Is $ f : \mathbb{Q} \to \mathbb{Q}$ via $ \ f(x) = x^{5} -3$ surjective? Question:
Is $ f : \mathbb{Q} \to \mathbb{Q}$ via  $ \ f(x) = x^{5} -3$ surjective?
My attempt:
No. Consider $ y = 2$. There doesn't exist $ x \in \mathbb{Q} \ such \ that \ y = f(x)$.
Am I correct? Is this the right approach to prove a function is not surjective?
 A: There exists no $\theta \in \Bbb Q$ such that
$f(\theta) = 0; \tag{1}$
if there were, then
$\theta^5 = 3; \tag{2}$
but
$\sqrt[5] \theta \notin \Bbb Q. \tag{3}$
To really finish things off, (3) should be proved:
If there were integers $r$, $s$ with $\gcd(r, s) = 1$ and
$\left (\dfrac{r}{s} \right )^5 = 3, \tag{4}$
then
$r^5 = 3s^5, \tag{5}$
so $3\mid r^5$, whence $3 \mid r$, whence
$r = 3t, \tag{6}$
so
$r^5 = 3^5 t^5,\tag{7}$
yielding
$3^5t^5 = 3s^5, \tag{8}$
whence
$3^4 t^5 = s^5, \tag{9}$
so $3 \mid s^5$, hence $3 \mid s$, contradicting $\gcd(r, s) = 1$.
This demonstration obviously parallels the classic proof that $\sqrt 2 \notin \Bbb Q$.
A: As an alternative to the "reduced form of a fraction" argument, you could apply eisenstein's criteria, where $p=5$, divides $5$, while $p^2$ does not divide $5$.
From this, we see that $x^5-5$ is irreducible over $\mathbb Q$, so none of the solutions $x^5=5$ are rational.
A: For yet another take on the non-surjectivity, consider the value $y = -1 \iff x^5-2=0\,$. The latter is a polynomial equation with integer coefficients, so by the rational root theorem the only potential rational roots could be $\pm1, \pm2\,$. It can be easily verified by inspection that none of those is actually a root, so $x^5-2=0$ has no rational roots, and therefore $-1 \not \in f(\mathbb{Q})\,$.
A: Suppose $\big(\frac ab\big)^5=5$, $a,b$ without a common divisor. Then $a^5=5b^5$.  Then $5$ divides the RHS so it divides the LHS so $5$ divides $a$.  So $5^5$ divides the LHS.  So $5^5$ divides the RHS.  So $5$ divides $b$.  But then $5$ divides both $a$ and $b$ a contradiction.
