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Next: 5.4 Logarithms. Up: 5. Area Previous: 5.2 Further Assumptions About   Index

5.3 Monotonic Functions

5.27   Definition (Partition.) Let $a,b$ be real numbers with $a\leq b$. A partition $P$ of the interval $[a,b]$ is a finite sequence of points

\begin{displaymath}P=\{x_0,x_1,\cdots ,x_n\}\end{displaymath}

with $a=x_0\leq x_1\leq x_2\cdots \leq x_n=b$. The intervals $[x_0,x_1],[x_1,x_2],\cdots$, $[x_{n-1},x_n]$ are called the subintervals of the partition $P$, and $[x_{j-1},x_j]$ is the $j^{th}$ subinterval of $P$ for $1\leq j\leq n$. The largest of the numbers $x_j-x_{j-1}$ is called the mesh of the partition $P$, and is denoted by $\mu (P)$. The partition

\begin{displaymath}\{ a,\; a+{{(b-a)}\over n},\; a+{{2(b-a)}\over n},\cdots , a+{{n(b-a)}\over
n}=b\}\end{displaymath}

is called the regular partition of $[a,b]$ into $n$ equal subintervals.

5.28   Example. Let

\begin{displaymath}P=\{ 0,{1\over {16}},{1\over 8},{1\over 4},{1\over 2},1\}\end{displaymath}

Then $P$ is a partition of $[0,1]$ into $5$ subintervals and $\displaystyle {\mu
(P)=1-{1\over
2}={1\over 2}}$.


The regular partition of $[1,2]$ into $5$ subintervals is $\displaystyle {\{1,{6\over
5},{7\over 5},{8\over 5},{9\over 5},2\}}$.


If $Q_n$ is the regular partition of $[a,b]$ into $n$ equal subintervals, then $\mu
(Q_n)={{b-a}\over n}$.

5.29   Exercise. Find a partition $P$ of $[0,1]$ into five subintervals, such that $\mu(P) = {4\over 5}$, or explain why no such partition exists.

5.30   Exercise. Find a partition $Q$ of $[0,1]$ into five subintervals, such that $\mu(Q) = {1\over 10}$, or explain why no such partition exists.

5.31   Definition (Monotonic function.) Let $J$ be an interval in $\mbox{{\bf R}}$, and let $f\colon J\to\mbox{{\bf R}}$ be a function. We say that $f$ is increasing on $J$ if
\begin{displaymath}\mbox{ for all }x,y \mbox{ in } J \Big[(x\leq y)\mbox{$\hspac...
...rightarrow\hspace{1ex}$}(f(x) \leq f(y))\Big]\mbox{{\nonumber}}\end{displaymath} (5.32)

and we say that $f$ is decreasing on $J$ if

\begin{displaymath}\mbox{ for all }x,y \mbox{ in } J \Big[(x\leq y)\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}(f(x)\geq f(y))\Big].\end{displaymath}

We say that $f$ is strictly increasing on $J$ if

\begin{displaymath}\mbox{ for all }x,y \mbox{ in } J \Big[(x < y) \mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}(f(x) < f(y))\Big]\end{displaymath}

and we say that $f$ is strictly decreasing on $J$ if

\begin{displaymath}\mbox{ for all }x,y \mbox{ in } J \Big[(x<y) \mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}(f(x) > f(y))\Big].\end{displaymath}

We say that $f$ is monotonic on J if $f$ is either increasing on $J$ or decreasing on $J$, and we say that $f$ is strictly monotonic on $J$ if $f$ is either strictly increasing or strictly decreasing on $J$.

\psfig{file=ch5b1.eps,width=1.5in} \psfig{file=ch5b2.eps,width=1.5in} \psfig{file=ch5b3.eps,width=1.5in}

A constant function on $J$ is both increasing and decreasing on $J$.


5.33   Notation ($S_a^bf$) Let $f$ be a function from the interval $[a,b]$ to the non-negative numbers. We will write

\begin{displaymath}S_a^b f=\{(x,y)\in\mbox{{\bf R}}^2\colon a\leq x\leq b \mbox{ and } 0\leq y\leq f(x)\},\end{displaymath}

i.e., $S_a^bf$ is the set of points under the graph of $f$ over the interval $[a,b]$.

