Set of Rationals Less than Root 2 has no Greatest Element

Theorem

Let $A$ be the set of all positive rational numbers $p$ such that $p^2 < 2$.

Then $A$ has no greatest element.

Aiming for a contradiction, suppose $p \in A$ is the greatest element of $A$.

Then by definition of $A$:

$p^2 < 2$

Let $h \in \Q$ be a rational number such that $0 < h < 1$ such that:

$h < \dfrac {2 - p^2} {2 p + 1}$

This is always possible, because by definition $2 - p^2 > 0$ and $2 p + 1 > 0$.

Let $q = p + h$.

Then $q > p$, and:

$\ds q^2$

$=$

$\ds p^2 + \paren {2 p + h} h$

$\ds $

$<$

$\ds p^2 + \paren {2 p + 1} h$

$\ds $

$<$

$\ds p^2 + \paren {2 - p^2}$

$\ds $

$=$

$\ds 2$

That means $q \in A$.

This contradicts our assertion that $p$ is the greatest element of $A$.

It follows that $A$ can have no greatest element.

$\blacksquare$

1964: Walter Rudin: Principles of Mathematical Analysis (2nd ed.) ... (previous) ... (next): Chapter $1$: The Real and Complex Number Systems: Introduction: $1.1$ Example