Editing Proof by infinite descent (section)

=== 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>