Editing Square root of 2 (section)

===Proof by infinite descent===
One proof of the number's irrationality is the following [[proof by infinite descent]]. It is also a [[Proof by contradiction#Refutation_by_contradiction|proof of a negation by refutation]]: it proves the statement "<math>\sqrt{2}</math> is not rational" by assuming that it is rational and then deriving a falsehood.

# Assume that <math>\sqrt{2}</math> is a rational number, meaning that there exists a pair of integers whose ratio is exactly <math>\sqrt{2}</math>.
# If the two integers have a common [[divisor|factor]], it can be eliminated using the [[Euclidean algorithm]].
# Then <math>\sqrt{2}</math> can be written as an [[irreducible fraction]] <math>\frac{a}{b}</math> such that {{math|''a''}} and {{math|''b''}}  are [[coprime integers]] (having no common factor) which additionally means that at least one of {{math|''a''}} or {{math|''b''}} must be [[parity (mathematics)|odd]].
# It follows that <math>\frac{a^2}{b^2}=2</math> and <math>a^2=2b^2</math>. &emsp; (&thinsp;{{math|[[Exponent#Identities and properties|({{sfrac|''a''|''b''}}){{sup|''n''}} {{=}} {{sfrac|''a''{{sup|''n''}}|''b''{{sup|''n''}}}}]]}}&thinsp;) &emsp; ( {{math|''a''{{sup|2}} and ''b''{{sup|2}}}} are integers)
# Therefore, {{math|''a''{{sup|2}}}} is [[parity (mathematics)|even]] because it is equal to {{math|2''b''{{sup|2}}}}. ({{math|2''b''{{sup|2}}}} is necessarily even because it is 2 times another whole number.)
# It follows that {{math|''a''}} must be even (as squares of odd integers are never even).
# Because {{math|''a''}} is even, there exists an integer {{math|''k''}} that fulfills <math>a = 2k</math>.
# Substituting {{math|2''k''}} from step 7 for {{math|''a''}} in the second equation of step 4: <math>2b^2 = a^2 = (2k)^2 = 4k^2</math>, which is equivalent to <math>b^2=2k^2</math>.
# Because {{math|2''k''{{sup|2}}}} is divisible by two and therefore even, and because <math>2k^2=b^2</math>, it follows that {{math|''b''{{sup|2}}}} is also even which means that {{math|''b''}} is even.
# By steps 5 and 8, {{math|''a''}} and {{math|''b''}} are both even, which contradicts step 3 (that <math>\frac{a}{b}</math> is irreducible).

Since we have derived a falsehood, the assumption (1) that <math>\sqrt{2}</math> is a rational number must be false. This means that <math>\sqrt{2}</math> is not a rational number; that is to say, <math>\sqrt{2}</math> is irrational.

This proof was hinted at by [[Aristotle]], in his ''[[Prior Analytics|Analytica Priora]]'', §I.23.<ref>All that Aristotle says, while writing about [[Proof by contradiction|proofs by contradiction]], is that "the diagonal of the square is incommensurate with the side, because odd numbers are equal to evens if it is supposed to be commensurate".</ref> It appeared first as a full proof in [[Euclid]]'s ''[[Euclid's Elements|Elements]]'', as proposition 117 of Book X. However, since the early 19th century, historians have agreed that this proof is an [[Interpolation (manuscripts)|interpolation]] and not attributable to Euclid.<ref>The edition of the Greek text of the ''Elements'' published by E. F. August in [[Berlin]] in 1826–1829 already relegates this proof to an Appendix. The same thing occurs with [[Johan Ludvig Heiberg (historian)|J. L. Heiberg's]] edition (1883–1888).</ref>