Rational Approximation of Real Numbers

From Density to Controlled Approximation

The Density Theorem for the Rational Numbers says that a rational number can be found between any two distinct real numbers. The Rational Approximation theorem gives the related conclusion that every real number can be approximated by a rational within any positive tolerance. Those statements guarantee that an approximation exists, but they do not prescribe a denominator or give a direct rule for constructing the rational.

The integer-part function supplies that rule. Fix a positive integer $n$. The numbers of the form $m/n$, where $m$ is an integer, divide the real line into equally spaced points, with spacing $1/n$. We will show that every real number lies within one such spacing of a point on this grid, and within half a spacing of a nearest grid point. The proofs use the Existence and Uniqueness of the Integer Part theorem from the tutorial on The Integer Part of a Real Number.

Rounding Down to a Prescribed Denominator

Theorem (Approximation by Rounding Down). Let $x\in\mathbb{R}$, and let $n$ be a positive integer. If $m=\lfloor nx\rfloor$, then $m\in\mathbb{Z}$ and $$ \frac{m}{n}\leq x<\frac{m+1}{n}. $$ Consequently, $$ 0\leq x-\frac{m}{n}<\frac{1}{n}. $$

Proof. By the defining property of the integer part, $m=\lfloor nx\rfloor$ is an integer and

m\leq nx<m+1.

Since $n>0$, division by $n$ preserves both inequalities. Thus $m/n\leq x<(m+1)/n$. Subtracting $m/n$ gives

0\leq x-\frac{m}{n}<\frac{m+1}{n}-\frac{m}{n}=\frac{1}{n}.

This proves both claims. $\square$

The rational $m/n$ is the grid point immediately at or below $x$. This description includes the case when $x$ is itself on the grid: then $nx$ is an integer, $m=nx$, and the approximation error is zero. The upper bound is strict because the defining integer-part inequality $nx<m+1$ is strict.

Worked Example: Rounding Down an Irrational Number

Approximate $x=\sqrt{2}/2$ from below using a denominator of $5$. We first verify which integer is $\lfloor 5x\rfloor$. Since both sides of each comparison are positive and

\left(\frac{7}{5}\right)^2=\frac{49}{25}<2<\frac{64}{25}=\left(\frac{8}{5}\right)^2,

we have $7/5<\sqrt{2}<8/5$. Multiplying by $5/2>0$ gives

\frac{7}{2}<\frac{5\sqrt{2}}{2}<4.

Therefore $\lfloor 5x\rfloor=3$. The theorem gives the rational approximation $3/5$, with

\frac{3}{5}\leq\frac{\sqrt{2}}{2}<\frac{4}{5}, \qquad 0\leq\frac{\sqrt{2}}{2}-\frac{3}{5}<\frac{1}{5}.

Thus the approximation is guaranteed to have error less than one fifth. The fraction $3/5$ is already in lowest terms, but the theorem does not require an approximation to be written in lowest terms.

The same method works for negative numbers because the integer-part theorem applies to every real number. In particular, rounding down does not mean rounding toward zero: for a negative noninteger, the integer part is the next integer to its left.

Worked Example: Rounding Down a Negative Number

Take $x=-\sqrt{2}/3$ and $n=10$. The bounds $7/5<\sqrt{2}<8/5$, verified by squaring in the preceding example, imply

-\frac{16}{3}<-\frac{10\sqrt{2}}{3}<-\frac{14}{3}.

Here $10x=-10\sqrt{2}/3$. Since $-16/3<-6$ and $-14/3<-4$, these bounds alone are not enough to determine its integer part. We refine the comparison using $7/5<\sqrt{2}<3/2$. The lower bound follows from $49/25<2$, and the upper bound follows from $2<9/4$. Multiplying by $-10/3$ reverses the inequalities, so

-5<-\frac{10\sqrt{2}}{3}<-\frac{14}{3}<-4.

