Real Number is not necessarily Rational Number

Theorem

Let $x$ be a real number.

Then it is not necessarily the case that $x$ is also a rational number.

Proof

By Proof by Counterexample:

Let $x = \sqrt 2$.

From Square Root of 2 is Irrational:

$\sqrt 2$ is an irrational number.

By definition:

$x \in \R \setminus \Q$

where:

$\R$ is the set of real numbers

$\Q$ is the set of rational numbers

$\setminus$ denotes the set difference.

Thus $x$, while being a real number, is not also a rational number.

Hence the result.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 1$: The Language of Set Theory