Question 654623
<font face="Times New Roman" size="+2">


Let *[tex \LARGE S] represent the set of irrational numbers.


Assume that for some *[tex \LARGE R\ \in\ \mathbb{Q}] and *[tex \LARGE S_i\ \in\ S] that *[tex \LARGE R\,\cdot\,S_i\ \in\ \mathbb{Q}]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \exists p,\,q\ \in\ \mathbb{Z}\ :\ R\ =\ \frac{p}{q}]


by definition of rationals.


Is it possible then that the product of *[tex \LARGE R] and *[tex \LARGE S_i] is rational, that is *[tex \LARGE \exists u,\,v\ \in\ \mathbb{Z}\ :\ \frac{p\,\cdot\,S_i}{q}\ =\ \frac{u}{v}]


But that means


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ S_i\ =\ \frac{u\,\cdot\,q}{v\,\cdot\,p}]


But since *[tex \LARGE u,\,v,\,p,\,&\,q] are all integers and the integers are closed under multiplication, *[tex \LARGE S_i] must be rational since it is the quotient of integers, contradicting the original assumption that it was irrational.


Therefore, reductio ad absurdum, the product of a non-zero rational number and an irrational number cannot be rational.  Q.E.D.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>