It follows that $\lfloor 10x\rfloor=-5$. Rounding down therefore gives $-5/10=-1/2$, and the theorem guarantees

-\frac{1}{2}\leq-\frac{\sqrt{2}}{3}<-\frac{4}{10}, \qquad 0\leq-\frac{\sqrt{2}}{3}+\frac{1}{2}<\frac{1}{10}.

The direction of the inequalities is important: $-1/2$ is below $-\sqrt{2}/3$, even though rounding $-1.414$ to an integer in the usual sense might suggest $-1$.

Rounding to the Nearest Grid Point

Rounding down gives an error less than $1/n$. A closer approximation is obtained by rounding to a nearest grid point. The integer-part function can implement this by adding $1/2$ before taking the integer part.

Theorem (Nearest-Grid Approximation). Let $x\in\mathbb{R}$, and let $n$ be a positive integer. Define $$ k=\left\lfloor nx+\frac{1}{2}\right\rfloor. $$ Then $k\in\mathbb{Z}$, and $$ \left|x-\frac{k}{n}\right|\leq\frac{1}{2n}. $$

Proof. By the integer-part property,

k\leq nx+\frac{1}{2}<k+1.

Subtracting $1/2$ gives $k-1/2\leq nx<k+1/2$, or equivalently

-\frac{1}{2}\leq nx-k<\frac{1}{2}.

The Absolute-Value Bound Criterion implies $|nx-k|\leq 1/2$. Since $n>0$, the Absolute Value of a Product theorem gives

\left|x-\frac{k}{n}\right| =\frac{|nx-k|}{n} \leq\frac{1}{2n}.

This proves the result. $\square$

At a tie, $x$ is exactly halfway between two adjacent grid points. The formula chooses the larger of those two points: if $nx=j+1/2$ for an integer $j$, then $k=\lfloor j+1\rfloor=j+1$. Either choice is a nearest point, and the stated error bound includes both.

Worked Example: A Nearest Approximation with a Negative Numerator

Use denominator $10$ to approximate $x=-\sqrt{2}/3$ by the nearest grid point. We determine $k=\lfloor 10x+1/2\rfloor$. From $7/5<\sqrt{2}<3/2$, multiplication by $10/3$ gives

\frac{14}{3}<\frac{10\sqrt{2}}{3}<5.

Negating and adding $1/2$ yields

-\frac{9}{2}<-\frac{10\sqrt{2}}{3}+\frac{1}{2}<-\frac{14}{3}+\frac{1}{2}=-\frac{25}{6}.

These bounds show that $10x+1/2$ lies between $-9/2$ and $-25/6$, hence in $(-5,-4)$, so $k=-5$. The following comparisons verify the same endpoints directly. The inequality $\sqrt{2}<3/2$ gives

-\frac{10\sqrt{2}}{3}+\frac{1}{2}> -\frac{10}{3}\cdot\frac{3}{2}+\frac{1}{2} =-5+\frac{1}{2}=-\frac{9}{2}>-5.

Also, $\sqrt{2}>7/5$ gives

-\frac{10\sqrt{2}}{3}+\frac{1}{2} < -\frac{10}{3}\cdot\frac{7}{5}+\frac{1}{2} =-\frac{14}{3}+\frac{1}{2} =-\frac{25}{6}<-4.

Consequently $-5<10x+1/2<-4$, so $k=-5$. The nearest approximation is $-5/10=-1/2$. Directly, its error satisfies

\left|-\frac{\sqrt{2}}{3}+\frac{1}{2}\right| =\frac{1}{2}-\frac{\sqrt{2}}{3} <\frac{1}{2}-\frac{7}{15} =\frac{1}{30}<\frac{1}{20}.

The expression inside the absolute value is positive because $\sqrt{2}<3/2$; the final bound uses $\sqrt{2}>7/5$. The theorem also gives the general bound $1/(2n)=1/20$, while this calculation verifies a stricter error bound for this particular $x$.

