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\title[Absolute Values and Intervals]
{Absolute Values and Intervals}

\institute[College of the Holy Cross]
{
   MATH 133 -- Calculus with Fundamentals, section 1, Prof. Little\\
}

%\date[September 4, 2015]
%{September 4, 2015}

\begin{document}

\begin{frame}
  \titlepage
\end{frame}

\begin{frame}
{\bf Absolute Value}

In everyday terms, the absolute value leaves nonnegative numbers unchanged, but 
``flips'' a negative number to the corresponding positive number.  The ``flipping''
can be done by multiplying by $-1$, so as a formula, 
if $x$ is a real number, then 
\[
|x| = \begin{cases} x & {\rm if }\ x \ge 0\\
                   -1\cdot x = -x & {\rm if }\ x < 0.
      \end{cases}
\]
For instance, 
\begin{itemize}
\item<2-> since 15.9 and 0 are both $\ge 0$, 
\[
|15.9| = 15.9 \quad {\rm and} \quad |0| = 0
\]
\item<3-> but since $-8.3 < 0$, 
\[
|-8.3| = -(-8.3) = +8.3.
\]
\end{itemize} 

\end{frame}

\begin{frame}
{\bf Geometric meaning}

We know that the collection of all real numbers can be seen as the set of 
points on the \emph{number line}.

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\put(50,5){\vector(-1,0){40}}
\put(50,4){\line(0,1){2}}
\put(49,8){$0$}
\end{picture}
\]

\begin{itemize}
\item<1->If $x \ge 0$, then $|x|$ is the distance from $0$ to $x$ along the line:
\[
\begin{picture}(100,12)
\put(50,5){\vector(1,0){40}}
\put(50,5){\vector(-1,0){40}}
\put(50,4){\line(0,1){2}}
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\put(49,8){$0$}
\put(69,8){$x$}
\put(50,6){\vector(1,0){20}}
\put(58,8){$|x|$}
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\]
\item<2->If $x < 0$, $|x|$ is still the \emph{positive} distance along the line:
\[
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\put(50,5){\vector(1,0){40}}
\put(50,5){\vector(-1,0){40}}
\put(50,4){\line(0,1){2}}
\put(30,4){\line(0,1){2}}
\put(49,8){$0$}
\put(29,8){$x$}
\put(30,6){\vector(1,0){20}}
\put(38,8){$|x|$}
\end{picture}
\]
\end{itemize}
\end{frame}

\begin{frame}
{\bf Similarly, ... }

If $a$ and $b$ are \emph{any} two real numbers, then $|a - b|$ is the (positive) distance along
the number line from $a$ to $b$ (or what is the same, from $b$ to $a$, if the numbers satisfy 
$b < a$ as in the following picture):
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\put(50,5){\vector(-1,0){40}}
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\put(29,8){$|a-b|$}
\end{picture}
\]
For instance, $|(-3) - 2| = |-5| = 5$ gives the distance from $b = -3$ to $a = 2$ along the 
number line.
\end{frame}

\begin{frame}
{\bf Intervals}

We will use this notation for intervals on the real number line:
\begin{itemize}
\item The {\bf closed interval} $[a,b]$ is the set of all real $x$ with $a \le x$ and $x \le b$.  It is more common to write
$a \le x \le b$ here, but you should always understand that this is the same as the two separate 
inequalities $a \le x$ and $x \le b$ given first.
\item The {\bf open interval} $(a,b)$ is the set of all real $x$ with $a < x$ and $x < b$, or $a < x < b$ (not including the
endpoints $x = a, b$).
\item The {\bf half-closed interval} $[a,b)$ includes the endpoint $a$ (and all $x$ 
strictly between $a$ and $b$), but not $b$.
\item Similarly, the {\bf half-closed interval} $(a,b]$ includes the endpoint $b$, but not $a$.
\end{itemize}

\end{frame}

\begin{frame}
{\bf An example}

\begin{itemize}
\item<1->Let's ask:  What set of real numbers $x$ is defined by the condition
$|x - 2| < 3$?
\item<2->In geometric terms, the distance $d$ from $x$ to $2$ along the number line is strictly less than 3.
So $x$ must be less than $5$ and greater than $-1$:
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\[
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\put(50,5){\vector(1,0){50}}
\put(50,5){\vector(-1,0){50}}
\put(70,4){\line(0,1){2}}
\put(69,7){$2$}
\put(70,6){\vector(1,0){30}}
\put(70,6){\vector(-1,0){30}}
\put(100,4){\line(0,1){2}}
\put(99,7){$5$}
\put(40,4){\line(0,1){2}}
\put(35,7){$-1$}
\put(80,7){$d = 3$}
\put(50,7){$d = 3$}
\end{picture}
\]
\item<3->That is, $|x - 2| < 3$ says the same thing as $-1 < x < 5$, so our set 
is the open interval $(-1,5)$. 
\end{itemize}
\end{frame}

\begin{frame}
{\bf Another way to see this with algebra}

For any real number $a$, the statement $|a| < r$ says the same thing as $-r < a < r$ (think of 
the distance from $a$ to $0$).  So

\begin{eqnarray*}
|x - 2| < 3 & {\rm is\ the\ same\ as} & -3 < x - 2 < 3,\\
& {\rm which\ is\ the\ same\ as} & (-3) + 2 < x < 3 + 2,\\
& {\rm which\ is\ the\ same\ as} & -1 < x < 5.
\end{eqnarray*}

So as before, the set of all such $x$ is the interval $(-1,5)$.  
\end{frame}

\begin{frame}
{\bf Any interval can be described with absolute values}

\begin{itemize}
\item<1-> In fact, any open interval $(a,b)$ can be described in the form we saw in the
last example.
\item<2-> To illustrate this, let's consider the example of $(4,7)$.  
\item<3-> We want to write ``the set of all $x$ with $4 < x < 7$'' in the form ``the 
set of all $x$ with $|x - c| < r$ for some $c$ and $r$.''
\item<4-> Thinking about what happened before, we see that $c$ should be exactly \emph{half-way
between} $4$ and $7$. 
\item<5-> Thus, $c$ is the \emph{midpoint} of the interval: $c = \frac{4 + 7}{2} = \frac{11}{2} = 5.5$.
\item<6-> Then, $r$ is the distance from the midpoint to either endpoint:
$|7 - 5.5| = 1.5$ and $|4 - 5.5| = 1.5$.  
\item<7-> Conclusion:  The interval $(4,7)$ is the set of $x$ with $|x - 5.5| < 1.5$.    
\end{itemize}
\end{frame}

\begin{frame}
{\bf The general pattern}

If $a < b$, then the open interval $(a,b)$:

\begin{itemize}
\item has midpoint $c = \frac{a + b}{2}$, and
\item the distance from $c$ to either endpoint is 
\[
\left|b - \frac{a+b}{2}\right| = \frac{b - a}{2} = \left|a - \frac{a+b}{2}\right|.
\]
\item So the interval is the set of all $x$ with 
\[
\left|x - \frac{a+b}{2}\right| < \frac{b - a}{2}.
\]
\item No need to memorize this formula, but understand the process of how we got it!
\end{itemize}
\end{frame}
\end{document}

