Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 2 Exercises 2.6

Our main tool is Corollary 2.3.

Solution: Since $b$ is rational, there exist integers $m$ and $n$ such that $b=\dfrac{n}{m}$ and $m\ne 0$. Then

$$

4-7b^2=4-7\frac{n^2}{m^2}=\frac{4m^2}{m^2}-\frac{7n^2}{m^2}=\frac{4m^2-7n^2}{m^2}.

$$ Note that $4m^2-7n^2$ and $m^2$ are integers as well. Moreover, $m^2\ne 0$. Therefore $4-7b^2$ is also rational.

In general, rational numbers are closed under addition, subtraction, multiplication and division. More precisely, performing such operations for rational numbers, the answer remains rational.