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A method of proof used by mathematicians of antiquity in order to determine areas and volumes. The name  "method of exhaustion"  was introduced in the 17th century.
 
A method of proof used by mathematicians of antiquity in order to determine areas and volumes. The name  "method of exhaustion"  was introduced in the 17th century.
  
The typical scheme of proof by the method of exhaustion can, in modern terms, be explained as follows. In order to determine a quantity <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e0368401.png" /> one constructs a certain sequence of quantities <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e0368402.png" /> such that
+
The typical scheme of proof by the method of exhaustion can, in modern terms, be explained as follows. In order to determine a quantity $  A $
 +
one constructs a certain sequence of quantities $  C _ {1} , C _ {2} \dots $
 +
such that
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e0368403.png" /></td> <td valign="top" style="width:5%;text-align:right;">(1)</td></tr></table>
+
$$ \tag{1 }
 +
C _ {n}  < A ;
 +
$$
  
one assumes that a <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e0368404.png" /> is known such that
+
one assumes that a $  B $
 +
is known such that
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e0368405.png" /></td> <td valign="top" style="width:5%;text-align:right;">(2)</td></tr></table>
+
$$ \tag{2 }
 +
C _ {n}  < B ,
 +
$$
  
and that for any integer <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e0368406.png" /> and all sufficiently large <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e0368407.png" /> the inequalities
+
and that for any integer $  K $
 +
and all sufficiently large $  n $
 +
the inequalities
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e0368408.png" /></td> <td valign="top" style="width:5%;text-align:right;">(3)</td></tr></table>
+
$$ \tag{3 }
 +
K ( A - C _ {n} )  < D ,\ \
 +
K ( B - C _ {n} ) < D
 +
$$
  
are fulfilled, with <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e0368409.png" /> a constant. From the modern point of view, to transfer (3) to
+
are fulfilled, with $  D $
 +
a constant. From the modern point of view, to transfer (3) to
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684010.png" /></td> <td valign="top" style="width:5%;text-align:right;">(4)</td></tr></table>
+
$$ \tag{4 }
 +
= B
 +
$$
  
 
one only has to notice that (1)–(3) imply
 
one only has to notice that (1)–(3) imply
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684011.png" /></td> </tr></table>
+
$$
 +
\lim\limits _ {n \rightarrow \infty }
 +
( A - C _ {n} )  = 0 ,\ \
 +
\lim\limits _ {n \rightarrow \infty }
 +
( B - C _ {n} )  = 0 ,
 +
$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684012.png" /></td> </tr></table>
+
$$
 +
= \lim\limits _ {n \rightarrow \infty }  C _ {n}  = B .
 +
$$
  
The mathematicians of antiquity, not having developed the theory of limits (cf. [[Limit|Limit]]), used a reductio ad absurdum argument here: they proved that neither of the inequalities <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684013.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684014.png" /> is possible. To disprove the first one, they established by the [[Archimedean axiom|Archimedean axiom]] that for <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684015.png" /> there exists a <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684016.png" /> such that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684017.png" />, and (1) then led to
+
The mathematicians of antiquity, not having developed the theory of limits (cf. [[Limit|Limit]]), used a reductio ad absurdum argument here: they proved that neither of the inequalities $  A < B $,  
 +
$  A > B $
 +
is possible. To disprove the first one, they established by the [[Archimedean axiom|Archimedean axiom]] that for $  R = B - A $
 +
there exists a $  K $
 +
such that $  K R > D $,  
 +
and (1) then led to
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684018.png" /></td> </tr></table>
+
$$
 +
K ( B - C _ {n} )  > \
 +
K ( B - A )  > D ,
 +
$$
  
 
which contradicts the second inequality in (3). The other assertion is disproved in a similar way. Hence (4) remains.
 
which contradicts the second inequality in (3). The other assertion is disproved in a similar way. Hence (4) remains.
  
The introduction of the method of exhaustion and of the axiom that lies at its foundation is ascribed to Eudoxus of Cnidus. The method was extensively used by Eudoxus, while Archimedes used it with extraordinary skill and variety. E.g., in order to determine the area <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684019.png" /> of a segment of a parabola, Archimedes constructs the areas <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684020.png" /> of segments that are stepwise  "exhausting"  the area <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684021.png" />.
+
The introduction of the method of exhaustion and of the axiom that lies at its foundation is ascribed to Eudoxus of Cnidus. The method was extensively used by Eudoxus, while Archimedes used it with extraordinary skill and variety. E.g., in order to determine the area $  A $
 +
of a segment of a parabola, Archimedes constructs the areas $  C _ {1} , C _ {2} \dots $
 +
of segments that are stepwise  "exhausting"  the area $  A $.
  
