Question 656388
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Seems reasonable.  Let's assume the opposite is possible; that is, assume that there is a rational number (other than zero) and an irrational number such that the product of the two is rational.


Let *[tex \LARGE S] be the set of irrational numbers and let *[tex \LARGE s] be a number such that *[tex \LARGE s\ \in\ S].  Let *[tex \LARGE r] be a number such that *[tex \LARGE r\ \in\ \mathbb{Q}].


*[tex \LARGE r\ \in\ \mathbb{Q}] means that *[tex \LARGE \exists p,\,q\ \in\ \mathbb{Z}\ :\ r\ =\ \frac{p}{q}]


If the product *[tex \LARGE s\,\cdot\,r\ \in\ \mathbb{Q}], then there must be integers *[tex \LARGE \exists u,\,v\ \in\ \mathbb{Z}\ :\ s\,\cdot\,r\ =\ \frac{u}{v}]


Substituting:


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


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s\ =\ \frac{uq}{vp}]


But since the integers are closed under multiplication, *[tex \LARGE s] must be the quotient of integers and therefore rational.  This contradicts the initial assumption that *[tex \LARGE s] is irrational.  Therefore, reductio ad absurdum, the product of a rational number and an irrational number cannot be rational.
 

John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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