# \( x^2 = 2 \). There exists a positive real whose square is 2. Proposition.

This proposition's proof is enabled by the existance of the least upper bound for sets of reals.### Proof

Let \( E \) be the set \( \{y \in R: y \ge 0 \text{ and } y^2 < 2 \} \). \( E \) has an upper bound of 2, as if \( y > 2 \) then \( y^2 > 4 > 2 \). As \( E \) is [...] and has [...], it must have a least upper bound. Let \( x = \sup E \) be the least upper bound of \( E \).

Exactly one of \( x^2 < 2, x^2 = 2 \text{ or } x^2 > 2 \) must be true. We proceed to show that [...] and [...] lead to contradictions and must be false.

#### \( x^2 < 2 \) must be false.

Assume \( x^2 < 2 \) is true. (This means that \( x \in E \), but we must go further to impose that \( x \) is not an [...] of \( E \)). Consider \( (x + \varepsilon)^2 \), where \( 0 < \varepsilon < 1 \) is a real:

\[\begin{align}(x + \varepsilon)^2 &= x^2 + 2x\varepsilon + \varepsilon^2 \\ &\le x^2 + 4\varepsilon + \varepsilon^2 && \text{as } [...] \\ &\le x^2 + 5\varepsilon && \text{as } [...] \\\end{align} \]

We can choose an \( \varepsilon \) such that \( x^2 + 5\varepsilon < 2 \). From above, \( (x + \varepsilon)^2 \le x^2 + 5\varepsilon \), so it must be the case that \( (x + \varepsilon)^2 < 2 \) is also true. This implies that \( x + \varepsilon \in E \). As \( x + \varepsilon > x \), \( x \) cannot be an upper bound of \( E \), which is a contradiction. Thus, \( x^2 < 2 \) is false.

#### \( x^2 > 2 \) must be false.

A similar argument to above. Assume \( x^2 > 2 \). Consider \( (x - \varepsilon)^2 \), where \( 0 < \varepsilon < 1 \) is a real:

\[\begin{align}(x - \varepsilon)^2 &= x^2 - 2x\varepsilon + \varepsilon^2 \\ &> x^2 - 4\varepsilon + \varepsilon^2 && \text{as } [...] \\ &> x^2 - 4\varepsilon && \text{as } [...] \\\end{align}\]

Since \( x^2 > 2 \), we can choose an \( \varepsilon \) such that \( x^2 - 4\varepsilon > 2 \) (why? This reasoning tripped me up when I first attempted the proof). In this case, \( (x-\varepsilon)^2 > 2 \) is also true.

We want \( x-\varepsilon \) to be an upper bound, and to show this, it must be true that \( x - \varepsilon \gt y \) for all \( y \in E \). We show this by contradiction: if \( x-\varepsilon < y \) for some \( y \in E \), then [...], a contradiction. Thus \( x - \varepsilon \) is an upper bound for E.

As \( x - \varepsilon \) is an upper bound for E and is less than \( x \), it cannot be true that \( x \) is a least upper bound. Thus, \( x^2 > 2 \) mus be false.

As both \( x^2 < 2 \) and \( x^2 > 2 \) are false, it must be the case that \( x^2 = 2 \) is true.