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## Rational numbers are closed under algebraic operations

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.