 
Here
 
Here
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684022.png" /></td> </tr></table>
+
$$
 +
C _ {2}  = C _ {1} +
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684023.png" /></td> </tr></table>
+
\frac{1}{4}
 +
C _ {1} ,
 +
$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684024.png" /></td> </tr></table>
+
$$
 +
{\dots \dots \dots \dots }
 +
$$
 +
 
 +
$$
 +
C _ {n}  = C _ {1} +
 +
\frac{1}{4}
 +
C _ {1} + \dots +
 +
\frac{1}{4  ^ {n-} 1 }
 +
C _ {1} .
 +
$$
  
 
Instead of the limit transition
 
Instead of the limit transition
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684025.png" /></td> </tr></table>
+
$$
 +
= \lim\limits _
 +
{n \rightarrow \infty } \
 +
C _ {n}  = \
 +
\left (
 +
1 +
 +
\frac{1}{4}
 +
+
 +
\frac{1}{16}
 +
+ \dots
 +
\right )
 +
C _ {1}  =
 +
\frac{4}{3}
 +
C _ {1} ,
 +
$$
  
Archimedes proves geometrically that for any <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684026.png" />,
+
Archimedes proves geometrically that for any $  n $,
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684027.png" /></td> </tr></table>
+
$$
 +
A - C _ {n}  < \
 +
 
 +
\frac{1}{4  ^ {n-} 1 }
 +
 
 +
C _ {1} .
 +
$$
  
 
Introducing the area
 
Introducing the area
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684028.png" /></td> </tr></table>
+
$$
 +
=
 +
\frac{4}{3}
 +
C _ {1} ,
 +
$$
  
 
he obtains
 
he obtains
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684029.png" /></td> </tr></table>
+
$$
 +
B - C _ {n}  = \
 +
 
 +
\frac{1}{3 \cdot 4  ^ {n-} 1 }
 +
 
 +
C _ {1} ,
 +
$$
  
 
and, following the reasoning explained above, finishes his proof with
 
and, following the reasoning explained above, finishes his proof with
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036840/e03684030.png" /></td> </tr></table>
+
$$
 
+
= = \
  
 +
\frac{4}{3}
 +
C _ {1} .
 +
$$
  
 
====Comments====
 
====Comments====
 
  
 
====References====
 
====References====
 
<table><TR><TD valign="top">[a1]</TD> <TD valign="top">  C.B. Boyer,  "A history of mathematics" , Wiley  (1968)  pp. 100; 142–146</TD></TR></table>
 
<table><TR><TD valign="top">[a1]</TD> <TD valign="top">  C.B. Boyer,  "A history of mathematics" , Wiley  (1968)  pp. 100; 142–146</TD></TR></table>

Latest revision as of 19:38, 5 June 2020


A method of proof used by mathematicians of antiquity in order to determine areas and volumes. The name "method of exhaustion" was introduced in the 17th century.

The typical scheme of proof by the method of exhaustion can, in modern terms, be explained as follows. In order to determine a quantity $ A $ one constructs a certain sequence of quantities $ C _ {1} , C _ {2} \dots $ such that

$$ \tag{1 } C _ {n} < A ; $$

one assumes that a $ B $ is known such that

$$ \tag{2 } C _ {n} < B , $$

and that for any integer $ K $ and all sufficiently large $ n $ the inequalities

$$ \tag{3 } K ( A - C _ {n} ) < D ,\ \ K ( B - C _ {n} ) < D $$

are fulfilled, with $ D $ a constant. From the modern point of view, to transfer (3) to

$$ \tag{4 } A = B $$

one only has to notice that (1)–(3) imply

$$ \lim\limits _ {n \rightarrow \infty } ( A - C _ {n} ) = 0 ,\ \ \lim\limits _ {n \rightarrow \infty } ( B - C _ {n} ) = 0 , $$

$$ A = \lim\limits _ {n \rightarrow \infty } C _ {n} = B . $$

The mathematicians of antiquity, not having developed the theory of limits (cf. Limit), used a reductio ad absurdum argument here: they proved that neither of the inequalities $ A < B $, $ A > B $ is possible. To disprove the first one, they established by the Archimedean axiom that for $ R = B - A $ there exists a $ K $ such that $ K R > D $, and (1) then led to

$$ K ( B - C _ {n} ) > \ K ( B - A ) > D , $$

which contradicts the second inequality in (3). The other assertion is disproved in a similar way. Hence (4) remains.

The introduction of the method of exhaustion and of the axiom that lies at its foundation is ascribed to Eudoxus of Cnidus. The method was extensively used by Eudoxus, while Archimedes used it with extraordinary skill and variety. E.g., in order to determine the area $ A $ of a segment of a parabola, Archimedes constructs the areas $ C _ {1} , C _ {2} \dots $ of segments that are stepwise "exhausting" the area $ A $.

Here

$$ C _ {2} = C _ {1} + \frac{1}{4} C _ {1} , $$

$$ {\dots \dots \dots \dots } $$

$$ C _ {n} = C _ {1} + \frac{1}{4} C _ {1} + \dots + \frac{1}{4 ^ {n-} 1 } C _ {1} . $$

Instead of the limit transition

$$ A = \lim\limits _ {n \rightarrow \infty } \ C _ {n} = \ \left ( 1 + \frac{1}{4} + \frac{1}{16} + \dots \right ) C _ {1} = \frac{4}{3} C _ {1} , $$

Archimedes proves geometrically that for any $ n $,

$$ A - C _ {n} < \ \frac{1}{4 ^ {n-} 1 } C _ {1} . $$

Introducing the area

$$ B = \frac{4}{3} C _ {1} , $$

he obtains

$$ B - C _ {n} = \ \frac{1}{3 \cdot 4 ^ {n-} 1 } C _ {1} , $$

and, following the reasoning explained above, finishes his proof with

$$ A = B = \ \frac{4}{3} C _ {1} . $$

Comments

References

[a1] C.B. Boyer, "A history of mathematics" , Wiley (1968) pp. 100; 142–146
How to Cite This Entry:
Exhaustion, method of. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Exhaustion,_method_of&oldid=46872
This article was adapted from an original article by BSE-3 (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article