In Stephan Abbot's understanding analysis, he proves that the square root of two exists and equals $\sup(T)$ where $T = \{t \in R: t^2<2\}$ (Theorem 1.4.5). I proved this differently then in the book and I was wondering if my solution is correct.
If $a \neq \sqrt 2$, $a^2 >2$ or $a^2 < 2$. If $a^2<2$, choose $k^2 = \frac{a^2+2}{2}$ (it will be assumed k is positive). Then $$a^2 < 2 \Rightarrow a^2 + 2 < 4 \Rightarrow \frac{a^2 +2}{2} < 2$$ or $k \in T$. Similarly, $$a^2<2 \Rightarrow 2a^2 < 2+a^2 \Rightarrow a^2 < \frac{a^2 + 2}{2}$$ which means a is not an upper bound for T. If $a^2 > 2$, again set $k^2 = \frac{a^2 + 2}{2}$. It follows that $$a^2 > 2 \Rightarrow a^2+2>4 \Rightarrow \frac{a^2 + 2}{2} > 2$$ so $k$ is an upper bound for T. Similarly, $$a^2>2 \Rightarrow 2a^2 >2+a^2 \Rightarrow a^2 > \frac{a^2 + 2}{2}$$ Which means k is an upper bound which is less than the supremum which is a contradiction. Since the axiom of Completeness requires a supremum exists for a non-empty bounded set, $a^2 = 2$ so $\sqrt 2$ must exist in R. $QED$