Theorem. There exists $x\in\mathbb {R}: $ with ${x^2=2} $.
Can someone please share a simple proof of the above theorem by defining,
$$\{y\in\mathbb R: \mbox{$y^2\leq {2}$}\}.$$
How do we proceed to use the completeness axiom to prove the theorem? The proof that my lecturer has provided uses contradictions. He also uses the same process to define the square root function. I'm finding it difficult to grasp. Thanks.
Let $S = \{q\in\mathbb R: \mbox{$q^2<2$}\}$. Since $S \subset \Bbb R$ and $\forall x\in S(x^2<2<4)$, $2$ is upper bound for this set. Since this is a subset of real numbers, it has some real least upper bound $L$. If $L^2 -2 > 0,$ choose a sufficiently large $m$ such that $L' = L - 1/m$ and $L'^2 = L^2 - 2L/m + 1/m^2 > 2$, which contradicts the minimality of $L$ (try it yourself using the Archimedian property of the reals). The same for $L^2 - 2 < 0$. Thus $L^2 = 2$ and $L = \sqrt 2 \in \Bbb R$.
Edit: Given the original paper, my answer doesn't really help because I used the exact same process, so I'll explain the idea: completeness is pretty much what makes $\Bbb R$ without "gaps". You'll want to construct, usually, sets that look like Dedekind cuts which will have an upper bound (possibly the least, and that's what you'll seek) as your desired number. Since the least upper bound is always real, prove it is what you want! In the given case, the choices for $\delta$ are not arbitrary: sketch a $\delta$ that would solve your problem (small numbers that will still keep the conditions) and, if your least upper bound isn't the number you desire, you'll have a contradiction! The process of construction is like solving a one-variable linear equation: you first assume there's a solution and then find what it would be, but, to formalize, you just show that there is a solution and then prove it's unique.