Proof that the Square Root of 2 is a Real Number

# Proof that the Square Root of 2 is a Real Number

Recall that earlier we proved on the Proof that the Square Root of 2 is Irrational page that $\sqrt{2} \not \in \mathbb{Q}$. We will now look into proving that the square root of $2$ is in fact a real number nevertheless.

Theorem 1: There is a positive real number $x = \sqrt{2}$ such that $x^2 = 2$. |

**Proof of Theorem:**Let the set $S$ be such that $S := \{ s \in \mathbb{R} : 0 ≤ x, s^2 < 2 \}$. We first note that this set is nonempty since clearly $1 \in S$, that is $0 ≤ 1$ and $1^2 = 1 < 2$. Furthermore, we note that this set is also bounded above by 2 since for any $t \in \mathbb{R}$, if $t > 2$ then $t^2 > 4$ so $t%2 \not < 2$.

- Therefore this set has a supremum in $\mathbb{R}$. Let $x = \sup S$ and $x > 1$. We will now show that our supremum $x = \sqrt{2}$ by showing that $x^2 < 2$ and $x^2 > 2$ are invalid cases.

- First consider the case where $x^2 < 2$. We want to find a natural number $n \in \mathbb{N}$ such that $x + \frac{1}{n} \in S$ which would imply that $x = \sup S$ is not an upper bound.

\begin{align} \quad \left ( x + \frac{1}{n} \right ) ^2 = x^2 + \frac{2x}{n} + \frac{1}{n^2} ≤ x^2 + \frac{2x}{n} + \frac{1}{n} = x^2 + \frac{1}{n} ( 2x + 1) \end{align}

- Now we want to choose $n$ such that $x^2 + \frac{1}{n} (2x + 1) < 2$ or rather $\frac{1}{n} (2x + 1) < 2 - x^2$ so that $\left ( x + \frac{1}{n} \right )^2 < 2$.

- Now we note that $\frac{2 - x^2}{2x + 1} > 0$ since the numerator is positive because $x^2 < 2$ so $0 < 2 - x^2$ and the denominator is positive since $x ≥ 0$. Therefore by The Archimedean Property there exists some natural number $n$ such that $\frac{1}{n} ≤ \frac{2 - x^2}{2x + 1}$. Therefore we have obtained such a number $n$ where $\left ( x + \frac{1}{n} \right ) \in S$ which contradicts the fact that $x = \sup S$ since $x < x + \frac{1}{n}$.

- Now consider the case where $x^2 > 2$. We want to show that it is possible to find a natural number $m \in \mathbb{N}$ such that $x - \frac{1}{m}$ is also an upper bound of $S$. Now note that:

\begin{align} \: \left ( x - \frac{1}{m} \right )^2 = x^2 - \frac{2x}{m} + \frac{1}{m^2} > x^2 - \frac{2x}{m} \end{align}

- So we want to choose $m$ such that $x^2 - \frac{2x}{m} > 2$ or equivalently $\frac{2x}{m} < x^2 - 2$ to ensure that $\left ( x - \frac{1}{m} \right )^2 > 2$. We know that $x^2 - 2 > 0$ and $2x > 0$ and so therefore:

\begin{align} \frac{1}{m} < \frac{x^2 - 2}{2x} \end{align}

- Once again by the Archimedean property we know that such a natural number $m$ exists.

- Now suppose that $s \in S$. Then $s^2 < 2 < x - \frac{1}{m}$ which implies that $x - \frac{1}{m}$ is an upper bound for $S$. But then we have a contradiction since $x = \sup S$ is not the supremum of $S$ since $x - \frac{1}{m} < x$.

- Therefore our only possibility is that $x^2 = 2$ and so $x = \sqrt{2} = \sup S$. Since the supremum of a set is a real number, we conclude $\sqrt{2} \in \mathbb{R}$. $\blacksquare$