# Is there a bijection between $(0,1)$ and $\mathbb{R}$ that preserves rationality?

While reading about cardinality, I've seen a few examples of bijections from the open unit interval $(0,1)$ to $\mathbb{R}$, one example being the function defined by $f(x)=\tan\pi(2x-1)/2$. Another geometric example is found by bending the unit interval into a semicircle with center $P$, and mapping a point to its projection from $P$ onto the real line.

My question is, is there a bijection between the open unit interval $(0,1)$ and $\mathbb{R}$ such that rationals are mapped to rationals and irrationals are mapped to irrationals?

I played around with mappings similar to $x\mapsto 1/x$, but found that this never really had the right range, and using google didn't yield any examples, at least none which I could find. Any examples would be most appreciated, thanks!

• For "explicit" examples mathoverflow.net/questions/48910/… is relevant Jan 15 '11 at 10:18
• Just to make it clear: if you are asking for a bijection with no further properties (e.g. no continuity etc. conditions) then this is true simply for cardinality reasons: $\mathbb Q$ and $\mathbb Q\cap (0,1)$ are both countably infinite, and so we can find a bijection $\phi$ from the first to the second. Also $\mathbb R\setminus \mathbb Q$ and $(0,1) \setminus \mathbb Q$ both have the cardinality of the continuum, so we can find a bijection $\psi$ from the first to the second. Gluing $\phi$ and $\psi$ gives a bijection from $\mathbb R$ to $(0,1)$ that takes (ir)rationals to (ir)rationals. ... Jan 15 '11 at 21:42
• ... This is discussed more carefully in Asaf Karagila's answer, and the accompanying comments. Jan 15 '11 at 21:42

$(1/x)-2$ on $(0,1/2]$ and $2-(1/(x-1/2))$ on $(1/2,1)$.

• Modify this to $2-1/(1-x)$ on $({1\over2},1)$, and it is even continuous. Jan 15 '11 at 12:55
• @ Christian. You can find a $C^k$ function composing by a well chosen polynomial. But I don't know if it is possible to find a smooth function. Jan 15 '11 at 18:11
• Thanks Nabyl Bod, this was nice and clear and easy for me to verify for myself. Jan 15 '11 at 19:40
• @Nabyl: Can could please tell me what kind of polynomial you think of? I can't figure it out. Jan 16 '11 at 14:22
• @Hendrik. I think of composing by a rational polynomial (taking non negative coefficients to make it one-to-one) with a degree greater than the constraints at point $1/2$ BUT this construction does not preserve irrationality. Thanks for your comment ;) But you can find a $C^1$ function: $1/(4x)-1/2$ on $(0,1/2)$ and $1/2-1/(4(1-x))$ on $[1/2,1)$ that answers the question. Jan 16 '11 at 16:50

$$f(x) = \frac{2x - 1}{1 - |2x - 1|}.$$

• Thanks for your response also, mjqxxxx. Jan 15 '11 at 19:42
• How do you make sure that if $x$ is irrational then so is $f(x)$? irrational divided by an irrational could be rational.. Jan 26 '15 at 0:02
• Solve for $x$ as a function of $f(x)$. You will find that if $f(x)$ is rational, then so is $x$. Jan 26 '15 at 3:35

With the axiom of choice we can find well orderings of $\mathbb{R}$ and of $(0,1)$ such that the first $\omega$ elements are all the rationals of the set, then we can define our map to go from one well ordering to another by preserving the index (that is $a_\alpha\mapsto b_\alpha$, for $\alpha<2^{\aleph_0}$)

As discussed in the comments below by Colin and Jason (and myself), one does not need the axiom of choice for that. Using the Cantor-Schroeder-Bernstein theorem one can have two bijections, one from $(0,1)\setminus\mathbb{Q}$ to $\mathbb{R}\setminus\mathbb{Q}$ and one from $(0,1)\cap\mathbb{Q}$ to $\mathbb{Q}$, and define a bijection as needed without the use of the axiom of choice.

• One can use the en.wikipedia.org/wiki/… to avoid using the axiom of choice Jan 15 '11 at 10:21
• Colin: How can you ensure that the map you have from CSB preserves rationals? Jan 15 '11 at 10:25
• @Asaf: Apply CBS theorem to injections between $(0,1) \setminus \mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ to get a bijection between the two sets and then extend by the identity function. Jan 15 '11 at 10:37
• @Jason: The identity function is not a bijection from $(0,1)\cap\mathbb{Q}$ to $\mathbb{Q}$. However you can in fact well order the rationals without AC, and then you can create the bijection. Jan 15 '11 at 10:44
• True, that was a lazy answer by me. Jan 15 '11 at 10:50

Consider the function $$f: (0,1) \rightarrow \mathbb{R}$$.

$$f(n) =\left\{ \begin{array}{ll} \dfrac{1}{x} - 2 & \text{if}\ 0 < x \leq \dfrac{1}{2}\\ \dfrac{1}{x-1} + 2 & \text{if} \ \dfrac{1}{2} < x < 1 \end{array} \right.$$

We claim that $$f$$ is a bijective function between the open unit interval $$(0,1)$$ and $$\mathbb{R}$$ that takes rationals to rationals and irrationals to irrationals.

First, we notice that $$f$$ is piece-wise defined in such a way that $$dom \ f$$ partitions the open unit interval into two sets $$S$$ and $$R$$ defined by $$S = \left \{ x \in \mathbb{R} \ | \ x \in (0,\dfrac{1}{2}] \right\}$$ and $$R = \left \{ x \in \mathbb{R} \ | \ x \in (\dfrac{1}{2},1) \right\}$$.

