Let $f:\Bbb R\rightarrow\Bbb R$ have the properties $\forall x,y\in\Bbb R,\space x<y\implies f(x)<f(y)$ and $\forall x\in\Bbb R,\space f(x)\notin\Bbb A$ where $\Bbb A$ is the set of algebraic numbers; i.e. $f$ is strictly increasing, but nowhere is $f(x)$ algebraic.

Does such a function exist? And if so, can one be explicitly constructed?

My thoughts are that such a function should exist, since the algebraic numbers are "small" compared to the reals; we can show that a bijection (or more weakly an injection) must exist from $\Bbb R$ to $\Bbb R\backslash\Bbb A$ because they have the same cardinality, but I'm not entirely sure how to show rigorously that a strictly increasing function exists, even if in principle this is just a special type of injection.

Replacing $\Bbb A$ by a set such as $\Bbb Z$ in the definition makes the question trivial, and these sets have the same cardinality, so clearly the difficulty arises because $\Bbb A$ is dense in the reals - any hints or answers would be appreciated.

  • $\begingroup$ @астонвіллаолофмэллбэрг Gelfond-Schneider only holds if $r$ and $f(x)$ are algebraic, in $r^{f(x)}$. But $f(x)$ cannot be algebraic for all $x$ as otherwise $f$ would be an injection from the reals to a set with smaller cardinality, which is not possible. $\endgroup$ Nov 28, 2018 at 17:49
  • $\begingroup$ @stanleydodds I see. Thank you for the point out. $\endgroup$ Nov 28, 2018 at 18:01
  • $\begingroup$ Do you know the answer to the easier question of finding an increasing $f$ which is never rational? Also, something which might be relevant is the fact that an increasing function is continuous off of a countable set. $\endgroup$ Nov 28, 2018 at 18:26
  • $\begingroup$ No I do not know the answer to that simpler question - from all I've been able to do so far, exchanging $\Bbb A$ with any other set that is dense in the reals but countable (e.g. $\Bbb Q$) makes another tricky question. $\endgroup$ Nov 28, 2018 at 18:32

3 Answers 3


A possible (I will explain why later) example could be ...

Let's take an $x \in \mathbb{R}$ and have its binary (for simplicity) representation $x=(x_nx_{n-1}...x_0\color{red}{,}x_{-1}x_{-2}...x_{-m}...)_2, x_k\in\{0,1\}, k\in\{-\infty,...,n\}$ or $$x=\sum\limits_{k=0}^nx_k2^k + \sum\limits_{m=1}\frac{x_{-m}}{2^{m}}$$ and build the function $$f(x)=f\left(\sum\limits_{k=0}^nx_k2^k + \sum\limits_{m=1}\frac{x_{-m}}{2^{\color{red}{m}}}\right)= \sum\limits_{k=0}^nx_k2^k + \sum\limits_{m=1}\frac{x_{-m}}{2^{\color{red}{m!}}}$$ i.e. $f(x)$ becomes

  • a Liouville number, if $x$ is irrational
  • a Liouville number, if $x$ is rational with periodic (never ending) fractional part
  • a rational, if $x$ is rational with finite fractional part
  • $f(x)=x$, if $x$ is integer

All the Liouville numbers are transcendentals, so this function never returns an algebraic number.

It's not too difficult to show it's strictly increasing, if $a < b$ or $$(a_na_{n-1}...a_0\color{red}{,}a_{-1}a_{-2}...a_{-m}...)_2 < (b_nb_{n-1}...b_0\color{red}{,}b_{-1}b_{-2}...b_{-m}...)_2$$ ($a_n,a_{n-1}, ...$ can be $0$, just to have a common upper index $n$ for both $a$ and $b$) means that $\exists k \in\{-\infty, ...,n\}$ such that $a_k<b_k$ while $a_t=b_t, t\in\{k+1,...,n\}$. With $f(x)$ we have $$(a_na_{n-1}...a_0\color{red}{,}a_{-1}a_{-2}\color{blue}{000}a_{-3}\color{blue}{00000000000000000}a_{-4}\color{blue}{00...}a_{-m}...)_2 < (b_nb_{n-1}...b_0\color{red}{,}b_{-1}b_{-2}\color{blue}{000}b_{-3}\color{blue}{00000000000000000}b_{-4}\color{blue}{00...}b_{-m}...)_2$$

