Exercises §2 — Ross, Elementary Analysis, 2nd Edition
Exercise 2.1. Show that \(\sqrt{3},\ \sqrt{5},\ \sqrt{7},\ \sqrt{24},\ \sqrt{31}\) are not rational numbers.
Proof.
\(\sqrt{3}\): Note \(\sqrt{3}\) is a solution to \(x^2 - 3 = 0\). By the corollary to the Rational Zeros Theorem, any rational solution must be an integer that divides \(-3\), which is one of \(\pm 1, \pm 3\). None of these squares to \(3\), hence \(\sqrt{3}\) is irrational.
The same argument applies to any square root of a prime number, hence \(\sqrt{5},\ \sqrt{7},\ \sqrt{31}\) are all irrational.
\(\sqrt{24}\): Here \(x^2 - 24 = 0\), so any rational candidate is an integer that divides \(-24\), namely
\[ \pm 1,\ \pm 24,\ \pm 2,\ \pm 12,\ \pm 3,\ \pm 8,\ \pm 4,\ \pm 6. \]
None of these squares to \(24\), hence \(\sqrt{24}\) is irrational. \(\square\)
Exercise 2.8. Find all rational solutions to the equation \(x^8 - 4x^5 + 13x^3 - 7x + 1 = 0\).
Solution. By the corollary to the Rational Zeros Theorem, the only rational candidates are the integers dividing the constant term \(1\), namely \(\pm 1\).
\[ \begin{aligned} x = 1: &\quad 1 - 4 + 13 - 7 + 1 = 15 - 11 = 4 \neq 0. \\ x = -1: &\quad 1 + 4 - 13 + 7 + 1 = 13 - 13 = 0. \end{aligned} \]
Hence \(-1\) is the only rational solution. \(\square\)