A rational root means?

Question

the set $F = \lbrace a:x^4 + ax -5 = 0\text{ has a rational root}\rbrace$, is $F$ countable or uncountable?

I have two expression with very different meanings,

First one, $a=\frac{5}{x}-x^3,x \in \mathbb{Q}\setminus \lbrace 0\rbrace \Rightarrow F\subseteq \mathbb{Q}$, $\therefore$ it is countable.

second one,

suppose $r_1,r_2,r_3,r_4$ are the roots of the equation, at least one of it is rational so call $r_1$ is rational. \begin{align*} x^4+ax-5=0 \Leftrightarrow (x-r_1)(x-r_2)(x-r_3)(x-r_4) &= 0\\x^4-(r_1+r_2+r_3+r_4)x^3 +(r_1r_2+r_1r_3+...)x^2\\-(r_1r_2r_3+r_1r_2r_4+r_2r_3r_4)x+r_1r_2r_3r_4&=0\\\therefore a&=-(r_1r_2r_3+r_1r_2r_4+r_2r_3r_4)\\ ,-5&=r_1r_2r_3r_4\\ r_2r_3r_4=\frac{-5}{r_1}&\Rightarrow r_2r_3r_4\quad rational\\ -(r_1r_2r_3+r_1r_2r_4+r_2r_3r_4)&=-[r_1(r_2r_3+r_2r_4)+r_2r_3r_4]\\(r_2r_3+r_2r_4)\in \mathbb{R}\setminus\mathbb{Q}&\Rightarrow r_1(r_2r_3+r_2r_4)\in \mathbb{R}\setminus\mathbb{Q}\\&\Rightarrow r_1(r_2r_3+r_2r_4)+r_2r_3r_4\in \mathbb{R}\setminus\mathbb{Q} \end{align*} $\therefore$it is a set collecting irrational number so it is uncountable. Which one is correct? Thank you.

So otherwise, I can get a subset $S_{r_1}$ to collect rational a and $S_{r_2}$ to say it is uncountable then the Union is also uncountable?

Answer 1

Your first reasoning is correct but you have wrong symbol in place. It should be $F \subseteq \Bbb{Q}$ and not $F \in \Bbb{Q}$. If the polynomial equation $x^4 + ax - 5$ has to have a rational root, then as you deduced correctly, the number $a$ SHOULD be of the form $\frac{5}{x} - x^3$ for some rational number $x$ without worrying about the other roots of the equation (Also, $x = 0$ can never be a solution of the equation for any value of $a$). We know that the total number of rational numbers is countable. Hence, $a$ can take only countably many values.

Answer 2

Consider the function $f:F\to \mathbb Q$ defined by $$ f(a) = \min\{x\in \mathbb Q \mid x^4+ax-5=0\} $$ (where the minimum always exists because the polynomial always has at most 4 roots).

Then $f$ is injective -- one easily sees that no nonzero $x$ can be a root for more than one $a$, and $x=0$ is never a root.

Since there is an injection from $F$ to $\mathbb Q$, $F$ is (at most) countable.

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It is not true that just because a set contains some irrational numbers, it is uncountable. For example $\{\sqrt x\mid x\in\mathbb N\}$ is countable

Answer 3

In the first one, you just proved that there exists a countable subset of $F$. You haven't proven that all of the elements of $F$ are covered by your mapping.