Question

Use the well-ordering property to show that \(\sqrt 2 \) is irrational. (Hint: Assume that \(\sqrt 2 \) is rational. Show that the set of positive integers of the form \(b\sqrt 2 \) has a least element \(a\).Then show that \(a\sqrt 2 - a\) is smaller positive integer of this form.)

Short answer

Thus, it proved that \(\sqrt 2 \) is irrational.

Step-by-step solution

Take \(\sqrt 2 \) is rational

Consider \(\sqrt 2 \) is rational. Then, there would exist positive integers \(a,b\) such that \(\sqrt 2 = \frac{a}{b}\). Consequently, the set \(S = \left\{ {k\sqrt 2 :k\;{\rm{and}}\;k\sqrt 2 \;{\rm{are}}\;{\rm{positive}}\;\,{\rm{integers}}} \right\}\) is non-empty set of positive integers,\(S\) has a smallest element, say \(s = t\sqrt 2 \).

Thus, solve as follows:

\(\begin{aligned}{c}s\sqrt 2 - s &= s\sqrt 2 - t\sqrt 2 \\ &= \left( {s - t} \right)\sqrt 2 \end{aligned}\)

Because, \(s\sqrt 2 = t\sqrt 2 \) and \(s\) are both integers, further solve as follows:

\(\begin{aligned}{c}s\sqrt 2 - s &= s\sqrt 2 - t\sqrt 2 \\ &= \left( {s - t} \right)\sqrt 2 \end{aligned}\)

This must also be an integer.

Showing that \(\sqrt 2 \) is irrational.

Furthermore, it is positive because \(s\sqrt 2 - s = s\left( {\sqrt 2 - 1} \right)\) and \(\sqrt 2 > 1\).

It is less than \(s\), because \(\sqrt 2 < 2\) so that \(\sqrt 2 - 1 < 1\). This contradicts the choice of \(s\) as the smallest positive integer in \(S\).

Hence, it follows that \(\sqrt 2 \) is irrational.