Proof: Let $T_1$ be the set of real numbers $x_0$ such that $x_{0}^2<2$. By theorem 5.5.9, $T_1$ has a least upper bound. Suppose $t_0$ is the least upper bound of $T_1$ then $t_0\in\Bbb{R}$ and $t_{0}^2\leq 2$. If $t_{0}^2< 2$ then by proposition 5.4.14., $\exists q\in\Bbb{Q^+}$ such that $t_{0}^2<q<2$. This implies that $q\in T_1$ and $t_0$ is not an upperbound which contradicts the premise. Therefore $x^2=2$ where $2$ is the upperbound of $T_1$.
[Math] Proof verification: There exists a positive real number $x$ such that $x^2=2$
proof-verificationreal-analysis
Best Answer
What is $\sqrt2$?
If you don't yet know there is a positive real number $x$ such that $x^2=2$, you can't identify a real number by the name of $\sqrt2$, since $\sqrt2$ is, by definition, the positive real number $x$ such that $x^2=2$.
Instead, try starting with letting $T_1$ be the set of real numbers $x_0$ such that $x_0^2 < 2$.