Противречност: Разлика помеѓу преработките
| [непроверена преработка] | [непроверена преработка] |
Избришана содржина Додадена содржина
Нова страница: ===Proof by contradiction=== {{main|Reductio ad absurdum}} In '''proof by contradiction''' (also known as ''reductio ad absurdum'', Latin for "reduction into the absurd"),... |
Нема опис на уредувањето |
||
Ред 1:
===Доказ со контрадикција (противречност)===
:Suppose that <math>\sqrt{2}</math> is rational, so <math>\sqrt{2} = {a\over b}</math> where ''a'' and ''b'' are non-zero integers with no common factor (definition of rational number). Thus, <math>b\sqrt{2} = a</math>. Squaring both sides yields 2''b''<sup>2</sup> = ''a''<sup>2</sup>. Since 2 divides the left hand side, 2 must also divide the right hand side (as they are equal and both integers). So ''a''<sup>2</sup> is even, which implies that ''a'' must also be even. 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 share a factor, namely 2. This contradicts our assumption, so we are forced to conclude that <math>\sqrt{2}</math> is irrational.
| |||