Why the Nearest-Grid Bound Is Sharp

The estimate $1/(2n)$ cannot be replaced by a smaller uniform bound if the denominator is fixed at $n$. This is not a weakness of the rounding rule: some real numbers really are exactly halfway between adjacent grid points.

Proposition (Sharpness of the Nearest-Grid Bound). Let $n$ be a positive integer and $j\in\mathbb{Z}$. For $$ x=\frac{j+1/2}{n}, $$ every rational of the form $p/n$, with $p\in\mathbb{Z}$, satisfies $$ \left|x-\frac{p}{n}\right|\geq\frac{1}{2n}. $$ Equality holds when $p=j$ or $p=j+1$.

Proof. For any integer $p$,

\left|x-\frac{p}{n}\right| =\frac{|j+1/2-p|}{n}.

The number $j-p$ is an integer. Its sum with $1/2$ is therefore a half-integer, whose absolute value is at least $1/2$. Hence $|j+1/2-p|/n\geq1/(2n)$. If $p=j$, then $|j+1/2-p|=1/2$; if $p=j+1$, then $|j+1/2-p|=1/2$ as well. In both cases equality follows. $\square$

Worked Example: An Exact Tie Between Two Grid Points

Let $n=5$, $j=3$, and $x=(3+1/2)/5=7/10$. The adjacent grid points are $3/5$ and $4/5$, and direct subtraction gives

\left|\frac{7}{10}-\frac{3}{5}\right| =\left|\frac{7}{10}-\frac{6}{10}\right| =\frac{1}{10}, \qquad \left|\frac{7}{10}-\frac{4}{5}\right| =\left|\frac{7}{10}-\frac{8}{10}\right| =\frac{1}{10}.

Since $1/(2n)=1/10$, both points attain the sharp bound. The nearest-grid theorem guarantees an error no greater than this value, but cannot promise a strictly smaller error for every real number when the denominator is fixed.

Interpreting the Denominator and the Error

The phrase “denominator $n$” here means that the approximation is written as $m/n$ or $k/n$. It does not assert that the fraction is in lowest terms. For example, $6/10=3/5$; the displayed denominator may simplify, but the approximation still came from a grid with spacing $1/10$. The error estimates concern the chosen grid, not the reduced denominator.

The two constructions serve slightly different purposes. Rounding down gives a one-sided estimate: the approximant never exceeds $x$, and its error is strictly less than $1/n$. Rounding to the nearest grid point gives a two-sided estimate with at most half that error. If an application requires an approximation from below, the first construction preserves that direction. If only the size of the error matters, the second is usually more efficient.

The estimates also explain how denominator size controls accuracy. Increasing $n$ makes the grid points more closely spaced: the guaranteed one-sided error is less than $1/n$, and the nearest-grid error is at most $1/(2n)$. The earlier Rational Approximation theorem guarantees approximations within any positive tolerance. The results here add a constructive choice of numerator and an explicit relation between denominator and error.

Key takeaway. For any real $x$ and positive integer $n$, $\lfloor nx\rfloor/n$ approximates $x$ from below with error less than $1/n$. The integer $\lfloor nx+1/2\rfloor$ instead gives a nearest-grid approximation with error at most $1/(2n)$, and that bound is sharp.

Check Your Understanding

Use the defining property of the integer part, the order rules for division by positive numbers, and the approximation results above to answer these questions.

Why does $m=\lfloor nx\rfloor$ give a rational $m/n$ that is at or below $x$, including when $x$ is negative?
What error bound does rounding down guarantee, and why is its upper inequality strict?
Why does adding $1/2$ before taking the integer part produce a nearest-grid approximation?
For a fixed positive integer $n$, which real numbers attain the error $1/(2n)$ for a nearest grid point?
Does writing an approximation as $m/n$ imply that its reduced denominator is exactly $n$? Explain.