The rational numbers \(\mathbb{Q}\) appear at first glance to be a continuous whole: between any two fractions, there is always another. But this impression is deceptive: there are sets of rational numbers that are indeed bounded, but whose supremum or infimum simply does not exist in \(\mathbb{Q}\) . The reason lies in the existence of irrational numbers like \(\sqrt{2}\) , which, in a sense, create invisible holes in the rational number line.
To show that \(\mathbb{Q}\) is not complete, we will specify two non-empty subsets of \(\mathbb{Q}\) : one that is bounded above but has no supremum in \(\mathbb{Q}\) , and another that is bounded below but has no infimum in \(\mathbb{Q}\) .
Consider the set \(X = \{x \in \mathbb{Q} \mid x \geq 0,\; x^2 < 2\}\) . The set \(X\) is not empty, since \(1 \in X\) . If \(x \geq 2\) , then \(x^2 \geq 4 > 2\) , which contradicts \(x^2 < 2\) . Therefore, \(x < 2\) holds for every \(x \in X\) . Thus \(X \subset [0, 2]\) , and \(2\) is one of the upper bounds of \(X\) . Therefore, \(X\) is bounded above.
Let \(x \in X\) . We show that there exists a \(n \in \mathbb{N}\) such that \(x + \frac{1}{n} \in X\) . It is found that
\[\left(x + \frac{1}{n}\right)^2 = x^2 + \frac{2x}{n} + \frac{1}{n^2} \leq x^2 + \frac{2x}{n} + \frac{1}{n} = x^2 + \frac{1}{n}(2x + 1).\]
Thus \(x^2 + \frac{1}{n}(2x + 1) < 2 \iff \frac{1}{n} < \frac{2 - x^2}{2x + 1}\) . Since \(2 - x^2 > 0\) and \(2x + 1 > 0\) , Archimedes' axiom guarantees the existence of \(n \in \mathbb{N}\) . Therefore, \(\left(x + \frac{1}{n}\right)^2 < 2\) , and thus \(x + \frac{1}{n} \in X\) .
Now consider the set \(Y = \{y \in \mathbb{Q} \mid y > 0,\; y^2 > 2\}\) . Obviously, \(Y \neq \varnothing\) . Furthermore, \(Y\) is bounded below and unbounded above, that is \(Y \subset (0, +\infty)\) . Let \(y \in Y\) . We show that there exists a \(m \in \mathbb{N}\) such that \(y - \frac{1}{m} \in Y\) . It is observed that
\[\left(y - \frac{1}{m}\right)^2 = y^2 - \frac{2y}{m} + \frac{1}{m^2} > y^2 - \frac{2y}{m}.\]
Thus \(y^2 - \frac{2y}{m} > 2 \iff \frac{1}{m} < \frac{y^2 - 2}{2y}\) . Since \(y^2 - 2 > 0\) and \(2y > 0\) , Archimedes' axiom again guarantees the existence of a \(m \in \mathbb{N}\) . Therefore, \(\left(y - \frac{1}{m}\right)^2 > 2\) , and thus \(y - \frac{1}{m} \in Y\) .
Suppose \(w = \sup X\) . It cannot be that \(w^2 < 2\) , because then \(w \in X\) , and there would be a \(n \in \mathbb{N}\) with \(w + \frac{1}{n} \in X\) and \(w < w + \frac{1}{n}\) , which contradicts the fact that \(w\) is an upper bound of \(X\) . It also cannot be true that \(w^2 > 2\) , because then \(w \in Y\) , and there would be a \(m \in \mathbb{N}\) with \(w - \frac{1}{m} \in Y\) and \(w - \frac{1}{m} < w\) , which contradicts the fact that \(w\) is the least upper bound of \(X\) .
Let us now assume that \(v = \inf Y\) . It cannot be true that \(v^2 > 2\) , because then \(v \in Y\) , and \(v\) would not be a lower bound of \(Y\) . Likewise, it cannot be true that \(v^2 < 2\) , because then \(v \in X\) , and thus \(v\) would not be the greatest lower bound of \(Y\) .
The only remaining possibilities are therefore \(w^2 = 2\) and \(v^2 = 2\) . However, there is no rational number whose square is equal to \(2\) . Therefore \(w = \sup X\) cannot exist in \(\mathbb{Q}\) . For the same reason, \(v = \inf Y\) cannot exist in \(\mathbb{Q}\) .
Therefore, \(\mathbb{Q}\) is not a completely ordered field. \(\square\)