Rational numbers are not closed under the square root
To show that the rational numbers are not closed under the square root, it is enough to examine a single, well-known example: the square root of two.
$$ \sqrt{2} $$
I partition the set of nonnegative rational numbers Q into two subsets.
- the set of apparent fractions, namely natural numbers
- the set of non-apparent fractions
I then analyze which of these subsets the square root of two could possibly belong to.
1] Apparent fractions
Apparent fractions coincide with the natural numbers.
Example. The fractions 4/2 and 2/1 are apparent fractions because both are equal to a natural number. $$ \frac{4}{2} = \frac{2}{1} = 2 $$
No natural number, when squared, yields the value 2.
$$ 1^2 = 1 \\ 2^2 = 4 \\ 3^2 = 9 \\ \vdots $$
Therefore, the square root of two, √2, cannot belong to the subset of apparent fractions.
2] Non-apparent fractions
By elimination, the square root of two would have to belong to the subset of non-apparent fractions.
$$ \sqrt{2} = \frac{a}{b} $$
Thus, the number a/b is assumed to be a non-apparent fraction.
Note. In a non-apparent fraction, the numerator is not a multiple of the denominator.
If a/b is equal to the square root of two, √2, then squaring a/b must give 2.
$$ \sqrt{2} = \frac{a}{b} \Leftrightarrow ( \frac{a}{b} )^2 = 2 $$
However, if a/b is a non-apparent fraction, then its square (a/b)2 is also a non-apparent fraction.
Example. Consider any non-apparent fraction, for example $$ \frac{3}{2} $$ Its square remains a non-apparent fraction: $$ ( \frac{3}{2} )^2 = \frac{3^2}{2^2} = \frac{9}{4} $$ because in non-apparent fractions the numerator and denominator are coprime numbers, meaning they have no prime factors in common.
From this observation, it follows that the following equality is impossible.
$$ (\frac{a}{b})^2 = 2 $$
This is because (a/b)2 is a non-apparent fraction, hence not a natural number, whereas the number 2 is an apparent fraction, that is, a natural number.
3] Conclusion
In conclusion, the square root of two is neither an apparent fraction nor a non-apparent fraction.
Therefore, the square root of two does not belong to the set of rational numbers.
$$ \sqrt{2} \notin Q $$
This suffices to conclude that the rational numbers are not closed under the square root.
Alternative proof
This proof is based on a sequence of approximations from below and from above.
The aim is to exhibit a square root that does not lie in the set of rational numbers.
Once again, I consider the square root of two.
$$ \sqrt{2} $$
I begin by identifying two integers whose squares give the best approximation of 2 from below and from above.
$$ 1^2 < 2 < 2^2 $$
As a second step, I look for two rational numbers with one decimal digit, one in the interval (12;2) and the other in (2;22), whose squares provide the closest approximations of 2 from below and from above.
$$ 1.4^2 < 2 < 1.5^2 $$
Note. To determine the best lower and upper approximations, I compute the squares of rational numbers with one decimal digit, increasing gradually from 1, the lower bound of the inequality, until the 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 approximation from below is 1.4, while the best approximation from above is 1.5.
I repeat the same procedure to obtain successive approximations, each time adding one more decimal digit.
$$ 1.41^2 < 2 < 1.42^2 $$
$$ 1.414^2 < 2 < 1.415^2 $$
$$ 1.4142^2 < 2 < 1.4143^2 $$
$$ 1.41421^2 < 2 < 1.41422^2 $$
The sequence of lower approximations is increasing.
$$ s = 1 \ , \ 1.41 \ , \ 1.414 \ , \ 1.4142 \ , \ 1.41421 \ , \ ... $$
The sequence of upper approximations is decreasing.
$$ s = 2 \ , \ 1.42 \ , \ 1.415 \ , \ 1.4143 \ , \ 1.41422 \ , \ ... $$
As the number of decimal digits grows, both sequences converge to the square root of two.
However, they do not converge to a terminating decimal, nor to a repeating infinite decimal. They converge to a non-terminating, non-repeating decimal number.
By definition, rational numbers include terminating decimals and repeating infinite decimals.
Since the square root of two is a non-terminating non-repeating decimal, it follows that the square root of two is not a rational number.
$$ \sqrt{2} \notin Q $$
Therefore, the rational numbers are not closed under the square root.
Note. The same reasoning applies to roots of any order, such as cube roots, fourth roots, and higher-order roots. In general, nth roots are not closed operations on the rational numbers.
And so on.
