Real Numbers

The Number Line and the Real Numbers

The previous tutorial used the number line as a setting in which rational and irrational numbers both occur. We now give that setting its standard name: the real numbers. The real numbers include the familiar counting numbers, their negatives, fractions, and numbers such as \(\sqrt{2}\) that cannot be expressed as fractions. They are the numbers represented by points on the ordinary number line.

We write \(\mathbb{R}\) for the set of real numbers. At this stage, we use the number line as an intuitive picture of \(\mathbb{R}\), rather than constructing the real numbers from a more basic system. The next tutorial begins to make their arithmetic structure precise through the field axioms. For now, the central task is to see how the familiar number systems fit together and how rational and irrational numbers divide the real numbers between them.

This definition recalls the meaning of irrationality from the previous tutorial. In particular, it is important that “irrational” does not mean “not a number” or “not on the number line.” An irrational number is a real number; it simply has no representation as a ratio of integers with nonzero denominator.

The Nested Number Systems

Several familiar number systems sit inside the real numbers. We use \(\mathbb{N}_0\) for the nonnegative integers, including zero; \(\mathbb{Z}\) for all integers; and \(\mathbb{Q}\) for all rational numbers. Every nonnegative integer is an integer, every integer is rational, and every rational number is real. Thus the sets are nested:

Each inclusion is strict. There are integers that are not nonnegative, rational numbers that are not integers, and real numbers that are not rational. The last kind are exactly the irrational numbers.

Proof. Each nonnegative integer is an integer, so \(\mathbb{N}_0\subseteq\mathbb{Z}\). But \(-1\) is an integer and is not in \(\mathbb{N}_0\); therefore \(\mathbb{N}_0\subsetneq\mathbb{Z}\).

For every integer \(n\), the equality \(n=n/1\) expresses \(n\) as a ratio of integers with nonzero denominator. Hence \(\mathbb{Z}\subseteq\mathbb{Q}\). But \(1/2\) is rational and is not an integer, so \(\mathbb{Z}\subsetneq\mathbb{Q}\).

By definition, every rational number is real, and therefore \(\mathbb{Q}\subseteq\mathbb{R}\). The previous tutorial proved that \(\sqrt{2}\) is a real number that is not rational. Thus \(\mathbb{Q}\subsetneq\mathbb{R}\). Combining these inclusions proves the claim. \(\square\)

The phrase “smallest set in the chain” is a convenient way to classify these examples, not a new type of number. For instance, every integer is also rational, but calling \(-11\) an integer gives more information than calling it merely rational.

Rational Numbers and Irrational Numbers Partition the Reals

A partition divides a set into non-overlapping pieces whose union is the whole set. The rational and irrational numbers provide exactly such a division of \(\mathbb{R}\). In set notation, the irrational numbers are \(\mathbb{R}\setminus\mathbb{Q}\): the real numbers left after the rational numbers are removed.

Proof. By the definition of irrationality, a number belongs to \(\mathbb{I}\) exactly when it is real and does not belong to \(\mathbb{Q}\). This is precisely the membership condition for \(\mathbb{R}\setminus\mathbb{Q}\), so \(\mathbb{I}=\mathbb{R}\setminus\mathbb{Q}\).

Now let \(x\in\mathbb{R}\). Either \(x\in\mathbb{Q}\) or \(x\notin\mathbb{Q}\). In the first case \(x\in\mathbb{Q}\); in the second, since \(x\) is real, \(x\in\mathbb{I}\). Thus every real number belongs to \(\mathbb{Q}\cup\mathbb{I}\). Both \(\mathbb{Q}\) and \(\mathbb{I}\) consist of real numbers, so their union is a subset of \(\mathbb{R}\) as well. Hence \(\mathbb{Q}\cup\mathbb{I}=\mathbb{R}\).

No number can be both rational and irrational: irrationality means being real and not rational. Therefore \(\mathbb{Q}\cap\mathbb{I}=\varnothing\). This proves all three claims. \(\square\)

How Many Real Numbers Are Irrational?

The partition theorem describes which numbers are rational and which are irrational, but it does not say how many numbers are in each part. Earlier in the course, the Countable Sets tutorial established tools for comparing the sizes of infinite sets, and the Uncountability and Cantor’s Diagonal Argument tutorial proved that the open interval \((0,1)\) is uncountable. Since \((0,1)\subseteq\mathbb{R}\), it follows that \(\mathbb{R}\) is uncountable: if \(\mathbb{R}\) were countable, each of its subsets would be countable, contradicting the result for \((0,1)\).

In contrast, the rational numbers are countable. The key idea is that a rational number can be represented by an integer pair \((a,b)\) with \(b\ne0\). There are countably many such pairs: the integers can be listed, the nonzero integers form a subset of that countable set, and the set of pairs of natural numbers is countable by an earlier theorem. The map that sends \((a,b)\) to \(a/b\) has the rational numbers as its range. A range of a countable set is countable, even though different pairs can represent the same rational number.

Together, these facts show that the irrational numbers are not merely present among the reals; they cannot be listed in a sequence. Here is the argument.

Proof. Suppose, for a contradiction, that \(\mathbb{I}\) is countable. The rational numbers are countable as described above. The Countable Union Theorem then implies that \(\mathbb{Q}\cup\mathbb{I}\) is countable. But the Rational–Irrational Partition of the Reals gives \(\mathbb{Q}\cup\mathbb{I}=\mathbb{R}\), so this would make \(\mathbb{R}\) countable. That contradicts the uncountability of \(\mathbb{R}\), which follows because it contains the uncountable interval \((0,1)\). Therefore \(\mathbb{I}\) is uncountable. \(\square\)

This argument uses a useful general strategy: if a set is uncountable and one part of a two-part partition is countable, the other part cannot also be countable. Otherwise the union of the two parts would be countable. The result also gives perspective on the number line: although fractions are dense in the familiar picture, the rational numbers do not account for all its points, or even for most of its size in the sense of countability.

Reading the Number Systems Carefully

A common source of confusion is treating “real” and “rational” as competing labels. They are not. Every rational number is real, so a number such as \(5/6\) is both rational and real. “Irrational” is the contrasting category within the reals: it means real but not rational. The partition theorem makes this distinction exact.

Another useful caution concerns decimal notation. A terminating decimal is rational because it is an integer divided by a power of \(10\), as in the worked example. But the absence of a terminating decimal is not, by itself, a complete proof of irrationality unless the properties of infinite decimal expansions have been established and handled carefully. Here, the definition by integer fractions and the results about irrationality provide a more direct basis for classification.

The number systems can therefore be remembered in two complementary ways. The inclusion chain places \(\mathbb{N}_0\), \(\mathbb{Z}\), and \(\mathbb{Q}\) inside \(\mathbb{R}\); the rational–irrational partition divides \(\mathbb{R}\) into two disjoint sets. The real line contains both kinds, and the irrational part is uncountable.