# Prove $\sqrt[m]{D}$ irrationality simply, without unique prime factorization

I apologize if this has been asked already. I am aware of proofs that $\sqrt[m]{D}$ is either integer or irrational for $m,D\in\mathbb{N}$, all of which that I recall and understand make use of the unique prime-factorization theorem (perfect example here; this one may not but I'm not familiar with the method). When I was young, I always disliked proofs that made use of unique prime factorization (for some reason) such as the standard proof of the irrationality of $\sqrt{2}$. However, there are other ways of proving $\sqrt{D}$ irrational, my personal favourite being the following proof (I have no idea where I got this from):

Let $D$ be a non-square integer. Suppose $\sqrt{D}$ is irrational and let $q$ be the smallest integer such that $q\sqrt{D}$ is integer, then for some $n\in\mathbb{N}$ we have: $$n^2<D<(n+1)^2$$ $$\therefore\;\;n<\sqrt{D}<n+1$$ $$\therefore\;\;0<\sqrt{D}-n<1$$ $$\therefore\;\;0<q\sqrt{D}-qn<q$$ But $q\sqrt{D}-qn$ is an integer and so $(q\sqrt{D}-qn)\sqrt{D}=qD-qn\sqrt{D}$ is also an integer; thus $q\sqrt{D}-qn$ is an integer that contradicts minimality of $q$. Thus we have a contradiction and hence $\sqrt{D}$ is irrational.

I like this proof since it uses only basic algebra and does not invoke 'heavy' theorems (unless I'm missing something). I have wondered whether there is a proof of irrationality of $\sqrt[m]{D}$ that is spiritually related to this proof, i.e. one that just uses basic algebra without having to invoke unique prime factorization or other similar results of basic number theory, but I don't see a simple way of extending this proof.

Thus I have the following question: Does anyone know a simple proof of the irrationality of $\sqrt[m]{D}$ for $m\in\mathbb{N}$ and $D\ne k^m$ for any $k\in\mathbb{N}$ that does not rely on the unique prime-factorization theorem (or other similarly strong theorems), and preferably just uses basic algebra like the above proof for $m=2$?

• Why the downvote? I've rolled back until I get more feedback. It would be great if someone could clarify why I'm getting downvotes for this question. I can't see what's wrong with the question, so I'd love to know. – Anon Feb 7 '17 at 22:10
• I see I should have asked the question more carefully since it hasn't received much love. Maybe I'll ask another question sometime to get what I really wanted. – Anon Feb 7 '17 at 22:23

I don't know if this is what is expected, anyway ...

Consider the equation $x^m-D=0$ (with $D\in\mathbb{N}$ not a perfect $m-$th power and $m\ge2$).

If $p/q$ is a rational root of this equation, with $(p,q)\in\mathbb{Z}\times\mathbb{N^\star}$ and $gcd(p,q)=1$,

then (by RRT) $q\mid 1$, so that $p/q$ is an integer, which is excluded.

Hence $\sqrt[m]{D}\not\in\mathbb{Q} \setminus \mathbb Z$.

• But is the rational root theorem any weaker than the fundamental theorem of arithmetic? I added a clarifying phrase to my question in case it wasn't clear - I was looking for a proof which followed by basic algebra. Perhaps RRT is simpler than I thought though?... – Anon Feb 7 '17 at 6:02
• @Anon: For the RRT (Rational Roots Theorem), you only need Gauss theorem, which asserts that, given $a,b,c\in\mathbb{N}$, if $a\mid bc$ and $gcd(a,b)=1$, then we have $a\mid c$. And Gauss theorem doesn't rely on the FTA, but can be view as a consequence of Bezout identity, which in turn can be viewed as a consequence of the definition of the greatest common divisor : $d=gcd(a,b)$ is the only positive integer such that $a\mathbb{Z}+b\mathbb{Z}=d\mathbb{Z}$. – Adren Feb 7 '17 at 6:18
• I see, the RRT is a bit simpler than I thought. Thank you for the answer; I've upvoted it. It wasn't exactly what I was hoping for but it probably answers the question as it stands. I'll wait before accepting though in case anyone posts a proof more like the proof in the question that doesn't use these number theory results. – Anon Feb 7 '17 at 6:34
• @Santiago : Did you notice that $D$ was supposed not to be a perfect $m-$th power ? – Adren Feb 7 '17 at 22:31