Let $f$ be an increasing function from the interval $[a,b]$ to the non-negative numbers. Let $P=\{x_0,\cdots ,x_n\}$ be a partition of $[a,b]$ and let

\begin{eqnarray*}
I_a^b (f,P)&=& \bigcup_{i=1}^n B(x_{i-1},x_i\colon 0,f(x_{i-1}))\\
O_a^b (f,P)&=& \bigcup_{i=1}^n B(x_{i-1},x_i\colon 0,f(x_i)).
\end{eqnarray*}



Then
\psfig{file=ch5c.eps,width=5.5in}

\begin{displaymath}
I_a^b (f,P)\subset S_a^bf\subset O_a^b(f,P).
\end{displaymath} (5.34)

To see this, observe that since $f$ is increasing

\begin{displaymath}x_{i-1}\leq x\leq x_i\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}f(x_{i-1})\leq f(x)\leq f(x_i),\end{displaymath}

so

\begin{eqnarray*}
(x,y) \in I_a^b(f,P) &\mbox{$\Longrightarrow$}& (x,y) \in B(x_...
...mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}(x,y) \in S_a^bf.
\end{eqnarray*}



and also

\begin{eqnarray*}
(x,y)\in S_a^b f &\mbox{$\Longrightarrow$}& x_{i-1}\leq x\leq ...
... for some } i\\
&\mbox{$\Longrightarrow$}& (x,y)\in O_a^b(f,P).
\end{eqnarray*}



By equation (5.34) and monotonicity of area, we have
\begin{displaymath}
\alpha \Big( I_a^b(f,P)\Big)\leq\alpha (S_a^b f)\leq \alpha \Big(
O_a^b(f,P)\Big).
\end{displaymath} (5.35)

Now
$\displaystyle {\alpha \Big(O_a^b(f,P)\Big)-\alpha \Big( I_a^b (f,P)\Big)}\mbox{{}}$
  $\textstyle =$ $\displaystyle \sum_{i=1}^n (x_i-x_{i-1})f(x_i)-\sum_{i=1}^n
(x_i-x_{i-1})f(x_{i-1})$  
  $\textstyle =$ $\displaystyle \sum_{i=1}^n (x_i-x_{i-1})\Big(f(x_i)-f(x_{i-1})\Big).$ (5.36)

Now let $\mu (P)$ be the mesh of $P$ (cf. definition 5.27) so that

\begin{displaymath}0\leq x_i-x_{i-1}\leq\mu (P) \mbox{ for } 1\leq i\leq n.\end{displaymath}

Since $f(x_i)-f(x_{i-1})\geq 0$ for all $i$, we have

\begin{displaymath}(x_i-x_{i-1})\Big(f(x_i)-f(x_{i-1})\Big)\leq\mu (P)\Big(
f(x_i)-f(x_{i-1})\Big)\end{displaymath}

for all $i$, and hence
$\displaystyle \sum_{i=1}^n(x_i-x_{i-1})\Big(f(x_i)-f(x_{i-1})\Big)$ $\textstyle \leq$ $\displaystyle \sum_{i=1}^n\mu
(P)\Big(f(x_i)-f(x_{i-1})\Big)$  
  $\textstyle =$ $\displaystyle \mu (P)\sum_{i=1}^n\Big( f(x_i)-f(x_{i-1})\Big).$ (5.37)

Now

\begin{eqnarray*}
\sum_{i=1}^n\Big(f(x_i)-f(x_{i-1})\Big)&=&\Big(f(x_n)-f(x_{n-1...
...f(x_1)-f(x_0)\Big)\\
&=&f(x_n)-f(x_0)=f(b)-f(a).\phantom{\Big)}
\end{eqnarray*}



so by equations (5.37) and (5.36), we have

\begin{displaymath}\alpha\Big(O_a^b(f,P)\Big)-\alpha\Big(I_a^b(f,P)\Big)\leq\mu(P)\Big(f(b)-f(a)\Big).\end{displaymath}

Now suppose that $A$ is any real number that satisfies

\begin{displaymath}\alpha(I_a^b(f,P)) \leq A \leq \alpha(O_a^b(f,P)) \mbox{ for every partition
$P$ of $[a,b]$.} \end{displaymath}

We will show that $A = \alpha(S_a^b f).$ We have

\begin{displaymath}-\alpha(O_a^b(f,P)) \leq -A \leq -\alpha(I_a^b(f,P)).\end{displaymath}

