Theorem 0.3 Irrationality of \(\sqrt{n}\)
Let \(n \in \mathbb{N}\) and \(n\) is not a perfect square (i.e. \(n \neq m^2\). Then \(\sqrt{n}\) is irrational.

Proof

Suppose that \(\sqrt{n}\) is rational. \(\sqrt{n} = \frac{p}{q}.\) Then let

$$ E = \{k \in \mathbb{N}: k\sqrt{n} \in \mathbb{N}\}. $$

Then \(E\) is non-empty since \(q\sqrt{n}=p\). This means that \(E\) has a least element by the well ordering principle. Let this element be \(n_0\). Now, let

$$ \begin{align*} A = \{k \in \mathbb{N} \mid k > \sqrt{n}\} \end{align*} $$

Then since \(A \subset \mathbb{N}\) and \(A\) is non-empty, then by the well ordering principle, \(A\) has a least element \(m_0\). By the definition of \(A\), we must have that \(m_0 > \sqrt{n}\). Hence, can find \(m_0 \in \mathbb{N}\) such that

$$ \begin{align*} m_0 - 1< \sqrt{n} < m_0 \quad\quad\quad (1) \end{align*} $$

Note here that if \(m_0 - 1 > \sqrt{n}\), then \(m_0 - 1 \in A\) contradicting the minimality of \(m_0\). Moreover, the above inequality is strict since \(\sqrt{n}\) is not an integer. Then if we subtract \(m_0-1\) from (1), we get that

$$ \begin{align*} 0 < (\sqrt{n} - (m_0-1)) < 1 \quad\quad\quad (2) \end{align*} $$

Now let \(x = n_0 (\sqrt{n} - (m_0-1))\). Then \((2)\) implies

$$ \begin{align*} 0 < x < n_0 \end{align*} $$

[TODO…………………………………………]

\(\ \blacksquare\)

References

  • Lecture Notes by Professor Chun Kit Lai