Editing Proof by exhaustion (section)

==Example==
Proof by exhaustion can be used to prove that if an [[integer]] is a [[perfect cube]], then it must be either a multiple of 9, 1 more than a multiple of 9, or 1 less than a multiple of 9.<ref>{{Cite web| url=https://www.m-a.org.uk/resources/2017-Sep-Proof-Glaisters-MiS.pdf| title=Mathematical argument, language and proof — AS/A Level 2017| last=Glaister| first=Elizabeth|last2=Glaister|first2=Paul|date=September 2017|website=Mathematical Association| access-date=October 25, 2019}}</ref>

'''Proof''':
<br>Each perfect cube is the cube of some integer ''n'', where ''n'' is either a multiple of 3, 1 more than a multiple of 3, or 1 less than a multiple of 3. So these three cases are exhaustive:
*Case 1: If ''n'' = 3''p'', then ''n''<sup>3</sup> = 27''p''<sup>3</sup>, which is a multiple of 9.
*Case 2: If ''n'' = 3''p''&nbsp;+&nbsp;1, then ''n''<sup>3</sup> = 27''p''<sup>3</sup>&nbsp;+&nbsp;27''p''<sup>2</sup>&nbsp;+&nbsp;9''p''&nbsp;+&nbsp;1, which is 1 more than a multiple of 9. For instance, if ''n''&nbsp;=&nbsp;4 then ''n''<sup>3</sup> = 64 = 9×7&nbsp;+&nbsp;1.
*Case 3: If ''n'' = 3''p''&nbsp;−&nbsp;1, then ''n''<sup>3</sup> =&nbsp;27''p''<sup>3</sup>&nbsp;−&nbsp;27''p''<sup>2</sup>&nbsp;+&nbsp;9''p''&nbsp;−&nbsp;1, which is 1 less than a multiple of&nbsp;9. For instance, if ''n'' =&nbsp;5 then ''n''<sup>3</sup> = 125 = 9×14&nbsp;−&nbsp;1. [[Q.E.D.]]