Rational numbers are not closed under taking square roots
To demonstrate that the rational numbers are not closed under the square root operation, I will focus on a concrete example, namely the square root of two.
$$ \sqrt{2} $$
I partition the set of non-negative rational numbers Q into two subsets.
- the subset of rational numbers that are integers, that is, fractions equal to natural numbers
- the subset of rational numbers that are not integers
I then examine which of these two subsets the square root of two could possibly belong to.
1] Rational numbers that are integers
These are rational numbers that, although written in fractional form, coincide with natural numbers.
Example. The fractions 4/2 and 2/1 belong to this subset because both are equal to a natural number. $$ \frac{4}{2} = \frac{2}{1} = 2 $$
No natural number, when squared, is equal to 2.
$$ 1^2 = 1 \\ 2^2 = 4 \\ 3^2 = 9 \\ \vdots $$
Therefore, the square root of two, √2, cannot belong to the subset of rational numbers that are integers.
2] Rational numbers that are not integers
By elimination, the square root of two would then have to belong to the subset of rational numbers that are not integers.
$$ \sqrt{2} = \frac{a}{b} $$
This amounts to assuming that the number a/b is a rational number that is not an integer.
Note. A rational number that is not an integer can be expressed as a fraction whose numerator is not a multiple of the denominator.
If a/b were equal to the square root of two, √2, then squaring a/b would necessarily yield the value 2.
$$ \sqrt{2} = \frac{a}{b} \Leftrightarrow ( \frac{a}{b} )^2 = 2 $$
However, if a/b is a rational number that is not an integer and is written in lowest terms, then its square (a/b)2 is also a rational number that is not an integer.
Example. Consider a rational number that is not an integer, for instance $$ \frac{3}{2} $$ Its square is still not an integer. $$ ( \frac{3}{2} )^2 = \frac{3^2}{2^2} = \frac{9}{4} $$ This follows from the fact that, when a rational number is written in lowest terms, its numerator and denominator are coprime numbers, meaning they share no common prime factors.
From this reasoning, it follows that the following equality cannot hold.
$$ (\frac{a}{b})^2 = 2 $$
This is because (a/b)2 is not an integer, whereas the number 2 is an integer.
3] Conclusion
In conclusion, the square root of two is neither an integer nor a rational number that is not an integer.
Therefore, the square root of two does not belong to the set of rational numbers.
$$ \sqrt{2} \notin Q $$
This is sufficient to conclude that the rational numbers are not closed under taking square roots.
Alternative proof
This alternative argument is based on a sequence of approximations from below and from above.
The aim is to exhibit explicitly a square root that does not belong to the rational numbers.
Once again, I consider the square root of two.
$$ \sqrt{2} $$
I first identify two integers whose squares provide lower and upper bounds for the number 2.
$$ 1^2 < 2 < 2^2 $$
As a second approximation, I look for two rational numbers with one decimal place, lying in the intervals (12;2) and (2;22), whose squares yield the best lower and upper approximations of 2.
$$ 1.4^2 < 2 < 1.5^2 $$
Note. To determine the best lower and upper approximations, I compute the squares of rational numbers with one decimal place, proceeding incrementally from 1, the lower bound of the inequality, until I encounter a value whose square exceeds 2. $$ 1.0^2 = 1 \\ 1.1^2 = 1.21 \\ 1.2^2 = 1.44 \\ 1.3^2 = 1.69 \\ 1.4^2 = 1.96 \\ 1.5^2 = \color{red}{2.25} $$ Hence, the best lower approximation is 1.4, while the best upper approximation is 1.5.
I then repeat the same procedure to obtain successive approximations, adding one additional decimal digit at each step.
$$ 1.41^2 < 2 < 1.42^2 $$
$$ 1.414^2 < 2 < 1.415^2 $$
$$ 1.4142^2 < 2 < 1.4143^2 $$
$$ 1.41421^2 < 2 < 1.41422^2 $$
The sequence of lower approximations is strictly increasing.
$$ s = 1 \ , \ 1.41 \ , \ 1.414 \ , \ 1.4142 \ , \ 1.41421 \ , \ ... $$
The sequence of upper approximations is strictly decreasing.
$$ s = 2 \ , \ 1.42 \ , \ 1.415 \ , \ 1.4143 \ , \ 1.41422 \ , \ ... $$
Both sequences converge to the square root of two as the number of decimal digits increases.
However, they do not converge to a terminating decimal, nor to a repeating infinite decimal. Instead, they converge to a non-terminating, non-repeating decimal number.
By definition, rational numbers are precisely those numbers that admit either a terminating decimal expansion or a repeating infinite decimal expansion.
Since the square root of two has a non-terminating non-repeating decimal expansion, it follows that the square root of two is not a rational number.
$$ \sqrt{2} \notin Q $$
Therefore, taking square roots is not a closed operation on the rational numbers.
Note. The same line of reasoning applies to roots of any order, such as cube roots, fourth roots, and higher-order roots. In general, taking n-th roots is not a closed operation on the rational numbers.
And so on.
