mathbalarka said:
Equivalent to using integer root theorem. I avoid it as much as possible because it usually reminds me of Gauss' lemma on quadratic reciprocity.
...which is why i said: ..."in the same vein"...
But my point was a bit subtler:
When asked to prove something, it is natural to ask: which tools can I use...or, as on these forums, which things has the poster already learned?
An argument can be made to appealing to facts known about polynomials as continuous functions in number theory on the grounds that someone taking a number theory course has likely already taken calculus.
Of course, invoking Euclid's proof begs the question: has the fact that the integers are a UFD been established (being "told" this is true isn't quite the same thing...even if you BELIEVE it...)?
It *is* true, however, that if gcd(a,b) = 1, and:
$\left(\dfrac{a}{b}\right)^n = m$ where $a,b,m \in \Bbb Z^+$
That $b = 1$.
To see this, suppose $p$ is a prime such that $p^k|a$ but $p^{k+1} \not\mid a$.
Since b is co-prime to a, we have that p necessarily divides m. Repeating this process with:
$\dfrac{a^n}{p} = \left(\dfrac{m}{p}\right)b^n$
and so on, for powers of $p$ up to $p^k$, we obtain that $p^k|m$, so that:
$\dfrac{a^n}{p^k} = \left(\dfrac{m}{p^k}\right)b^n$
where this is a relation among 3 integers: $\dfrac{a^n}{p^k},\dfrac{m}{p^k},b^n$.
Now $p$ is not a factor on either side.
Repeating this for every prime factor of a, eventually leads to:
$1 = \left(\dfrac{m}{a^n}\right)b^n$ so that $b^n|1 \implies b^n = 1 \implies b = 1$.
This (perhaps) is unexpected:
every rational $n$-th root of an integer is also an integer.
We can thus conclude:
If $m$ is an $n$-th power-free integer, $\sqrt[n]{m} \not \in \Bbb Q$.
So an even SHORTER proof of the irrationality of $\sqrt[5]{672}$ is:
21 is 5th-power free.
This reduces checking the irrationality of many $n$-th roots of integers to a purely mechanical process of looking at the prime factorization.
We can, without any additional effort, actually go a bit further:
There is a natural isomorphism between $(\Bbb Z^+,\ast)$ and the abelian monoid:
$\displaystyle \bigoplus_{i = 1}^{\infty} \Bbb N$ under addition (this is actually the free abelian monoid on infinitely many generators),
where we can represent any positive integer as a finite $k$-tuple, for example:
$24 \mapsto (3,1)$
So writing $m \mapsto (k_1,\dots,k_r)$ we have:
$\sqrt[n]{m}$ is rational if and only if $n|k_j$ for all $j = 1,\dots,r$.
In other words the "real" reason $\sqrt[5]{672}$ is irrational is because 5 does not divide 1 and:
$672 \mapsto (5,1,0,1)$.
(I'm about 95% sure this is correct, but I never really thought about this problem in this way, so I would appreciate it if someone could confirm I haven't made some egregious error).
(Additonal musing: is the Grothendieck groupification of this monoid just $\Bbb Q^{\ast}$? If so, can we devise a similar mechanical check on the irrationality of $\sqrt[n]{q}$ for any $q \in \Bbb Q$?).
