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\begin{document}

\noindent\parbox[b]{1.525in}{\Huge MATH 101} \
\parbox[b]{2in}{\normalsize {Assignment 1 \\[-3pt]
January 1, 2011}} \ \hfill Name \rule{.25in}{.5pt} Branko \'{C}urgus

\normalsize

\begin{problem}
Prove that for all $a \in (-1,1)$ and all $b \in [-1,1]$ we have
\begin{equation} \label{eq-to-prove}
\frac{a+b}{1+ab} \in [-1,1].
\end{equation}
\end{problem}

\begin{proof}[Solution]
Assume that $a \in (-1,1)$ and $b \in [-1,1]$. Then $|a| \in [0,1)$ and $|b| \in [0,1]$. Consequently, $|ab| \in [0,1)$. That is $ab \in (-1,1)$. Therefore $1+ab > 0$.

Since $a \in (-1,1)$, we have $1+a > 0$ and $1-a > 0$. Similarly, since $b \in [-1,1]$, we have $1+b \geq 0$ and $1-b \geq 0$.

Since $1+a > 0$ and $1+b \geq 0$, we have $(1+a)(1+b) \geq 0$. Consequently, $1+a+b+ab \geq 0$ and hence
\begin{equation*}
-1-ab \leq a+b.
\end{equation*}
As $1+ab > 0$, dividing both sides of the last inequality by $1+ab$ we get
\begin{equation} \label{eq-ineq1}
-1 \leq \frac{a+b}{1+ab}.
\end{equation}


Since $1-a > 0$ and $1-b \geq 0$, we have $(1-a)(1-b) \geq 0$. Consequently, $1-a-b+ab \geq 0$ and hence
\begin{equation*}
 a+b \leq 1+ab.
\end{equation*}
As $1+ab > 0$, dividing both sides of the last inequality by $1+ab$ we get
\begin{equation} \label{eq-ineq2}
 \frac{a+b}{1+ab} \leq 1.
\end{equation}

Inequalities \eqref{eq-ineq1} and \eqref{eq-ineq2} prove inequality \eqref{eq-to-prove}.
\end{proof}


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