I wish to make a proof that $sqrt[n]{m} quad forall quad m,n in Bbb N$ is irrational if $m$ can’t be expressed as an ideal $n^textual content{th}$ energy of another quantity $z in Bbb N$.

Start by saying it is unconditionally rational and therefore:

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

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

# $mq^n=p^n$

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

On this case $b$ and $d$ might be $0$

What you get now could be:

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

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

Trying on the exponents of $m$, one is all the time divisible by $n$ and one other isn’t. Therefore it’s a contradiction to say that in a common case they’d be equal.

In an try to resolve this, say that certainly $m=z^n$.

Then:

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

And it’s now potential, for the reason that exponents at the moment are each divisible by $n$.

To finish, say that $q=cm^d=1$. Then it could suffice to point out 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 one manner this equation might be happy is that if $n=0$ or $a=z, b=0$. For the primary it implies that $sqrt[n]{m}$ has a couple of technical issues are aren’t all that attention-grabbing. For the second, it implies:

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

Q.E.D

Is that this a reputable argument or is there some missing within the generality I used to be aiming for? More information and recommendation could be appreciated.