Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 2 Exercises 2.1 & 2.2

Our main tool is Corollary 2.3.

Solution:

$\sqrt{3}$: It is clear that $\sqrt 3$ is a root of the polynomial $x^2-3=0$. By Corollary 2.3, if $\sqrt 3$ is rational, then it must be a divisor of $3$. That is, it must be $\pm 1$ or $\pm 3$. But this is impossible because $\sqrt 3\ne 1$ and $\sqrt 3\ne 3$. Hence $\sqrt 3$ cannot be rational.

$\sqrt{5}, \sqrt 7, \sqrt{31}$ are the same as that of (a) because 5, 7, 31 are prime numbers.

$\sqrt{24}$: It is clear that $\sqrt{24}$ is a root of the polynomial $x^2-24=0$. By Corollary 2.3, if $\sqrt{24}$ is rational, then it must be a divisor of $24$. That is, it must be $\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, $\pm 8$, $\pm 12$, $\pm 24$. But this is impossible because $4<\sqrt{24}<5$. Hence $\sqrt{24}$ cannot be rational.