Editing Proof by infinite descent (section)

== Application examples ==

=== Irrationality of {{radic|2}} ===
The proof that the [[square root of 2]] ({{radic|2}}) is [[irrational number|irrational]] (i.e. cannot be expressed as a fraction of two whole numbers) was discovered by the [[ancient Greek]]s, and is perhaps the earliest known example of a proof by infinite descent. [[Pythagoreanism|Pythagoreans]] discovered that the diagonal of a square is incommensurable with its side, or in modern language, that the square root of two is [[irrational number|irrational]]. Little is known with certainty about the time or circumstances of this discovery, but the name of [[Hippasus]] of Metapontum is often mentioned. For a while, the Pythagoreans treated as an official secret the discovery that the square root of two is irrational, and, according to legend, Hippasus was murdered for divulging it.<ref>Stephanie J. Morris, [http://jwilson.coe.uga.edu/emt669/student.folders/morris.stephanie/emt.669/essay.1/pythagorean.html "The Pythagorean Theorem"], Dept. of Math. Ed., [[University of Georgia]].</ref><ref>Brian Clegg, [http://nrich.maths.org/2671 "The Dangerous Ratio ..."], Nrich.org, November 2004.</ref><ref>Kurt von Fritz, [https://www.jstor.org/pss/1969021 "The discovery of incommensurability by Hippasus of Metapontum"], Annals of Mathematics, 1945.</ref> The square root of two is occasionally called "Pythagoras' number" or "Pythagoras' Constant", for example {{harvtxt|Conway|Guy|1996}}.<ref>{{citation
 | last1 = Conway | first1 = John H. | author1-link = John H. Conway
 | last2 = Guy | first2 = Richard K. | author2-link = Richard K. Guy
 | page = 25
 | publisher = Copernicus
 | title = The Book of Numbers
 | year = 1996}}</ref>

The [[ancient Greek]]s, not having [[algebra]], worked out a [[Square root of 2#Geometric proof|geometric proof]] by infinite descent ([[John Horton Conway]] presented another geometric proof by infinite descent that may be more accessible<ref>{{Cite web|url=http://www.cut-the-knot.org/proofs/sq_root.shtml|title=Square root of 2 is irrational (Proof 8)|website=www.cut-the-knot.org|access-date=2019-12-10}}</ref>). The following is an [[algebra]]ic proof along similar lines:

Suppose that {{radic|2}} were [[rational number|rational]]. Then it could be written as

:<math>\sqrt{2} = \frac{p}{q}</math>

for two natural numbers, {{math|''p''}} and {{math|''q''}}. Then squaring would give

:<math>2 = \frac{p^2}{q^2}, </math>
:<math>2q^2 = p^2, </math>

so 2 must divide ''p''<sup>2</sup>. Because 2 is a [[prime number]], it must also divide ''p'', by [[Euclid's lemma]]. So ''p'' = 2''r'', for some integer ''r''.

But then,

:<math>2q^2 = (2r)^2 = 4r^2, </math>
:<math>q^2 = 2r^2, </math>

which shows that 2 must divide ''q'' as well. So ''q'' = 2''s'' for some integer ''s''.

This gives

:<math>\frac{p}{q}=\frac{2r}{2s}=\frac{r}{s}</math>.

Therefore, if {{radic|2}} could be written as a rational number, then it could always be written as a rational number with smaller parts, which itself could be written with yet-smaller parts, ''[[ad infinitum]]''. But [[Well-ordering principle|this is impossible in the set of natural numbers]]. Since {{radic|2}} is a [[real number]], which can be either rational or irrational, the only option left is for {{radic|2}} to be irrational.<ref>{{Cite web|url=https://kconrad.math.uconn.edu/ross2008/descent1.pdf|title=Infinite Descent|last=Conrad|first=Keith|date=August 6, 2008|website=kconrad.math.uconn.edu|access-date=2019-12-10}}</ref>

(Alternatively, this proves that if {{radic|2}} were rational, no "smallest" representation as a fraction could exist, as any attempt to find a "smallest" representation ''p''/''q'' would imply that a smaller one existed, which is a similar contradiction.)

=== Irrationality of {{radic|''k''}} if it is not an integer ===

For positive integer ''k'', suppose that {{radic|''k''}} is not an integer, but is rational and can be expressed as {{sfrac|''m''|''n''}} for natural numbers ''m'' and ''n'', and let ''q'' be the largest integer less than {{radic|''k''}} (that is, ''q'' is the [[Floor and ceiling functions|floor]] of {{radic|''k''}}).  Then

:<math>\begin{align}
\sqrt k &=\frac mn\\
[6pt] &=\frac{m\left(\sqrt k-q\right)}{n\left(\sqrt k-q\right)}\\
[6pt] &=\frac{m\sqrt k-mq}{n\sqrt k-nq}\\
[6pt] &=\frac{\left(n \sqrt k\right)\sqrt k-mq}{n \left(\frac{m}{n}\right) -nq}\\
[6pt] &=\frac{nk-mq}{m-nq}
\end{align}</math>

