Rational Equivalence Classes on the Unit Interval

Rational Differences on the Unit Interval

Points in $[0,1]$ can be grouped according to whether their difference is rational. This simple relation produces classes with two properties that may at first seem incompatible: each class is countable, yet every class is dense in the whole interval. The restriction to $[0,1]$ matters throughout: a rational shift gives a member of a class only when the shifted point remains in the interval.

We work on the set $[0,1]$, and equip it with the restricted Lebesgue measure space $([0,1],\mathcal{L}|_{[0,1]},m)$. This measure space is finite, complete, and has total measure $1$. Write $\mathbb{Q}$ for the rational numbers. Define a relation on $[0,1]$ by $x\sim y$ exactly when $x-y\in\mathbb{Q}$. The equivalence class of $x$ is denoted by $[x]_{\mathbb{Q}}$; it contains only points of $[0,1]$.

Definition: For $x\in[0,1]$, its rational-difference equivalence class is $$ [x]_{\mathbb{Q}}=\{y\in[0,1]:y-x\in\mathbb{Q}\} =\{x+q:q\in\mathbb{Q},\ -x\le q\le 1-x\}. $$ The inequalities in the second description enforce the requirement $x+q\in[0,1]$.

The relation is defined only for points in the unit interval, but the difference $x-y$ is computed in $\mathbb{R}$. In particular, it is not a relation saying that the points themselves are rational. If $x$ is irrational, its class consists of irrational points: adding a rational number to an irrational number remains irrational.

Why This Is an Equivalence Relation

Proposition: The relation $\sim$ on $[0,1]$, defined by $x\sim y$ if and only if $x-y\in\mathbb{Q}$, is an equivalence relation. Moreover, for $x,y\in[0,1]$, the classes $[x]_{\mathbb{Q}}$ and $[y]_{\mathbb{Q}}$ are equal if $x-y\in\mathbb{Q}$, and are disjoint otherwise.

Proof. To check reflexivity, let $x\in[0,1]$. Then $x-x=0\in\mathbb{Q}$, so $x\sim x$. For symmetry, if $x\sim y$, then $x-y\in\mathbb{Q}$, and therefore $y-x=-(x-y)\in\mathbb{Q}$; thus $y\sim x$. For transitivity, if $x\sim y$ and $y\sim z$, then $x-y$ and $y-z$ are rational, so $x-z=(x-y)+(y-z)\in\mathbb{Q}$, and $x\sim z$. These calculations use $x,y,z\in[0,1]$, so every relation considered is defined on the stated set.

Now suppose $x-y\in\mathbb{Q}$. If $u\in[x]_{\mathbb{Q}}$, then $u-x\in\mathbb{Q}$, and $u-y=(u-x)+(x-y)\in\mathbb{Q}$. Hence $u\in[y]_{\mathbb{Q}}$, proving $[x]_{\mathbb{Q}}\subseteq[y]_{\mathbb{Q}}$. The same reasoning with $x$ and $y$ reversed gives the reverse inclusion. Thus the classes are equal. Conversely, if the classes intersect, choose $u\in[x]_{\mathbb{Q}}\cap[y]_{\mathbb{Q}}$. Then $x-u$ and $u-y$ are rational, so $x-y=(x-u)+(u-y)$ is rational. Therefore, if $x-y$ is not rational, the classes cannot intersect and are disjoint. $\square$

Consequently the classes form a partition of $[0,1]$: every point belongs to its own class, and any two classes are either equal or disjoint. In computations, it is useful to test equality of classes by checking the difference between the proposed representatives, rather than by trying to list all points in either class.

Worked Example: The Class of a Rational Point

Compute $[1/3]_{\mathbb{Q}}$. By the definition, a point $y\in[0,1]$ is in this class exactly when $y-1/3$ is rational. Since $1/3$ is rational, this is equivalent to $y$ being rational. Therefore

[1/3]_{\mathbb{Q}}=\mathbb{Q}\cap[0,1].

For example, $0$, $1/3$, and $1$ belong to the class: their differences from $1/3$ are respectively $-1/3$, $0$, and $2/3$, all rational, and all three points lie in $[0,1]$. As a check, every rational $y\in[0,1]$ has rational difference $y-1/3$; no irrational $y$ can have rational difference, since $y=(y-1/3)+1/3$ would then be rational.

Rational Shifts and Interval Constraints

The class formula describes all valid shifts at once. For a fixed representative $x$, the shift $q$ must be rational and satisfy $-x\le q\le1-x$. The first inequality ensures $x+q\ge0$; the second ensures $x+q\le1$. Both endpoints are included because the class is taken in the closed interval $[0,1]$.

Worked Example: Shifting an Irrational Representative

Let $x=\sqrt{3}/2$. Check three proposed rational shifts and compute the corresponding points. The shifts $-1/4$, $-1/2$, and $1/10$ are rational. They give

\frac{\sqrt{3}}2-\frac14=\frac{2\sqrt{3}-1}{4},\qquad \frac{\sqrt{3}}2-\frac12=\frac{\sqrt{3}-1}{2},\qquad \frac{\sqrt{3}}2+\frac1{10}=\frac{5\sqrt{3}+1}{10}.

For the first point, $\sqrt{3}/2>1/4$, since $\sqrt{3}>1/2$, so it is positive; also $\sqrt{3}/2<1$, since $\sqrt{3}<2$, so it is less than $1$. For the second point, $\sqrt{3}>1$ makes it positive, and $\sqrt{3}<2$ makes it less than $1$. For the third point, positivity is immediate, and it is less than $1$ because $\sqrt{3}/2<9/10$: indeed, both sides are positive and $3<81/25$. Thus all three points lie in $[0,1]$, and their differences from $x$ are the stated rational shifts, so all belong to $[x]_{\mathbb{Q}}$.

