Exists Integer Below Any Real Number

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Theorem

Let $x$ be a real number.


Then there exists an integer less than $x$:

$\forall x \in \R: \exists n \in \Z: n < x$


Proof

Clearly we may assume without loss of generality that $x < 0$.

From the Archimedean Principle:

$\exists m \in \N: m > -x$

By Real Numbers form Totally Ordered Field, we have that $\R$ is a totally ordered field.

Therefore by property $(3)$ of Properties of Totally Ordered Field, $\Z \owns -m < x$.

$\blacksquare$