How can I prove: if $p $ is prime and $n>1$, then $ p^{\frac1n} $ is irrational?

Question

Please see this question's title.

Answer 1

Suppose $p^{1/n}=\frac{a}{b}$, a fraction in its lowest terms. Then $pb^n=a^n$, so $p$ divides $a^n$, so $p$ divides $a$, so $p^n$ divides $a^n$. This means $p$ also divides $b$, a contradiction.

Answer 2

Hint $\ $ By the rational root test, any rational root of $\rm\ x^n - a\ $ is integral, so every prime in the unique prime factorization of $\rm\ a\ $ occurs to a power divisible by $\rm\:n.$

Answer 3

You've added the number theory tag but I'm trying to solve it using ring theory.

We know if $p$ is a prime then $n>1$ implies $x^n-p \in \mathbb{Z} [x]$ is irreducible. Hence it is irreducible in $\mathbb{Q} [x]$. In particular, it has no rational roots.

Answer 4

Even simpler: By the rational roots test, the polynomial $x^n - p$ has rational roots only if they are integers dividing $p$; and $p^{1/n}$ can't be an integer.

Answer 5

As shown in this answer, any algebraic integer (a number satisfying $(2)$) which is rational is an integer. In this case, the monic polynomial is $x^n-p$. Since $x$ is an algebraic integer, but cannot be an integer, $x$ must be irrational.

Answer 6

Their is a more general theorem that may be applied. Let $r=\frac{a}{b}$, written in lowest terms. Then $r$ has a rational $n$-th root if and only if $a$ and $b$ are the $n$-th power of integers. Indeed suppose that their is a rational number $\frac{a'}{b'}$, also in lowest terms such that $(\frac{a'}{b'})^n=\frac{a}{b}$. This means that $\frac{a'^n}{b'^n}$ is also in lowest terms Note that both fractions are in lowest terms. At this point (to avoid dealing with signs, we will assume everything to be positive). If we have two fractions written in lowest terms and the two are equal, then they are written in the same way. More explicitly, Suppose that we have $\frac{x}{y}=\frac{x'}{y'}$ are both written in lowest terms, and all is assumed positive (so we do not have to say things like up to an even number of sign changes). This will follow from prime factorization and some fiddling with the fractions. Thus we may conclude that $a'^n=a$, and $b'^n=b$, which shows the proposition.

Edit 1 Note that the main thing in this proof that makes it work is prime factorization. I would be interested to see how far one can take this line of reasoning if we move to more general number fields and rings of integers.

Edit 2 Their is a nice group theoretic interpretation of the above proof. It is that the positive rationals, $(\mathbb{Q}_+, *)$ is isomorphic to a countable sum of copies of $\mathbb{Z}$. With the formal symbols $<p>$, one for each prime generating the group. Explicitly, their is an isomorphism which we will call $$div:\mathbb{Q}_+\to\mathbb{Z}<2>\oplus\mathbb{Z}<3>\oplus\mathbb{Z}<5>\oplus\cdots$$. We will notationally conflate the symbol $<p>$ with the element that is zero in every factor except the the $<p>$ factor, which will one in that factor. We defined the map $div$ to be the following. $$div(p^k)=k<p>$$. We use unique factorization to show that this induces an isomorphism of groups. Now note that a rational number $r$ has an $n$-root if and only if $div(r)$ has the property that each factor is divisible by $n$.