Second, we show that $$f$$ takes rationals only to rationals and irrationals only to irrationals. There are a total of four cases to consider.

$$\bf{Case \ \#1}$$: Say $$x \in S \cap \mathbb{Q}$$. Then, by the definition of our function $$f$$, $$\exists a,b \in \mathbb{Z}$$ such that $$x = \dfrac{a}{b}$$ and $$f(x) = \dfrac{b}{a} - 2$$ and $$b \neq 0$$ because every rational $$x$$ can be represented as a ratio of two integers, with the denominator non-zero. Since rationals are closed under division and subtraction, we know that $$\dfrac{b}{a} - 2$$ is rational.

$$\bf{Case \ \#2}$$: Say $$x \in R \cap \mathbb{Q}$$. Then, by the definition of our function $$f$$, $$\exists a,b \in \mathbb{Z}$$ such that $$x = \dfrac{a}{b}$$ and $$f(x) = \dfrac{1}{\dfrac{a}{b}-1} + 2$$. Since rationals are closed under division, subtraction, and addition, we know that $$\dfrac{1}{x-1} + 2$$ is rational.

This shows that $$\forall x \in (0,1) \cap \mathbb{Q}, f(x) \in \mathbb{Q} \subseteq \mathbb{R}$$. So every rational in the open interval is mapped to a rational.

$$\bf{Case \ \#3}$$: Say $$x \in S \cap (\mathbb{R} - \mathbb{Q}$$). Then, by the definition of $$f$$, $$\sim \exists a,b \in \mathbb{Z}$$ such that $$f(x) = \dfrac{b}{a} - 2$$. Since both the subtraction and division of an irrational number by a rational number still produces an irrational number, we know that $$\dfrac{b}{a} - 2$$ is irrational.

$$\bf{Case \ \#4}$$: Say $$x \in R \cap (\mathbb{R} - \mathbb{Q}$$). Then, by the definition of $$f$$, $$\sim \exists a,b \in \mathbb{Z}$$ such that $$f(x) = \dfrac{1}{\dfrac{a}{b}-1} + 2$$. Since the subtraction, division, and addition of an irrational number with a rational number produces an irrational number, we know that $$\dfrac{1}{\dfrac{a}{b}-1} + 2$$ is irrational.

Therefore, $$f$$ maps irrationals only to irrationals.

Thirdly, we must show that $$f$$ is a bijective function.

To prove injectivity, there are two cases:

$$\bf{Case \ \#1}$$: Let $$f(x) = f(y)$$, where $$x,y \in S \cap \mathbb{Q}$$. Then, $$\exists a,b,c,d \in \mathbb{Z}$$, where $$b,d \neq 0$$, and $$x = \dfrac{a}{b}$$ and $$y = \dfrac{c}{d}$$ such that $$\dfrac{b}{a} - 2 = \dfrac{d}{c} -2$$. Adding both sides by two, we get $$\dfrac{b}{a} = \dfrac{d}{c}$$ which implies that $$\dfrac{a}{b} = \dfrac{c}{d}$$. So $$x = y$$ and $$f$$ is injective.

$$\bf{Case \ \#2}$$: Let $$f(x) = f(y)$$, where $$x,y \in R \cap \mathbb{Q}$$. Then, $$\exists a,b,c,d \in \mathbb{Z}$$, where $$b,d \neq 0$$, and $$x = \dfrac{a}{b}$$ and $$y = \dfrac{c}{d}$$ such that $$\dfrac{1}{\dfrac{a}{b}-1} + 2 = \dfrac{1}{\dfrac{c}{d}-1} + 2$$. Subtracting both sides by 2 we get $$\dfrac{1}{\dfrac{a}{b}-1} = \dfrac{1}{\dfrac{c}{d}-1}$$. This implies that $$\dfrac{c}{d} - 1 = \dfrac{a}{b} -1$$. Adding both sides by 1, we get $$y = x$$ so $$f$$ is injective.

To prove surjectivity, there are four cases:

$$\bf{Case \ \#1}$$: We show that $$\forall y \in \mathbb{Q}$$, $$y = f(x)$$ for some $$x \in S \cap \mathbb{Q}$$. Let $$f(x) = y$$; so $$y = \dfrac{1}{x} - 2$$. Then, $$x = \dfrac{1}{y+ 2}$$ and we are done.

$$\bf{Case \ \#2}$$: We show that $$\forall y \in \mathbb{Q}$$, $$y = f(x)$$ for some $$x \in R \cap \mathbb{Q}$$. Let $$f(x) = y$$; so $$y = \dfrac{1}{x-1} + 2$$. Then, $$x = \dfrac{1}{y-2} + 1$$ and we are done.

$$\bf{Case \ \#3}$$: We show that $$\forall y \in (\mathbb{R} - \mathbb{Q})$$, $$y = f(x)$$ for some $$x \in S \cap (\mathbb{R} - \mathbb{Q})$$. Let $$f(x) = y$$; so $$y = \dfrac{1}{x} - 2$$. Then, $$x = \dfrac{1}{y+ 2}$$ and we are done.

$$\bf{Case \ \#4}$$: We show that $$\forall y \in (\mathbb{R} - \mathbb{Q})$$, $$y = f(x)$$ for some $$x \in R \cap (\mathbb{R} - \mathbb{Q})$$. Let $$f(x) = y$$; so $$y = \dfrac{1}{x-1} + 2$$. Then, $$x = \dfrac{1}{y-2} + 1$$ and we are done.

This completes our proof that $$f$$ is a bijective function between the open unit interval $$(0,1)$$ and $$\mathbb{R}$$ that takes rationals to rationals and irrationals to irrationals.