Let $n$ be contained in $\mathbb N$. Show that $ n^{1/2}$ is either an integer or it is irrational.

Question

My textbook shows the proof of $\sqrt{2}$ but I'm having trouble doing the same exact thing for $\sqrt{n}$. So, I have $A:= \{x$ contained in $\mathbb R: x^2 < n\}$. First if $x^2 < n$ then $x < n$. We know that $A$ is nonempty because $1$ is contained in $A$.... I'm completely lost after this. I know it involves taking an $r$ and so on...

Answer 1

Here's the proof for any (nth) root: The initial steps just establish the "unique factorisation theorem"

(1) 2^n > n for any Positive Integer n.

The proof is trivial by induction. 2^1 = 2 > 1

If 2^n > n then 2^(n+1) = 2.2^n = (2^n) + (2^n) > n + n > n+1

(2) Any Product of n Positive Integers Each ≥ 2 is > n.

If P = {p_i,with all p_i ≥ 2} for i = 1 to n, then ∏p_i ≥ 2^n > n

(3) Finite Factorisation of Positive Integers

If n is a positive integer then it is either prime or for some prime p, p|n in which case n = n'.p We can repeat this for n', but the process must end before n steps because any product of n primes > n.

(3) Unique Factorisation of Positive Integers

Suppose n factors (finitely) into P = {p_i,with all p_i ≥ 2} and Q = {q_i,with all q_i ≥ 2}. We allow factors to be repeated rather than summarising into powers. Then n=p_1.(p_2.p_3…..)=∏q_i (or n=p_1 if there is only one p_i) So, for some q_i, q_i = p_1. Re-label the set Q so that this q_i is q_1 Now, continue this process on-by-one for the remainder of the p_i and conclude that P ⊆ Q. We could alternatively proceed the other way round and conclude that Q ⊆ P. Therefore, P = Q.

(4) The Nth Root of any Positive Integer is either an Integer or Not a Rational Number

Let n√(M) = p/q where M is an integer and p and q have no common factors. Then q is either = 1, or has a unique finite factorisation in primes {q_i,with all q_i ≥ 2}; q=∏q_i , and no q_i divides p.

So M =p^N/q^N = p^N/∏q_i ^N. But no q_i divides p, so unless q=1 then p^N/∏q_i ^N cannot be an integer (M). Therefore M must be of the form p^N, and if it isn’t then n√(M) can’t be rational.