Density of the Irrationals

From Rational Density to Irrational Density

The Density of the Rationals Theorem from “Density of the Rationals” says that every nonempty open interval contains a rational number. A related question is whether every such interval also contains an irrational number. The rational density theorem gives a direct way to answer it: shift a suitable rational number by a fixed irrational number. The shift preserves the interval we want to reach, and the resulting number cannot be rational.

We will use \(\sqrt{2}\) as the fixed irrational shift. Here \(\sqrt{2}\) denotes the positive real number whose square is \(2\). We first verify that it is irrational, so the construction rests on a proved fact.

Proof. Suppose, to the contrary, that \(\sqrt{2}\) is rational. Then \(\sqrt{2}=m/n\) for integers \(m,n\) with \(n>0\), chosen so that the fraction is in lowest terms. Squaring gives \(m^2=2n^2\), so \(m^2\) is even. An odd integer has the form \(2j+1\), whose square is \(4j^2+4j+1\), an odd integer. Therefore an integer whose square is even must itself be even. Write \(m=2k\) for an integer \(k\). Substituting into \(m^2=2n^2\) gives \(4k^2=2n^2\), hence \(n^2=2k^2\). The same parity argument shows that \(n\) is even. Thus \(m\) and \(n\) are both divisible by \(2\), contradicting that \(m/n\) is in lowest terms. This contradiction proves that \(\sqrt{2}\) is irrational. \(\square\)

Translating the Rationals

If \(q\) is rational, then \(q+\sqrt{2}\) is irrational. Indeed, if the sum were rational, subtracting the rational number \(q\) would make \(\sqrt{2}\) rational, contrary to the lemma. This observation converts a rational selected from one interval into an irrational in a translated interval.

Proof. Let \(a,b\in\mathbb{R}\) with \(a<b\). Subtract \(\sqrt{2}\) from the endpoints. Since subtraction preserves strict inequalities, \(a-\sqrt{2}<b-\sqrt{2}\). By the Density of the Rationals Theorem, there is a rational number \(q\) such that

Adding \(\sqrt{2}\) throughout preserves the inequalities and gives

The number \(q+\sqrt{2}\) is irrational: if it were rational, then subtracting \(q\in\mathbb{Q}\) would imply that \(\sqrt{2}\) is rational. Thus the interval \((a,b)\) contains an irrational number. Since \(a\) and \(b\) were arbitrary with \(a<b\), the irrationals are dense in \(\mathbb{R}\). \(\square\)

The translation is the essential step. We do not need to find an irrational number directly by manipulating the endpoints. Instead, we move the interval by \(-\sqrt{2}\), use rational density there, and move the chosen rational back by \(+\sqrt{2}\). The argument works for every interval, regardless of whether its endpoints are rational or irrational.

Infinitely Many Irrationals in Every Open Interval

Density guarantees at least one irrational in each nonempty open interval. In fact, every such interval contains infinitely many distinct irrationals. We can obtain this stronger conclusion from the earlier result that every nonempty open interval contains infinitely many rational numbers, again using translation by \(\sqrt{2}\).

Proof. Fix \(a<b\). The translated interval \((a-\sqrt{2},b-\sqrt{2})\) is nonempty because \(a-\sqrt{2}<b-\sqrt{2}\). By the Theorem (Infinitely Many Rationals in Every Open Interval) from “Density of the Rationals,” this translated interval contains infinitely many distinct rational numbers. Choose distinct rational numbers \(q_1,q_2,\ldots\) in it. For every positive integer \(j\), define \(x_j=q_j+\sqrt{2}\).

Because \(a-\sqrt{2}<q_j<b-\sqrt{2}\), adding \(\sqrt{2}\) gives \(a<x_j<b\). Each \(x_j\) is irrational, by the irrational-shift argument used in the Density of the Irrationals Theorem. Finally, if \(j\neq k\) and \(x_j=x_k\), subtracting \(\sqrt{2}\) from both sides would give \(q_j=q_k\), contrary to the choice of distinct rationals. Therefore the interval contains infinitely many distinct irrationals \(x_1,x_2,\ldots\). \(\square\)

The injectivity of translation matters in this proof: it ensures that distinct rational choices do not collapse to the same irrational after the shift. More generally, adding a fixed real number to distinct numbers always gives distinct results, because equality of the sums can be cancelled by subtracting that fixed number.

Irrational Approximations to Any Real Number

The Density of the Irrationals Theorem also gives an approximation statement parallel to Rational Approximation from “Density of the Rationals.” For any target \(x\in\mathbb{R}\) and tolerance \(\varepsilon>0\), apply density to the interval \((x-\varepsilon,x+\varepsilon)\). The resulting irrational lies within \(\varepsilon\) of \(x\).

Proof. Fix \(x\in\mathbb{R}\) and \(\varepsilon>0\). Since \(x-\varepsilon<x+\varepsilon\), the Density of the Irrationals Theorem gives an irrational \(y\) with \(x-\varepsilon<y<x+\varepsilon\). Subtracting \(x\) gives \(-\varepsilon<y-x<\varepsilon\), which is equivalent to \(|x-y|<\varepsilon\). \(\square\)

In particular, for any real \(x\), apply the corollary with \(\varepsilon=1/n\) for each positive integer \(n\). Choose an irrational \(y_n\) satisfying \(|x-y_n|<1/n\). Since \(1/n\) tends to zero, these inequalities imply \(y_n\to x\). This is also an instance of the Sequential Characterization of Density from “Dense Sets.” Thus every real number is the limit of a sequence of irrationals, whether or not the real number itself is irrational.

Density Does Not Mean Membership

A dense set need not contain every real number. The irrationals are dense, but a rational number is still not irrational. Density says that every neighborhood of every real number meets the set; it does not say that the center of the neighborhood belongs to the set. The Irrational Approximation Corollary makes the distinction precise: the approximating irrational may depend on both the target and the chosen tolerance.

A common mistake is to reason that the sum of any irrational and any real number must be irrational. That statement is false: for example, \(\sqrt{2}+(-\sqrt{2})=0\) is rational. The translation proof requires that the number being added, \(q\), be rational. If \(q+\sqrt{2}\) were rational, subtracting that rational \(q\) would force \(\sqrt{2}\) to be rational, a contradiction. Keeping track of which term is rational is what makes the argument valid.

The result also links the current topic to closure. By the Density and Closure Theorem from “Dense Sets,” density of the irrationals is equivalent to \(\overline{\mathbb{R}\setminus\mathbb{Q}}=\mathbb{R}\). The sequence constructed above gives the same approximation idea in sequential form: each real number can be approached by irrational terms. These formulations will be useful when studying sequences and closure.