   SEARCH HOME Math Central Quandaries & Queries  Question from John: mathcentral.uregina.ca/QQ/database/QQ.09.06/sylvia1.html In the initial assumption of that proof, root 6 is assumed to be a/b where a and b have no common factors, but why does having a common factor make it irrational? Hi John,

The assumption is that $\sqrt 6$ is rational and hence can be written as a fraction $\frac{c}{d}$ where $c$ and $d$ are integers. From your knowledge of equivalent fractions you know there is a fraction $\frac{a}{b}$ which is equivalent to $\frac{c}{d}$ where $a$ and $b$ have no common factors. Hence if $\sqrt 6$ is rational than it can be written as $\frac{a}{b}$ where $a$ and $b$ have no common factors.

The remainder of the proof shows from this assumption you can prove that $a$ and $b$ do have a common factor. Hence the assumption that $\sqrt 6$ is rational must be false.

I hope this helps,
Penny       * Registered trade mark of Imperial Oil Limited. Used under license. Math Central is supported by the University of Regina and the Imperial Oil Foundation.