The numerator and denominator were each multiplied by the expression ({{radic|''k''}}&nbsp;−&nbsp;''q'')—which is positive but less than 1—and then simplified independently. So, the resulting products, say ''m′'' and ''n′'', are themselves integers, and are less than ''m'' and ''n'' respectively. Therefore, no matter what natural numbers ''m'' and ''n'' are used to express {{radic|''k''}}, there exist smaller natural numbers ''m′''&nbsp;<&nbsp;''m'' and ''n′''&nbsp;<&nbsp;''n'' that have the same ratio.  But infinite descent on the natural numbers is impossible, so this disproves the original assumption that {{radic|''k''}} could be expressed as a ratio of natural numbers.<ref>{{Citation | last = Sagher | first = Yoram |date=February 1988 | journal = [[American Mathematical Monthly]] | volume = 95 | page = 117 | title = What Pythagoras could have done | issue = 2 | doi=10.2307/2323064| jstor = 2323064 }}</ref>

===Non-solvability of ''r''<sup>2</sup> + ''s''<sup>4</sup> = ''t''<sup>4</sup> and its permutations===
{{See also|Fermat's right triangle theorem#Fermat's proof}}

The non-solvability of <math>r^2 + s^4 =t^4</math> in integers is sufficient to show the non-solvability of <math>q^4 + s^4 =t^4</math> in integers, which is a special case of [[Fermat's Last Theorem]], and the historical proofs of the latter proceeded by more broadly proving the former using infinite descent. The following more recent proof demonstrates both of these impossibilities by proving still more broadly that a [[Pythagorean triangle]] cannot have any two of its sides each either a square or twice a square, since there is no smallest such triangle:<ref>Dolan, Stan, "Fermat's method of ''descente infinie''", ''[[Mathematical Gazette]]'' 95, July 2011, 269–271.</ref>

Suppose there exists such a Pythagorean triangle. Then it can be scaled down to give a primitive (i.e., with no common factors other than 1) Pythagorean triangle with the same property. Primitive Pythagorean triangles' sides can be written as <math>x=2ab,</math> <math>y=a^2-b^2,</math>  <math>z=a^2+b^2</math>, with ''a'' and ''b'' [[coprime|relatively prime]] and with ''a+b'' odd and hence ''y'' and ''z'' both odd. The property that ''y'' and ''z'' are each odd means that neither  ''y'' nor ''z'' can be twice a square. Furthermore, if ''x'' is a square or twice a square, then each of ''a'' and ''b'' is a square or twice a square. There are three cases, depending on which two sides are postulated to each be a square or twice a square:

*'''''y'' and ''z''''': In this case, ''y'' and ''z'' are both squares. But then the right triangle with legs <math>\sqrt{yz}</math> and <math>b^2</math> and hypotenuse <math>a^2</math> also would have integer sides including a square leg (<math>b^2</math>) and a square hypotenuse (<math>a^2</math>), and would have a smaller hypotenuse (<math>a^2</math> compared to <math>z=a^2+b^2</math>).
*'''''z'' and ''x''''': ''z'' is a square. The integer right triangle with legs <math>a</math> and <math>b</math> and hypotenuse <math>\sqrt{z}</math> also would have two sides (<math>a</math> and <math>b</math>) each of which is a square or twice a square, and a smaller hypotenuse (<math>\sqrt{z}</math> compared to {{nowrap|<math>z</math>)}}.
*'''''y'' and ''x''''': ''y'' is a square. The integer right triangle with legs <math>b</math> and <math>\sqrt{y}</math> and hypotenuse <math>a</math> would have two sides (''b'' and ''a'') each of which is a square or twice a square, with a smaller hypotenuse than the original triangle (<math>a</math> compared to <math>z=a^2+b^2</math>).

In any of these cases, one Pythagorean triangle with two sides each of which is a square or twice a square has led to a smaller one, which in turn would lead to a smaller one, etc.; since such a sequence cannot go on infinitely, the original premise that such a triangle exists must be wrong.

This implies that the equations
:<math>r^2 + s^4 = t^4,</math>
:<math>r^4 + s^2 =t^4,</math> and
:<math>r^4 + s^4 =t^2</math>
cannot have non-trivial solutions, since non-trivial solutions would give Pythagorean triangles with two sides being squares.

For other similar proofs by infinite descent for the ''n'' = 4 case of Fermat's Theorem, see the articles by Grant and Perella<ref>Grant, Mike, and Perella, Malcolm, "Descending to the irrational", ''Mathematical Gazette'' 83, July 1999, pp. 263–267.</ref> and Barbara.<ref>Barbara, Roy, "Fermat's last theorem in the case ''n''&nbsp;=&nbsp;4", ''Mathematical Gazette'' 91, July 2007, 260–262.</ref>