A Sequence of Irrational Numbers with a Rational Limit

A sequence of irrational numbers can converge to a rational number. The key point is that the limit of a sequence does not have to belong to the set containing all of its terms.

A simple example is

$$a_n=\frac{\sqrt{2}}{n}, \qquad n=1,2,3,\ldots$$

Every term is irrational, while the sequence converges to $0$, which is rational.

Why every term is irrational

For every positive integer $n$, the number $1/n$ is a nonzero rational number. If $\sqrt{2}/n$ were rational, multiplying it by the rational number $n$ would imply that $\sqrt{2}$ was rational. That is impossible.

Therefore,

$$\frac{\sqrt{2}}{n}\notin\mathbb{Q}$$

for every positive integer $n$. The first few terms are

$$\sqrt{2},\quad \frac{\sqrt{2}}{2},\quad \frac{\sqrt{2}}{3},\quad \frac{\sqrt{2}}{4},\ldots$$

and all of them are irrational.

Why the limit is rational

Because $\sqrt{2}$ is a fixed constant,

$$\lim_{n\to\infty}\frac{\sqrt{2}}{n} =\sqrt{2}\lim_{n\to\infty}\frac{1}{n} =\sqrt{2}\cdot 0 =0.$$

The number $0$ is rational because $0=0/1$.

An epsilon proof says the same thing without relying only on limit laws. Given any $\varepsilon>0$, choose an integer $N$ such that

$$N>\frac{\sqrt{2}}{\varepsilon}.$$

Then, whenever $n\geq N$,

$$\left|a_n-0\right|=\frac{\sqrt{2}}{n} \leq \frac{\sqrt{2}}{N}<\varepsilon.$$

Thus $a_n\to 0$ by the definition of convergence.

A sequence converging to any rational number

The same construction works for every rational limit. Let $q\in\mathbb{Q}$ and define

$$a_n=q+\frac{\sqrt{2}}{n}.$$

The term $\sqrt{2}/n$ is irrational. Adding a rational number to an irrational number produces an irrational number: if $q+\sqrt{2}/n$ were rational, subtracting $q$ would make $\sqrt{2}/n$ rational, a contradiction.

On the other hand,

$$\lim_{n\to\infty}a_n =\lim_{n\to\infty}\left(q+\frac{\sqrt{2}}{n}\right) =q+0=q.$$

For example, with $q=3/5$,

$$a_n=\frac{3}{5}+\frac{\sqrt{2}}{n}$$

has only irrational terms and converges to the rational number $3/5$.

The set-theoretic reason

The rational numbers $\mathbb{Q}$ and irrational numbers $\mathbb{R}\setminus\mathbb{Q}$ are both dense in the real numbers. This means that every open interval around every real number contains both rational and irrational numbers.

In particular, irrational numbers occur arbitrarily close to a rational number. A sequence can choose irrational numbers closer and closer to a rational target. The example $\sqrt{2}/n$ chooses irrational numbers approaching $0$.

In terms of closed sets, the irrational numbers are not a closed subset of $\mathbb{R}$. A closed set must contain the limits of all convergent sequences whose terms lie in that set. Since the sequence $\sqrt{2}/n$ lies entirely among the irrationals but has limit $0$, which is not irrational, the irrational numbers are not closed.

By contrast, the rational numbers are also not closed in $\mathbb{R}$. For example, the sequence of rational decimal truncations of $\sqrt{2}$ converges to the irrational number $\sqrt{2}$.

A concrete numerical example

Take $a_n=\sqrt{2}/n$. Since $\sqrt{2}\approx1.4142$, the first terms are approximately

$n$	$a_n$
1	1.4142
2	0.7071
5	0.2828
10	0.1414
100	0.0141
1000	0.0014

The values never equal $0$, and none is rational, but they can be made as close to $0$ as desired. That is exactly what convergence to $0$ means.

Common mistakes

Assuming the limit must have the same type as every term

This is false unless the set of possible terms is closed. A sequence of integers that converges in $\mathbb{R}$ must eventually be constant, so its limit is an integer. But irrational numbers do not have this property.

Thinking irrational plus rational is always irrational without checking the rational is nonzero

The statement is true: rational plus irrational is irrational, including when the rational number is zero. For multiplication, however, the rational multiplier must be nonzero. Multiplying an irrational number by $0$ gives the rational number $0$.

Using a sequence that includes rational terms

For instance, $1/n+\sqrt{2}/n$ is fine because it equals $(1+\sqrt{2})/n$ and is irrational. But a sequence such as $1/n$ does not answer the question: its terms are rational, not irrational.

The essential example is therefore

$$\boxed{a_n=\frac{\sqrt{2}}{n},\qquad a_n\in\mathbb{R}\setminus\mathbb{Q},\qquad a_n\to 0\in\mathbb{Q}.}$$