It follows from (5.35) that

\begin{displaymath}\alpha(I_a^b(f,P))-\alpha(O_a^b(f,P)) \leq \alpha(S_a^b f) - A
\leq \alpha(O_a^b(f,P))-\alpha(I_a^b(f,P)).
\end{displaymath}

Thus
\begin{displaymath}
-\mu(P)(f(b) - f(a)) \leq
\alpha(S_a^b f) - A \leq
\mu(P)(f(b) - f(a))
\end{displaymath} (5.38)

for every partition $P$ of $[a,b]$. Since we can find partitions $P$ with $\mu (P)$ smaller than any preassigned number, it follows that
\begin{displaymath}
A = \alpha(S_a^bf).
\end{displaymath} (5.39)

(After we have discussed the notion of limit, we will come back and reconsider how (5.39) follows from (5.38). For the present, I will just say that the implication is intuitively clear.) We have now proved the following theorem:

5.40   Theorem. Let $f$ be an increasing function from the interval $[a,b]$ to $\mbox{{\bf R}}_{\geq 0}$, and let $P=\{x_0,x_1,\cdots ,x_n\}$ be a partition of $[a,b]$. Let

\begin{displaymath}S_a^b f=\{(x,y)\in\mbox{{\bf R}}^2\colon a\leq x\leq b \mbox{ and } 0\leq y\leq f(x)\},\end{displaymath}


$\displaystyle I_a^b(f,P)$ $\textstyle =$ $\displaystyle \bigcup_{i=1}^n B\Big(x_{i-1},x_i\colon
0,f(x_{i-1})\Big),$ (5.41)
$\displaystyle O_a^b(f,P)$ $\textstyle =$ $\displaystyle \bigcup_{i=1}^n B\Big(x_{i-1},x_i\colon
0,f(x_i)\Big),$ (5.42)
$\displaystyle A_a^b f$ $\textstyle =$ $\displaystyle \alpha(S_a^bf).$  

Then
\begin{displaymath}
\alpha\Big( I_a^b(f,P)\Big)\leq A_a^b f\leq\alpha\Big(
O_a^b(f,P)\Big)
\end{displaymath} (5.43)

and
\begin{displaymath}\alpha\Big(O_a^b(f,P)\Big)-\alpha\Big(I_a^b(f,P)\Big)\leq\mu(P)\Big(f(b)-f(a)\Big).\end{displaymath} (5.44)

Also
$\displaystyle \alpha\Big(
I_a^b(f,P)\Big)$ $\textstyle =$ $\displaystyle \sum_{i=1}^nf(x_{i-1})(x_i-x_{i-1})$ (5.45)
$\displaystyle \alpha\Big( O_a^b(f,P)\Big)$ $\textstyle =$ $\displaystyle \sum_{i=1}^nf(x_i)(x_i-x_{i-1}).$ (5.46)

If $A$ is any real number such that

\begin{displaymath}\alpha(I_a^b(f,P)) \leq A \leq \alpha(O_a^b(f,P)) \mbox{ for every partition
$P$ of $[a,b]$}, \end{displaymath}

then

\begin{displaymath}A = A_a^b(f). \end{displaymath}

The following picture illustrates the previous theorem.
\psfig{file=ch5f.eps,width=5in}

5.47   Exercise. A version of theorem 5.40 for decreasing functions is also valid. To get this version you should replace the word ``increasing" by ``decreasing" and change lines (5.41), (5.42), (5.44), (5.45) and (5.46). Write down the proper versions of the altered lines. As usual, use $I$ to denote areas inside $S_a^bf$ and $O$ to denote sets containing $S_a^bf$. Draw a picture corresponding to the above figure for a decreasing function.

5.48   Definition (Right triangle $T_{\mathbf{{\bf c}}}$) Let $a$ and $b$ be non-zero real numbers, and let $\mathbf{{\bf c}}=(a,b)$. We define the triangle $T_{\mathbf{{\bf c}}} = T_{(a,b)}$ to be the set of points between the line segment $[{\bf0}\mathbf{{\bf c}}]$ and the $x$-axis. By example 4.8, we know that the equation of the line through ${\bf0}$ and $\mathbf{{\bf c}}$ is $y= {b\over a}x$. Hence we have:

