\( x^2 = 2 \). 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 a non-empty set of reals and has an upper bound, 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 \( x^2 < 2 \) and \( x^2 > 2 \) 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 upper bound of \( E \)). Consider \( (x + \varepsilon)^2 \), where \( 0 < \varepsilon < 1 \) is a real:
\[\begin{align}(x + \varepsilon)^2 &= x^2 + 2x\varepsilon + \varepsilon^2 \\ &\le x^2 + 4\varepsilon + \varepsilon^2 && \text{as } x \le 2 \\ &\le x^2 + 5\varepsilon && \text{as } \varepsilon < 1 \\\end{align} \]
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:
\[\begin{align}(x - \varepsilon)^2 &= x^2 - 2x\varepsilon + \varepsilon^2 \\ &> x^2 - 4\varepsilon + \varepsilon^2 && \text{as } x < 2 \\ &> x^2 - 4\varepsilon && \text{as } \varepsilon > 0 \\\end{align}\]
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 \( (x-\varepsilon)^2 < y^2 < 2 \) , 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.