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?
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to\cdot
for multiplication to improve readability. $\endgroup$ – 6005 Sep 14 '13 at 16:45