If $a>0$ and $b>0$, then $T_{(a,b)} = \{(x,y): 0 \leq x \leq a
\mbox{ and }0 \leq y \leq {b\over a}x \}$


\begin{picture}(5,1.5)(-2.8,-.5)
\put(-.25,0){\vector(1,0){1.75}}
\put(0,-.25){\...
...-.8,-.25){$(0,0)$}
\put(1.2,1.01){$(a,b)$}
\put(1.2,-.25){$(a,0)$}
\end{picture}

If $a>0$ and $b<0$, then $T_{(a,b)} = \{(x,y): 0 \leq x \leq a
\mbox{ and }{b\over a}x \leq y \leq 0\}$


\begin{picture}(5,1.5)(-2.8,-.5)
\put(-.25,1.){\vector(1,0){1.75}}
\put(0,-.25){...
...t(-.8,.75){$(0,0)$}
\put(1.2,.75){$(a,0)$}
\put(1.2,-.25){$(a,b)$}
\end{picture}

If $a<0$ and $b<0$, then $T_{(a,b)} = \{(x,y): a \leq x \leq 0
\mbox{ and }{b\over a}x \leq y \leq 0\}$


\begin{picture}(5,1.5)(-2.8,-.5)
\put(-1.5,1.){\vector(1,0){1.75}}
\put(0,-.25){...
...\put(.1,.75){$(0,0)$}
\put(-2,.75){$(a,0)$}
\put(-2,-.15){$(a,b)$}
\end{picture}

If $a<0$ and $b>0$, then $T_{(a,b)} = \{(x,y): a \leq x \leq 0
\mbox{ and }0 \leq y \leq {b\over a}x \}$


\begin{picture}(5,1.5)(-2.8,-.5)
\put(-1.5,0.){\vector(1,0){1.75}}
\put(0,-.25){...
...put(.1,-.25){$(0,0)$}
\put(-2,.75){$(a,b)$}
\put(-2,-.25){$(a,0)$}
\end{picture}

5.49   Remark. We know from Euclidean geometry that
\begin{displaymath}
\alpha(T_{(a,b)}) = {1\over 2} \vert a\vert\vert b\vert.
\end{displaymath} (5.50)

I would like to show that this relation follows from our assumptions about area. If $H$, $V$ and $R_\pi$ are the reflections and rotation defined in definition 4.9, then we can show without difficulty that for $a>0$ and $b>0$

\begin{eqnarray*}
T_{(-a,b)} &=& H(T{(a,b)}),\\
T_{(a,-b)} &=& V(T_{(a,b)}), \mbox{ and }\\
T_{(-a,-b)} &=& R_\pi(T_{(a,b)})
\end{eqnarray*}



so by invariance of area under symmetry,

\begin{displaymath}\alpha(T_{(a,b)}) = \alpha(T_{(-a,b)})
=\alpha(T_{(a,-b)}) = \alpha(T_{(-a,-b)})\end{displaymath}

when $a$ and $b$ are positive. It follows that if we prove formula (5.50) when $a$ and $b$ are positive, then the formula holds in all cases. For example if $a$ and $b$ are positive, and we know that (5.50) holds when $a$ and $b$ are positive, we get

\begin{displaymath}\alpha(T_{(-a,b)}) =\alpha(T_{(a,b)}) = {1\over 2}\vert a\vert\vert b\vert
= {1\over 2}\vert-a\vert\vert b\vert,\end{displaymath}

and thus our formula holds when $a$ is negative and $b$ is positive.

5.51   Theorem. Let $a$ and $b$ be non-zero real numbers, and let $T_{(a,b)}$ be the set defined in definition 5.48. Then

\begin{displaymath}
\alpha(T_{(a,b)}) = {1\over 2} \vert a\vert\vert b\vert.
\end{displaymath}

Proof: By the previous remark, if is sufficient to prove the theorem for the case when $a$ and $b$ are positive. So suppose that $a$ and $b$ are positive.


\begin{picture}(5,3)(-2.8,-1.5)
\put(-1.5,0){\vector(1,0){3}}
\put(0,-1){\vector...
....5,1){$(0,b)$}
\put(.1,.7){$E$}
\put(-1.2,-.3){$R_\pi(T_{(a,b)})$}
\end{picture}
Let $E = (a,b) + R_\pi(T_{(a,b)})$. It appears from the figure, and is straightforward to show, that

\begin{displaymath}E = \{(x,y):0\leq x\leq a \mbox{ and }{b\over a}x \leq y \leq b\}.\end{displaymath}

