Rational Numbers and the Least Upper Bound Property

Show that the set of rational numbers $Q$ does not have the least upper bound property by considering the set $A = \{ p \in \mathbb{Q} : p^2 < 2 \}$ and demonstrating that the supremum of this set is $\sqrt{2}$, which is not a rational number.

The concept of the least upper bound, or supremum, is central in real analysis, particularly in understanding the structure of ordered sets. The problem at hand explores the least upper bound property by examining the set of rational numbers. Rational numbers, although dense in the real numbers, lack this property, which is a key hallmark of the real number system. The challenge is to show that there exists a set of rational numbers that does not have a least upper bound within the rationals themselves, emphasizing the necessity of irrational numbers to complete the real line.

In attempting to demonstrate this, we're considering the set of all rational numbers whose squares are less than two. Intuitively, you'd expect this set to have an upper bound, and indeed it does, but not within the rational numbers. The least upper bound is actually the square root of two, an irrational number. This shows that while the rationals can approximate reals arbitrarily closely, they fall short when dealing with limits and bounds in certain contexts.

Understanding this distinction is crucial for grasping the completeness of the reals, contrasting with the incompleteness of the rationals. It highlights why certain real analysis concepts, such as convergence and limits, are so fundamentally anchored in the real number system. By endeavoring through this problem, one appreciates the elegance and necessity of irrational numbers in forming a complete and ordered field, encapsulating the essence of real numbers.