Check: adding back the respective shifts recovers $\sqrt{3}/2$ in each case. The inequalities also check the endpoint condition directly, rather than assuming that a rational shift must stay in the interval.

Worked Example: Two Representatives of the Same Class

Let $x=\sqrt{2}/3$ and $z=x+1/5$. First verify that both representatives are in $[0,1]$. We have $x>0$, and $x<1$ because $\sqrt{2}<3$. Also $$ z=\frac{\sqrt{2}}3+\frac15=\frac{5\sqrt{2}+3}{15}>0. $$ To check $z<1$, it suffices to check $\sqrt{2}/3<4/5$. Both sides are positive, and this follows from $\sqrt{2}<12/5$, whose square is $2<144/25$. Thus $x,z\in[0,1]$. Their difference is $z-x=1/5\in\mathbb{Q}$, so the proposition gives $[z]_{\mathbb{Q}}=[x]_{\mathbb{Q}}$.

The shift description gives the same conclusion point by point. If $y\in[x]_{\mathbb{Q}}$, then $y-x\in\mathbb{Q}$, and $y-z=(y-x)-1/5\in\mathbb{Q}$, so $y\in[z]_{\mathbb{Q}}$. Conversely, if $y\in[z]_{\mathbb{Q}}$, then $y-x=(y-z)+1/5\in\mathbb{Q}$, so $y\in[x]_{\mathbb{Q}}$. In both directions $y$ remains in $[0,1]$, as required by the definition. This verifies the equality of the classes without treating points outside the interval as class members.

Countable, Dense, and Null Classes

Proposition: In the finite complete measure space $([0,1],\mathcal{L}|_{[0,1]},m)$, every rational-difference class is countable, dense in $[0,1]$ in the relative topology, Lebesgue measurable, and of measure zero.

Proof. Fix $x\in[0,1]$. The map $q\mapsto x+q$ sends the rational numbers in $[-x,1-x]$ onto $[x]_{\mathbb{Q}}$. It is one-to-one, since $x+q_1=x+q_2$ implies $q_1=q_2$. The rational numbers are countable, so $[x]_{\mathbb{Q}}$ is countable.

To prove density, take any nonempty relatively open subset $U$ of $[0,1]$. Such a set contains a nonempty open interval $(a,b)$ with $0\le a<b\le1$. The interval $(a-x,b-x)$ is nonempty. By density of $\mathbb{Q}$ in $\mathbb{R}$, choose $q\in\mathbb{Q}\cap(a-x,b-x)$. Then $a<x+q<b$, so $x+q\in U$, and the inequalities $0\le a<x+q<b\le1$ show that $x+q$ is a valid member of $[x]_{\mathbb{Q}}$. Thus every nonempty relatively open subset meets the class, which is exactly density in $[0,1]$.

For measurability and measure, each singleton $\{y\}\subseteq[0,1]$ is Lebesgue measurable and has measure zero. Since $[x]_{\mathbb{Q}}$ is countable, it is a countable union of such singletons. Countable unions of Lebesgue measurable sets are measurable, and countable subadditivity gives $m([x]_{\mathbb{Q}})\le\sum_{y\in[x]_{\mathbb{Q}}}m(\{y\})=0$. Measures are nonnegative, so $m([x]_{\mathbb{Q}})=0$. This uses the restricted Lebesgue measure on $[0,1]$; in particular, the class is a measurable null set even though it is dense in that interval. $\square$

The density assertion has a practical interpretation: no matter how small a relative neighborhood in the interval is chosen, it contains a point of the class. Countability and density do not conflict. A countable set can be dense, just as $\mathbb{Q}\cap[0,1]$ is dense in $[0,1]$. Here the same density holds for every representative class, while the measure calculation shows that each such class has measure zero.

A Common Shift Error

A frequent mistake is to check that a shift is rational but forget to check that the shifted point lies in the interval defining the class. For instance, take $x=\sqrt{3}/2$ and propose the shift $+1/4$. Although $1/4$ is rational, the resulting point is greater than $1$: since $\sqrt{3}>3/2$, as follows by squaring the positive quantities and observing $3>9/4$, we have

\frac{\sqrt{3}}2+\frac14> \frac34+\frac14=1.

Therefore $\sqrt{3}/2+1/4$ is not in $[0,1]$ and is not a member of $[\sqrt{3}/2]_{\mathbb{Q}}$, despite having rational difference from the representative. A valid correction is the shift $-1/4$: the resulting point $\sqrt{3}/2-1/4$ is positive because $\sqrt{3}/2>1/4$, and is less than $1$ because $\sqrt{3}/2<1$. Its difference from $\sqrt{3}/2$ is $-1/4\in\mathbb{Q}$, so it does belong to the class.

The test for membership has two parts: the candidate point must lie in the underlying set $[0,1]$, and its difference from the representative must be rational. Neither condition replaces the other.

Check Your Understanding

For each question, keep both rationality of the difference and membership in $[0,1]$ in view.

Write $[2/5]_{\mathbb{Q}}$ as a familiar subset of $[0,1]$, and verify whether $7/10$ belongs to it.
For $x=\sqrt{2}/2$, test the shifts $-1/3$ and $1/3$. Which resulting points lie in $[0,1]$?
Prove directly from the definition that if $x-y\in\mathbb{Q}$, then every point of $[x]_{\mathbb{Q}}$ belongs to $[y]_{\mathbb{Q}}$.
Explain why a rational-difference class is measurable and has measure zero in the restricted Lebesgue measure space on $[0,1]$.
Give an example of a rational shift that is invalid for a chosen representative because the shifted point lies outside $[0,1]$, and replace it by a valid rational shift.