Note 1: I restricted the function to $\mathbb{R^+}\rightarrow \mathbb{R^+}$, but it can be extended, taking into account the sign of $x$.

Note 2 As per the comments below, integers and rationals are algebraic numbers. To overcome this part, we can apply these tricks $$(x_nx_{n-1}...x_0)_2=((x_nx_{n-1}...x_0-1)\color{red}{,}11111...)_2$$ and $$(x_nx_{n-1}...x_0\color{red}{,}x_{-1}x_{-2}...x_{-m})_2=(x_nx_{n-1}...x_0\color{red}{,}x_{-1}x_{-2}...(x_{-m}-1)11111...)_2$$ leading to Liouville numbers in all the cases.

Now why possible, because not all reals are computable.

  • 5
    $\begingroup$ This is a good answer, but I should mention that rationals and integers are also algebraic numbers, so for integer or rational with finite fractional part $x$ we still have an algebraic $f(x)$. This is not a problem though, since both of these $x$ can instead be written with infinite trailing 1's in base 2 form, and defining $f$ to use this form in these cases again gives a Liouville number. I'm not sure if you didn't notice, or if you knew this already. $\endgroup$ Nov 28, 2018 at 18:58
  • $\begingroup$ @stanleydodds oh yes, you are right! I ignored (or forgot) that part :( $\endgroup$
    – rtybase
    Nov 28, 2018 at 19:14
  • $\begingroup$ It's amusing that Liouville numbers come up here, because this solution versus mine seems somewhat like Liouville's proof of the existence of transcendental numbers versus Cantor's.... $\endgroup$ Dec 6, 2018 at 13:20

It's actually very simple; the same result holds with any countable set in place of the algebraic numbers. Since $\Bbb R$ is order-isomorphic to $(0,\infty)$ it's enough to prove this:

If $C\subset(0,\infty)$ is countable there exists a strictly increasing function $f:(0,\infty)\to(0,\infty)\setminus C$.

Since a countable set is contained in an open set of finite measure this follows from the stronger result (where $m$ is Lebesgue measure):

Suppose $V\subset(0,\infty)$ is open, let $E=(0,\infty)\setminus V$ and assume $m(E)=\infty$. There exists a strictly increasing function $f:(0,\infty)\to E$.

Proof: Define $\phi:[0,\infty)\to[0,\infty)$ by $$\phi(y)=m(E\cap[0,y)).$$Then $\phi$ is continuous, $\phi(0)=0$ and $\phi(\infty)=\infty$, so $$\phi((0,\infty))=(0,\infty).$$

Suppose $y\in V$. Say $y\in(a,b)$, where $(a,b)$ is a connected component of $V$. Then $\phi(y)=\phi(b)$ and $b\in E$. Hence $$\phi(E)=\phi((0,\infty))=(0,\infty).$$So for every $t>0$ there exists $f(t)\in E$ with $$\phi(f(t))=t.$$If $0<s<t$ it follows that $$f(t)-f(s)\ge m([f(s),f(t))\cap E)=\phi(f(t))-\phi(f(s))= t-s>0;$$hence $f$ is strictly increasing.


Let $q \colon \mathbb{N} \to \mathbb{Q}$ an enumeration of the rationals and $$f(x) = \sum_{q_n < x} \frac{1}{2^{n!}}$$ Then $f$ is strictly increasing and $f(x)$ is a Liouville number for every $x\in \mathbb{R}$.


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