# Algebraic numbers proof

I was just wondering if anyone knew how one would prove that every $k$-th root of a rational number $r$ is algebraic? My thought would be that some $r$ is equal to $\frac{p}{q}$ but since you are taking the $n$-th root $p$ and $q$ aren't necessarily integers anymore which is a requirement for algebraic- integer coefficients.

• It is an algebraic number for sure but not necessarily an algebraic integer Oct 23, 2016 at 5:00
• Uh, how would you prove ever even number is an integer? K-th root of a rational number solves the polynomial $x^k=r$ so it is algebraic by definition. Oct 23, 2016 at 5:37
• Okay, if you have the definition of integer coefficient (which is NOT nescessary ) then you have $nx^k=m$ where $r=m/n$. Integer coefficients are not nescessary as if you have rational coefficients you just multiply both sides by a common denominator. Oct 23, 2016 at 5:41

If $r$ is a rational number, then $r=\frac{p}{q}$ for integers $p,q$. Then a $k$th root of $r$ is a root of the polynomial $f(X)=qX^k-p$, which has integer coefficients.