Editing Proof by infinite descent (section)

=== 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.)