Editing Mathematical proof (section)

===Proof by contradiction===
{{Main|Proof by contradiction}}
In proof by contradiction, also known by the Latin phrase ''[[reductio ad absurdum]]'' (by reduction to the absurd), it is shown that if some statement is assumed true, a [[contradiction|logical contradiction]] occurs, hence the statement must be false.  A famous example involves the proof that <math>\sqrt{2}</math> is an [[irrational number]]:

:Suppose that <math>\sqrt{2}</math> were a rational number. Then it could be written in lowest terms as <math>\sqrt{2} = {a\over b}</math> where ''a'' and ''b'' are non-zero integers with [[coprime|no common factor]]. Thus, <math>b\sqrt{2} = a</math>. Squaring both sides yields 2''b''<sup>2</sup> = ''a''<sup>2</sup>. Since the expression on the left is an integer multiple of 2, the right expression is by definition divisible by 2. That is, ''a''<sup>2</sup> is even, which implies that ''a'' must also be even, as seen in the proposition above (in [[#Proof by contraposition]]). So we can write ''a'' = 2''c'', where ''c'' is also an integer. Substitution into the original equation yields 2''b''<sup>2</sup> = (2''c'')<sup>2</sup> = 4''c''<sup>2</sup>. Dividing both sides by 2 yields ''b''<sup>2</sup> = 2''c''<sup>2</sup>. But then, by the same argument as before, 2 divides ''b''<sup>2</sup>, so ''b'' must be even. However, if ''a'' and ''b'' are both even, they have 2 as a common factor. This contradicts our previous statement that ''a'' and ''b'' have no common factor, so we must conclude that <math>\sqrt{2}</math> is an irrational number.

To paraphrase: if one could write <math>\sqrt{2}</math> as a [[fraction]], this fraction could never be written in lowest terms, since 2 could always be factored from numerator and denominator.