Let $A=\{x\in \mathbb{R}:x^2\leq{2}\}$ with $\sup A=\alpha$.
When proving that the $\sqrt{2}$ exists in $\mathbb{R}$ using the method of contradiction, do we show that a contradiction arises for the case $\alpha^2 \lt{2}$ by negating condition $1$ of the definition for least upper bound, and a contradiction for the case $\alpha^2 \gt{2}$ by negating condition $2$? leading to the conclusion that $\alpha^2={2}$ is the only possibility?
Definition: Let $A\subseteq\mathbb{R}$ and let $\alpha\in \mathbb{R}$. Define $\alpha$ to be the least upper bound for $A$ if and only if the following two conditions are satisfied.
- $\alpha$ is an upper bound for $A$; and
- for all upper bounds $\alpha'$ for $A$, we have that $\alpha \leq{\alpha'}$