Can an irrational always be found by multiplying irrationals? I was thinking about the function $\ f(a,b) = a/b $ where $a$ and $b$ where both irrational. It quickly stood out to me that the codomain of that function would include every rational number. But, does it include every irrational number as well (in other words, is the codomain of the function $\mathbb{R}$)?
Then I thought that if we establish that $a = n \cdot m$  and $b = m$, then if for every irrational number $n$ there exists at least another irrational number $m$ (which could be itself) such that $n \cdot m$ is also irrational, every irrational could be represented by $a/b$ (as $m$ cancels out), and so the function would eventually "spit out" all the reals given only irrational input.
So my questions are: For every irrational $n$, does there always exist another irrational $m$ such that $n \cdot m$ is also irrational? If this is true as I suspect, what is the simplest proof for it?
 A: Yes.
If $a$ is an algebraic number then $a\cdot\pi$ is irrational.
If $a$ is transcendental then $a\cdot\sqrt2$ is irrational.
A: Given an irrational $a$, if $\sqrt 2 a$ is irrational, we are done; and if $\sqrt 2 a$ is rational, then $\sqrt 3 a$ is irrational.
A: First, I want to point out that when you say that $f$ contains every rational number in its image, you are forgetting about zero!  Of course $a/b$ cannot equal zero for $a$ and $b$ irrational because that would imply $a = 0$.
To answer your question (elaborating along the lines of TonyK's answer), let $x$ be any irrational number.  $\sqrt{2},\sqrt{3},$ and $\sqrt{6}$ are all irrational.  Moreover, $\sqrt{2}x$ and $\sqrt{6}x$ cannot both be rational or else $\frac{\sqrt{6}x}{\sqrt{2}x} = \sqrt{3}$ would be irrational.  So one of the two products $x \cdot \sqrt{2} = x\sqrt{2}$ and $x \cdot \sqrt{6} = x\sqrt{6}$ works.
A: A non-computational answer: let $n$ be irrational (therefore not zero) and multiply by all irrationals $m$.  This gives you uncountably many different numbers: they can't all be rational.
