Question 974888
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Pick a set that you know has countably infinite elements, for example the set *[tex \Large \mathbb{N}] of positive integers.  Then pick your favorite irrational number, say *[tex \Large \sqrt{2}].  Since we know that the product of a rational number and an irrational number is irrational, every product of the form *[tex \Large n\sqrt{2}] where *[tex \Large n\ \in\ \mathbb{N}] is irrational and the number of such products is infinite because *[tex \Large \mathbb{N}] has infinite elements. Since a subset of the set of irrational numbers has infinite elements, the entire set must, perforce, be infinite.


By the way it is also true that between any two rational numbers there is an infinity of irrational numbers.


If you are in doubt about the assertion made as to the irrationality of a product of a rational and an irrational number, then read: <a href="http://mathforum.org/library/drmath/view/51594.html">Dr. Math Proof that the product of a rational and an irrational is irrational</a>


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \