Proving that $\sqrt[n]{m}$ is irrational

Question

I want to make a proof that $\sqrt[n]{m} \quad \forall \quad m,n \in \Bbb N$ is irrational if $m$ cannot be expressed as a perfect $n^\text{th}$ power of some other number $z \in \Bbb N$.

Begin by saying it's unconditionally rational and hence:

$\sqrt[n]{m}=\dfrac{p}{q} \iff p, q \in \Bbb N$

$m=\dfrac{p^n}{q^n} \tag 1$

$mq^n=p^n$

Now say that $p=am^b , q=cm^d \iff a, b, c, d \in \Bbb N$

In this case $b$ and $d$ can be $0$

What you get now is:

$m(cm^d)^n=(am^b)^n$

$c^n \cdot m^{nd+1}=a^n \cdot m^{nb} \tag 2$

Looking at the exponents of $m$, one is always divisible by $n$ and another is not. Hence it is a contradiction to say that in a general case they'd be equal.

In an attempt to resolve this, say that indeed $m=z^n$.

Then:

$c^n \cdot z^{n(nd+1)}=a^n \cdot z^{n^2b} \tag 3$

And it is now possible, since the exponents are now both divisible by $n$.

To end, say that $q=cm^d=1$. Then it would suffice to show that $\sqrt[n]{m}=z=p$

If $q=1$ then $c=1, d=0$

And:

$1^n \cdot z^n=a^n \cdot z^{n^2b}$

$z^n=a^n \cdot z^{n^2b}$

The only way this equation can be satisfied is if $n=0$ or $a=z, b=0$. For the first it means that $\sqrt[n]{m}$ has a few technical problems are aren't all that interesting. For the second, it implies:

$p=am^b=zm^0=z$

Q.E.D

Is this a legitimate argument or is there some lacking in the generality I was aiming for? Additional info and advice would be appreciated.

score 0 · Accepted Answer · answered Apr 23 '20 at 18:21

You overlook simplifying assumptions. First if $r=(k^n\cdot m)$, then $\sqrt[n]{r}=k\sqrt[n]{m}$. Since an integer ($k$) times an irrational ($\sqrt[n]{m}$) is still irrational, you can assume that if you solve the question for all $m$ that have no factors which are $n$th powers, you have solved the entire problem. So you can assume $m$ has no $n$th power factors. Second, you can assume that the fraction $\frac{p}{q}$ is in its lowest terms, i.e. $\gcd (p,q)=1$. If it were not to begin with, there is no reason not to extract common factors to reduce the fraction to lowest terms.

With that, you are essentially done. You give the rearranged equation (1), $mq^n=p^n$. By FTA, any factor of $p$ must be a factor of $mq^n$. By the second assumption, that factor cannot be a factor of $q$, so it must be a factor of $m$. But any factor of $p$ must be present in ($p^n$) $n$ times, so it must be present in ($m$) $n$ times. But by the first assumption, $m$ has no $n$th power factors, so that can't be the case either.

score 0 · Answer 2 · answered Apr 23 '20 at 18:38

It seems overcomplicated.

Start with this equivalent statement, which is easier to read and understand (less negations):

If $\sqrt[n]{m}$ is rational, then $m$ is a perfect $n^\text{th}$ power of some natural number.

We have $mq^n=p^n$. Let $r$ be any prime. Then, by looking at the exponent of $r$ in the factorization of both sides, we get $$ v_r(m) + n v_r(q) = n v_r(p) $$ This implies that $v_r(m)$ is a multiple of $n$. And this is equivalent to $m$ being a perfect $n^\text{th}$ power.