By translation invariance of area,

\begin{displaymath}\alpha(E) = \alpha(R_\pi(T_{(a,b)})) = \alpha(T_{(a,b)}).
\end{displaymath}

We have

\begin{displaymath}E \cup T_{(a,b)} = B(0,a:0,b),\end{displaymath}

and

\begin{displaymath}E \cap T_{(a,b)} = [{\bf0}\mathbf{{\bf c}}] \mbox{ where } \mathbf{{\bf c}}= (a,b).\end{displaymath}

By the addition rule for area (assumption 5.14) we have

\begin{eqnarray*}
ab &=& \alpha(B(0,a:0,b))\\
&=& \alpha( E \cup T_{(a,b)} ) ...
...)})\\
&=& 2\alpha(T_{(a,b)}) -\alpha([{\bf0}\mathbf{{\bf c}}]),
\end{eqnarray*}



i.e.,

\begin{displaymath}\alpha(T_{(a,b)}) = {1\over 2}ab +{1\over 2}\alpha([{\bf0}\mathbf{{\bf c}}]).\end{displaymath}

Thus our theorem will follow if we can show that the segment $[{\bf0}\mathbf{{\bf c}}]$ is a zero-area set. We will prove this as the next theorem.

5.52   Theorem. Let $\mathbf{{\bf c}}=(a,b)$ be a point in $\mbox{{\bf R}}^2$. Then

\begin{displaymath}\alpha([\bf0\mathbf{{\bf c}}]) = 0. \end{displaymath}

Proof: If $a=0$ or $b=0$, then ${[{\bf0}\mathbf{{\bf c}}]}$ is a box with width equal to zero, or height equal to zero, so the theorem holds in this case. Hence we only need to consider the case where $a$ and $b$ are non-zero. Since any segment $[{\bf0}\mathbf{{\bf c}}]$ can be rotated or reflected to a segment $[{\bf0}\mbox{{\bf q}}]$ where $\mbox{{\bf q}}$ is in the first quadrant, we may further assume that $a$ and $b$ are both positive. Let $n$ be a positive integer, and for $1\leq j\leq n$ let

\begin{displaymath}B^n_j = B\left({a(j-1)\over n},{aj\over n}:
{b(j-1)\over n},{bj\over n}\right).\end{displaymath}


\begin{picture}(14,10)(-2,-2)
\put(0,0){\line(1,0){10}}
\put(0,1){\line(1,0){10}...
...\line(2,1){10}}
\thicklines\multiput(0,0)(2,1){5}{\framebox (2,1)}
\end{picture}
Then
\begin{displaymath}
{[}{\bf0}\mathbf{{\bf c}}] \subset \bigcup_{j=1}^n B^n_j,
\end{displaymath} (5.53)

since

\begin{eqnarray*}
\mbox{{\bf x}}\in [{\bf0}\mathbf{{\bf c}}] &\mbox{$\Longrighta...
...box{{\bf x}}\in B^n_j \mbox{ for some $j$ with } 1\leq j\leq n.
\end{eqnarray*}



For each $j$ we have

\begin{displaymath}\alpha(B^n_j) = {a\over n}\cdot{b\over n} = {ab\over n^2}.\end{displaymath}

Also the sets $B^n_j$ and $B^n_k$ are almost disjoint whenever $1 \leq j,k \leq n$ and $j\neq k$. (If $j$ and $k$ differ by more than 1, then $B^n_j$ and $B^n_k$ are disjoint, and if $j$ and $k$ differ by 1, then $B^n_j\cap B^n_k$ consists of a single point.) By additivity for almost-disjoint sets (assumption 5.25), it follows that

\begin{displaymath}\alpha(\bigcup_{j=1}^n B^n_j) = \sum_{j=1}^n\alpha(B^n_j)
=\sum_{j=1}^n {ab\over n^2} = {nab\over n^2} ={ab\over n}.\end{displaymath}

By (5.53) and monotonicity of area we have
\begin{displaymath}
\alpha([0\mathbf{{\bf c}}]) \leq \alpha(\bigcup_{j=1}^n B^n_j) = {ab\over n}
\mbox{ for every positive integer } n.
\end{displaymath} (5.54)

