t+t)+(m+p)]$
which, in turn, implies $(s+u)+(n+p)>(t+u)+(m+p)$
Then, by Theorem II’, Chapter $3,(s+n)>(t+m)$ or $[s, m]>[t, n]$ and $a>b$, as required.
Suppose next that $a>b$ or $[s, m]>[t, n] ;$ then $(s+n)>(t+m) .$ Now to compare

Take $a \leftrightarrow[s, m], b \leftrightarrow[t, n]$, and $c \leftrightarrow[u, p]$ in which $ut+m \
a &>b
\end{aligned}
$$
(b) Suppose $a>b$. By simply reversing the steps in $(a)$, we obtain $a \cdot c<b \cdot c$, as required.

Suppose $a>0$ and $b>0$; then $|a|=a$ and $|b|=b$. Then $|a \cdot b|=a \cdot b=|a| \cdot|b|$. Suppose $a<0$ and $b<0$; then $|a|=-a$ and $|b|=-b$. Now $a \cdot b>0$; hence, $|a \cdot b|=a \cdot b=(-a) \cdot(-b)=|a| \cdot|b|$. Suppose $a>0$ and $b<0$; then $|a|=a$ and $|b|=-b$. Since $a \cdot b<0,|a \cdot b|=-(a \cdot b)=a \cdot(-b)=|a| \cdot|b|$. The case $a<0$ and $b>0$ is left as an exercise.