Prove there is no strictly increasing function $f$ from irrationals to reals. Prove there does not exist a strictly increasing function $f\colon \mathbb{R} - \mathbb{Q} \rightarrow \mathbb{R}$ such that $f(\mathbb{R} - \mathbb{Q}) = \mathbb{R}$.
I imagine the best way to go about this is by contradiction. So suppose there does exist a strictly increasing function. Then I think I have to use the fact that since $f$ is strictly increasing, it is one-to-one. However, I am not sure.
Any help is appreciated.
 A: Consider the sets $f(\Bbb I^-)$ and $f(\Bbb I^+)$, where $\Bbb I=\Bbb R\setminus\Bbb Q$. For any $x\in\Bbb I^-, y\in\Bbb I^+$, we have $x<0<y\implies f(x)<f(y)$, so each element of $f(\Bbb I^-)$ is less than each element of $f(\Bbb I^+)$. Now both sets are nonempty, so they partition $\Bbb R$ into two pieces, and there is a point $\alpha$ in the middle which is equal to the supremum of $f(\Bbb I^-)$ and the infimum of $f(\Bbb I^+)$. But since $f(\Bbb I^-)\cup f(\Bbb I^+)=\Bbb R$, $\alpha$ is in one of the two sets, and so  $\alpha=f(\beta)$ for some $\beta\in\Bbb I^+$ or $\Bbb I^-$. In the first case, $f(\beta/2)$ is a number less than the infimum; in the second case it is greater than the supremum - a contradiction.
Note: this is a direct application of the following simple topological property: $f$ is an order isomorphism, so under the order topologies, $\Bbb R$ and $\Bbb I$ must be homeomorphic. But $\Bbb I$ is disconnected (in particular, at 0) and $\Bbb R$ is not.
A: Another approach following your idea: $f^{-1}$ is also strictly increasing, so it is continuous almost everywhere.  But all of its values are irrational, so...
A: All we really need is the least upper bound property for real numbers.  
Let $u_1\lt u_2\lt u_3\lt\ldots$ be a strictly increasing sequence of irrational numbers, converging to a rational number $q$.  (For example, let $u_n=-\sqrt2/n$.)  Then $f(u_1)\lt f(u_2)\lt f(u_3)\lt\ldots$ is another increasing sequence of reals, bounded above by $f(U)$, for any irrational $U$ greater than $q$.  Hence it has a limit, $L=\lim_{n\to\infty}f(u_n)$.  
Now if $f(\mathbb{R}-\mathbb{Q})=\mathbb{R}$, then $L=f(u^*)$ for some irrational $u^*$.  Clearly $f(u_n)\lt L$ for all $n$, hence $u_n\lt u^*$ for all $n$.  Since the $u_n$'s converge to $q$, we must have $q\le u^*$ as well, and since $q$ is rational while $u^*$ is irrational, we must, in fact, have $q\lt u^*$.  
But now comes the contradiction:  Let $u'$ be any irrational number between $q$ and $u^*$ (say $u'=(q+u^*)/2$).  From $u_1\lt u_2\lt u_3\lt\ldots\lt q\lt u'\lt u^*$, we have $f(u_1)\lt f(u_2)\lt\ldots\lt f(u')\lt f(u^*)$, hence
$$f(u^*)=\lim_{n\to\infty}f(u_n)\le f(u')\lt f(u^*)$$
