Two ideas for proving irrationality I want to "construct" a proof showing that $\sqrt{n}$ is irrational and two ideas entered my mind. But i have doubts regarding the soundness of the subsequent reasoning that may spring forth based on these ideas. The ideas are the following:


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*Considering that an irrational number is not rational and that the integers can be mapped with the rationals, the irrationality of a number can be proved by showing that there doesn't exist a mapping to this number. (One of) the three definitions of bijection, injection and surjection can be used. 

*Considering that the decimal expansion of an irrational number is non-periodic, irrationality can be proved using the limit concept as the limit of a periodic expansion does not exist. In this sense, a limit in a repeating decimal expansion means the last digit of the period. Limit and supremum can be considered equivalent in this context of periodicity.
Now, i don't know if these are good ideas, so some output is appreciated
 A: Concerning 1, the real numbers have many countable subsets that contain irrational numbers: even quite nice ones like the subfield generated $\mathbb{Q}[x_1, x_2, \ldots]$ generated over the rational numbers $\mathbb{Q}$ by any finite or countably infinite sequence of irrational numbers $x_1, x_2 \ldots$. So this line of argument isn't promising.
Conerning 2: it might be interesting to see if you can analyse an algorithm for calculating the decimal expansion of the square root of a number and show that it it doees not terminate (i.e., if $n$ is not a perfect square) then its output is not periodic, but I don't know of such a proof, other than to use one of the usual proofs that integers that are not perfect squares are irrational (e.g., by considering the prime factorisation).
A: One can prove that the decimal expansion of $\sqrt n$ is non-periodic by first proving that $\sqrt n$ is irrational and also separately proving that decimal expansions of rational numbers are periodic.  Proving more directly that the decimal expansion is non-periodic I've never seen.  I won't say that it can't be done, but I would be a bit surprised.
To make your first idea work, I suspect you first need to specify which enumeration of the rational numbers you want to use.  I don't know where such an argument would go from there. One goes like this: Suppose we have an enumeration $a_1,a_2,a_3,\ldots$ of all rational numbers.  Let $b_1=a_1$ and let $c_1$ be the first term further along the sequence $\{a_n\}$ that is bigger than $b_1$.  We will seek an irrational nunmber in the interval $(b_1,c_1)$.  Now let $b_2$ be the first term further along the sequence $\{a_n\}$ that $c_1$ that is in that interval, and let $c_2$ be the first term further along the sequence $\{a_n\}$ that is between $b_2$ and $c_1$.  Now we will seek our irrational number in this narrower interval $(b_2,c_2)$. Define $b_3$ and $c_3$ similarly, and so on.  We have $b_1<b_2<b_3<\cdots<c_3<c_2<c_2$, and just one number is between the $b$s and the $c$s.  That number must be irrational because all rational numbers were excluded by our way of creating the sequence of successively narrower intervals. But can you do all of this in some way that assures that $\sqrt n$ will be the irrational number that you find?  Making sure that all rational numbers are included in a sequence used for this sort of purpose seems wasteful if all you're trying to prove is that one specified number is irrational.  And even if there were uncountably many rationals, you'd only need countably many of them for a proof like this, so the countability of the rationals doesn't seem important here.
Another way to show that $\sqrt n$ is irrational is to show that its simple continued fraction expansion does not terminate.  That can be done by showing that that expansion is periodic.  That doesn't work for cube roots or other higher-degree roots.
