\(\sqrt{2}\)
Theorem. There is no rational number \(r\) such that \(r^2=2\).
Proof. Assume toward a contradiction that there exists \(r=\frac{m}{n}\) with \(m\in\Z\) and \(n\in\N\) such that \(r^2=2\). We may assume that \(m\) and \(n\) are not both even, since we could cancel factors of \(2\) until at least one is odd.
The assumption that \(r^2=2\) implies \(m^2=2n^2\). This shows that \(m\) is even, that is \(m=2k\) for some \(k\in\Z\). Plugging this in we have
\[(2k)^2=2n^2\]
which implies \(2k^2=n^2\). This implies that \(n\) is even. This contradiction shows that our initial assumption was false, and hence the number \(r\in\mathbb{Q}\) does not exist. \(\Box\)
Theorem. There is a real number \(x\) such that \(x^2=2\).
Proof. Define the set \[S = \{s\in\R : s\geq 0\text{ and }s^2\leq 2\}.\]
The set \(S\) is not empty, since \(0\in S\). Also note that if \(t>2\), then \(t^2>4\), and hence \(t\notin S\). Therefore, \(2\) is an upper bound for \(S\). By the Completeness Property of \(\R\) we conclude that \(S\) has a supremum. Set \(x = \sup S\).
Theorem. There is a real number \(x\) such that \(x^2=2\).
This Shows that \(x+\frac{1}{n}\in S\) and hence \(x\) is not an upper bound for \(S\). \(\Rightarrow\Leftarrow\)
\[<x^2+2-x^2 = 2.\]
Proof continued. Assume toward a contradiction that \(x^2\neq 2\).
Case 1. Suppose \(x^2<2\). By the Archimedean Property there is a number \(n\in\N\) such that \[n>\frac{2x+1}{2-x^2},\]
and hence \[\frac{1}{n}(2x+1)<2-x^2.\] Since \(n\geq1\) we see that \(n^2\geq n\), and hence \(\frac{1}{n^2}\leq \frac{1}{n}\). We use this to see
\[\left(x+\frac{1}{n}\right)^2 = x^2+\frac{2x}{n} + \frac{1}{n^2}\leq x^2+\frac{2x}{n}+\frac{1}{n} = x^2+\frac{1}{n}(2x+1)\]
Theorem. There is a real number \(x\) such that \(x^2=2\).
Since \(x-\frac{1}{m}<x\) we conclude that \(x-\frac{1}{m}\) is not an upper bound for \(S\). Hence there is some \(s\in S\) such that \(s>x-\frac{1}{m}\). This implies
\[s^2>\left(x-\frac{1}{m}\right)^2>2\]
and hence \(s\notin S\). \(\Rightarrow\Leftarrow\)
Proof continued.
Case 2. Suppose \(x^2>2\). By the Archimedean Property there is a number \(m\in\N\) such that \[m>\frac{2x}{x^2-2},\text{ and hence }\frac{2x}{m}<x^2-2.\] We use this to see
\[\left(x-\frac{1}{m}\right)^2 = x^2-\frac{2x}{m} + \frac{1}{m^2}> x^2-\frac{2x}{m}>x^2-(x^2-2) = 2\]