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\title{Math 165: Honors Calculus I\\
Assignment 1 Solutions}

\begin{document}

\maketitle
\thispagestyle{empty}

\textbf{Page 19.}

\problem{4} Zero has no reciprocal.

\begin{proof}
Suppose $0$ has a reciprocal, $0^{-1}$, so that $0 \cdot 0^{-1} = 1$.
This immediately contradicts Theorem~I.6, which implies that $0 \cdot
0^{-1} = 0$ ($0 \ne 1$ by Axiom~4). Therefore our assumption that $0$
has a reciprocal must be false.
\end{proof}

\problem{5} $-(a+b) = -a - b$

\begin{proof}
By definition (I.2), $-a-b$ is the unique number such that $b + (-a-b) =
-a$. Adding $a$ to both sides gives $a+b + (-a-b) = a + (-a) = 0$.
This equation shows that $-a-b$ is the unique negative of $a+b$, and so
$-a-b = -(a+b)$.

\end{proof}

\problem{8} If $a \ne 0$ and $b \ne 0$, then $(ab)^{-1} = a^{-1}b^{-1}$.

\begin{proof}
By definition (I.8), $(ab)^{-1}$ is the unique number such that
$(ab)(ab)^{-1} = 1$. But $a^{-1}b^{-1}$ also has this property:
$(ab)a^{-1}b^{-1} = aa^{-1}bb^{-1} = 1\cdot1 = 1$ (Axioms~1, 2 and 6).
Therefore, $(ab)^{-1} = a^{-1}b^{-1}$.
\end{proof}

\problem{9} $-(a/b) = (-a)/b = a/(-b)$ if $b\ne0$.

\begin{proof}
By definition (I.2), $-(a/b)$ is the unique number such that $a/b +
(-(a/b)) = 0$. We now show $(-a)/b$ also has this property:
\[
\begin{array}{rcll}
a/b + (-a)/b &=& a b^{-1} + (-a)b^{-1}&\mbox{(I.9)}\\
             &=& (a + (-a))b^{-1}&\mbox{(Axiom~3)}\\
             &=& 0\cdot b^{-1}&\mbox{(I.2)}\\
             &=& 0&\mbox{(I.6)}\\
\end{array}
\]
Therefore, $(-a)/b = -(a/b)$.

Finally, we prove the same result for $a/(-b)$. First observe that
$(-(b^{-1}))(-b) = b^{-1}b = 1$ by I.12 and I.8, so that $(-b)^{-1} =
-(b^{-1})$. Then
\[
\begin{array}{rcll}
a/b + a/(-b) &=& a b^{-1} + a(-b)^{-1}&\mbox{(I.9)}\\
             &=& a(b^{-1} + (-b)^{-1})&\mbox{(Axiom~3)}\\
             &=& a(b^{-1} - b^{-1})&\mbox{(just shown above)}\\
             &=& a\cdot0&\mbox{(I.2)}\\
             &=& 0&\mbox{(I.6)}\\
\end{array}
\]
\end{proof}

\problem{10} $(a/b) - (c/d) = (ad -bc)/(bd)$ if $b \ne 0$ and $d \ne 0$.

\begin{proof}
The proof I gave in class was the following: $(ad-bc)/(bd)$, is,
by definition (I.8), the unique number that when multiplied by $bd$
gives $ad - bc$.  But $(a/b) - (c/d)$ also has this property:
\[
\begin{array}{rcll}
bd((a/b) - (c/d)) &=& bdab^{-1} - bdcd^{-1} &\mbox{(I.5 and I.9)}\\
              &=& adbb^{-1} - bcdd^{-1} &\mbox{(Axiom 1)}\\
              &=& ad - bc &\mbox{(I.12)}|\
\end{array}
\]
Therefore, $(a/b) - (c/d) = (ad-bc)/(bd)$.

Here's a different proof:
\[
\begin{array}{rcll}
(a/b) - (c/d) &=& a/b + (-c)/d &\mbox{(Problem 9)}\\
              &=& (ad + (-c)b)/(bd) &\mbox{(I.13)}\\
              &=& (ad - cd)/(bd) &\mbox{(I.12)}|\
\end{array}
\]
\end{proof}

\textbf{Page 21.}

\problem{2}  There is no real number $x$ such that $x^2 + 1 = 0$.

\begin{proof}
By I.20, $x^2 > 0$ for any real number~$x$. Then, by I.18, Axiom~4,
and I.21, $x^2 + 1 > 0 + 1 = 1 > 0$. In particular, by I.16, $x^2 +
1 \ne 0$ for any real number~$x$.
\end{proof}

\problem{4} If $a > 0$, then $1/a > 0$; if $ a < 0$, then $1/a < 0$.

\begin{proof}
By definition (I.8) and I.21, $a (1/a) = 1 > 0$. If $a>0$, then I.24
implies that $1/a > 0$, and if $a<0$, then I.24 implies that $1/a < 0$.
\end{proof}

\problem{5}  If $ 0 < a < b$, then $0 < b^{-1} < a^{-1}$.

\begin{proof}
By assumption, $a,b\in\R^+$ and $b-a\in\R^+$. Axiom~7 says that
$ab\in\R^+$ and Problem~4 implies that $1/b > 0$ and $1/(ab) = (ab)^{-1}
> 0$.  Since $1/a - 1/b = (b-a)/(ab)$ by Problem~10 on p.19, and since
$(b-a)/(ab) = (b-a) (ab)^{-1} >0$  by I.9 and Axiom~7, we conclude that
$1/a - 1/b > 0$. Therefore, $0 < 1/b < 1/a$.
\end{proof}

\problem{9} There is no real number $a$ such that $x \le a$ for all
real~$x$.

\begin{proof}
(By contradiction.) Suppose there really is a real number $a$ such
that $x \le a$ for all real numbers~$x$ (note that $a$ is fixed and the
inequality must hold no matter what $x$ we choose).  Let us consider $x
= a+1$. Since $1>0$ by I.21, we find that $x = a+1 > a$ by I.18. This
contradicts our assumption that $x \le a$ (see I.16). Therefore our
assumption can never be true and there is no real number $a$ such that
$x \le a$ for all~$x$.
\end{proof}

\problem{10}  If $x$ has the property that $0 \le x < h$ for \emph{every}
positive real number $h$, then $x = 0$.

\begin{proof}
Let $x$ be a (fixed) real number that has the above property.
By assumption, $x \ge 0$. Suppose $x > 0$, and consider the positive real
number $x/2$. We are assuming that $x$ is less than \emph{every}
positive real number, including $x/2$. But $x < x/2$ implies that $x < 0$
(use I.18 and I.19). This contradiction implies our assumption that $x
>0$ is false and hence $x = 0$.
\end{proof}

\end{document}