In order to conclude from this that $\alpha([{\bf0}\mathbf{{\bf c}}]) = 0$ We now make use of the Archimedean property of the real numbers, which says that for any real number $x$ there is a positive integer $n$ with $n> x$. We know $\alpha([{\bf0}\mathbf{{\bf c}}])\geq 0$, since all areas are non-negative. Suppose (in order to get a contradiction) that $\alpha([{\bf0}\mathbf{{\bf c}}])$ is positive. Then by the Archimedean property, there is a positive integer $N$ such that $N > {ab\over \alpha([{\bf0}\mathbf{{\bf c}}])}$. This implies that $\alpha([{\bf0}\mathbf{{\bf c}}]) > {ab\over N}$, and this contradicts (5.54). Hence $\alpha([{\bf0}\mathbf{{\bf c}}])$ is not positive, and we conclude that $\alpha([{\bf0}\mathbf{{\bf c}}]) = 0$. $\diamondsuit$


Archimedes' statement of the Archimedean property differs from our statement. He assumes that

Further, of unequal lines, unequal surfaces, and unequal solids, the greater exceeds the less by such a magnitude as, when added to itself, can be made to exceed any assigned magnitude among those which are comparable with [it and with] one another.[2, page 4]

5.55   Exercise. Let $\mbox{{\bf a}}$ and $\mbox{{\bf b}}$ be points in $\mbox{{\bf R}}^2$. Show that segment $[\mbox{{\bf a}}\mbox{{\bf b}}]$ is a zero are set. (Use theorem 5.52. Do not reprove theorem 5.52).

5.56   Entertainment (Area of a triangle) Let $\mbox{{\bf x}}_1 = (x_1,y_1)$, $\mbox{{\bf x}}_2 = (x_2,y_2)$ and $\mbox{{\bf x}}_3 = (x_3,y_3)$ be three points in $\mbox{{\bf R}}^2$, and let $T$ be the triangle with vertices $\mbox{{\bf x}}_1$, $\mbox{{\bf x}}_2$ and $\mbox{{\bf x}}_3$. Let

\begin{eqnarray*}
x_s &=& \mbox{smallest of $x_1$, $x_2$ and $x_3$}\\
x_l &=& ...
...and $y_3$}\\
y_l &=& \mbox{largest of $y_1$, $y_2$ and $y_3$}.
\end{eqnarray*}



Then the box $B(x_s,x_l:y_s,y_l)$ is an almost-disjoint union of $T$ and three triangles which are translates of triangles of the form $T_{\mathbf{{\bf c}}}$. Since you know how to find the area of a box and of a triangle $T_{\mathbf{{\bf c}}}$, you can find the area of $T$.

\begin{picture}(10,8)(-2,-2)
% put(0,-1)\{ vector(0,1)\{6\}\}
\put(0,-1){\line(0...
...}
\put(0,4){\line(1,0){5}}
\put(5,0){\line(0,1){4}}
\put(2,2){$T$}
\end{picture}
Using this remark show that for the triangles pictured below, $\alpha(T^1) = {1\over 2} (a_1b_2- a_2 b_1)$, and $\alpha(T^2) ={1\over 2} (a_2b_1-a_1b_2)$.

\begin{picture}( 10,7)(-2,-2)
\put(0,0){\line(1,0){5}}
\put(0,0){\line(0,1){4}}
...
...,b_2)$}
\put(-1,-1){$0<b_1<a_1$ and $0<b_2<a_2$}
\put(2,2){$T^1$}
\end{picture}

\begin{picture}(10,7)(-4,-2)
\put(0,0){\line(3,2){3}}
\put(0,0){\line(-1,2){2}}
...
...,a_2)$}
\put(0,2){$T^2$}
\put(-2,-1){$a_1<0<b_1$ and $0<b_2<a_2$}
\end{picture}
Then choose another triangle $T^3$ with vertices ${\bf0},\mbox{{\bf a}}$ and $\mbox{{\bf b}}$, where the coordinates of the points are related in a way different from the ways shown for $T^1$ and $T^2$, and calculate the area of $T^3$. You should find that

\begin{displaymath}\alpha(T^3) = {1\over 2} \vert a_1b_2-a_2b_1\vert\end{displaymath}

in all cases. Notice that if some coordinate is zero, the formula agrees with theorem 5.51.


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Next: 5.4 Logarithms. Up: 5. Area Previous: 5.2 Further Assumptions About   Index
Ray Mayer 2007-09-07