If $a$ and $b$ are distinct positive integers, is $\sqrt[a+b]{\dfrac ab}$ irrational?

Question

If $a,b \in \mathbb Z^+ , a \neq b$ then is it true that $\sqrt[a+b]{\dfrac ab}$ is irrational?

This question actually popped up from seeing whether there exists a non-trivial homomorphism from $(\mathbb Q,+)$ to $(\mathbb Q^+,.)$ for suppose a nontrivial homomorphism exists , then $\exists r \in \mathbb Q , 1 \ne \dfrac ab \in \mathbb Q^+$ such that $\bigg(f\Big(\dfrac rn\Big)\bigg)^n=f(r)=\dfrac ab , \forall n \in \mathbb Z$ , so $\sqrt[n]{\dfrac ab} \in \mathbb Q^+ , \forall n \in \mathbb Z$ , I am hoping to get a contradiction from here ...

Answer 1

Let $f$ be that homomorphism.

$$f(1/n)=\sqrt[n]{f(1)}$$

So the $n$th root of $f(1)$ is rational for all $n$. Therefore, $f(1)=1$.

EDIT: Since $f(1)=1$, then $f(n)=f(1)^n=1$ for $b\in \Bbb N$. Also, $1=f(0)=f(-1+1)=f(-1)f(1)=f(-1)$, so $f(-n)=1$ for all $n\in\Bbb N$. We already know that $f(1/n)=f(1)=1$ and $f(m/n)=f(1/n)^m=1$

That is, the only homomorphism is the trivial one.

Answer 2

$(a/b)^{1/(a+b)}$ cannot be rational, except for $a=b$.

Suppose that $a\ne b$ and $r = (\frac{a}{b})^{1/(a+b)}$ is rational. Then there is some prime $p$ that appears in the prime factorization of $r$. Obviously, $$ v_p(r) = \frac{v_p(a)-v_p(b)}{a+b}. $$

From $2^{v_p(a)}\le p^{v_p(a)}\le a<2^a$ we can see that $0\le v_p(a)<a$ and similarly $0\le v_p(b)<b$. But then $$ 0 < |v_p(r)| = \frac{|v_p(a)-v_p(b)|}{a+b} < \frac{a+b}{a+b} = 1. $$ This is a contradiction because $v_p(r)$ should be an integer.

Answer 3

This is just a long version of Hurkyl's suggestion. I'll try to be as informative, and as clear as I can.

I'll use the fact that for all natural number $x > 1$ (clearly 0, and 1 cannot be expressed as a product of primes), there can only be 1 way to express $x$ as the product of primes, i.e if $x = \frak{p}_1^{\alpha_1}.\frak{p}_2^{\alpha_2}\dots \frak{p}_n^{\alpha_n} = \frak{q}_1^{\beta_1}.\frak{q}_2^{\beta_2}\dots \frak{q}_m^{\beta_m}$ ($\frak{p}_i$, and $\frak{q}_i$ are all primes), then $n = m$, and for each index $i$, there's one and only one index $j$ such that $\frak{p}_i = \frak{q}_j$ and $\alpha_i = \beta_j$.

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Claim 1

Let $\mathbb{Q}^{+} \ni \alpha = \frac{{\frak{p}}^\beta.A}{B}$, where $\frak{p}$ is prime, $\beta > 0$, and $A$, $B$ contains no factor of $\frak{p}$ (i.e, I pull all factor of $\frak{p}$ out from the numerator of $x$). Then $\sqrt[n]{\alpha}$ is irrational if $\beta \ \not{\vdots} \ \ n$

Proof

Assume that $\sqrt[n]\alpha$ is rational, so $\sqrt[n]{\alpha} = \frac{a}{b}$ for some natural number $a, b$, then $\frac{{\frak{p}}^\beta.A}{B} = \frac{a^n}{b^n} \Leftrightarrow {\frak{p}}^\beta.A b^n = a^n B$. Notice on the LHS, some more factors of $\frak{p}$ may shown up in $b$, hence on the LHS there's $\beta$ plus a multiple of $n$ factors of $\frak{p}$, while on the RHS, there are just some multiple of $n$ factors of $\frak{p}$ (the only place where some factors of $\frak{p}$ may shown up is in $a$). The number of factor of $\frak{p}$ on both sides cannot be equal, since $b$ is not a multiple of $n$, so it violates the fact that each integer can only be expressed in 1 and only 1 way through the product of primes.

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Claim 2

Let $\mathbb{Q}^{+} \ni \alpha = \frac{A}{{\frak{p}}^\beta.B}$, where $\frak{p}$ is prime, $\beta > 0$, and $A$, $B$ contains no factor of $\frak{p}$ (i.e, I pull all factor of $\frak{p}$ out from the denominator of $x$). Then $\sqrt[n]{\alpha}$ is irrational if $\beta \ \not{\vdots} \ \ n$

Proof

The same as above.

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Claim 3

For every positive rational $x \neq 1$, there exists at least one $n$ such that $\sqrt[n]{x}$ is irrational.

Proof

Say we have $x = \frac{\frak{p}_1^{\alpha_1}\frak{p}_2^{\alpha_2} \dots \frak{p}_n^{\alpha_n}}{\frak{q}_i^{\beta_1}\frak{q}_2^{\beta_2} \dots \frak{q}_n^{\beta_n}}$, where all $\frak{p}_i$, and $\frak{q}_i$ are distinct primes. If the numerator of $x$ is not 1, choosing $n = \alpha_1 + 1$ should work (Claim 1); if the denominator of $x$ is not 1, then choosing $n = \beta_1 + 1$ will do (Claim 2). Of course both numerator and denominator of $x$ cannot be 1 at the same time, since it would force $x = 1$.