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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?

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"An element of the set of real numbers minus the rational numbers" is just one element. You have to actually count (or prove uncountable) the set as a whole. –  abiessu Feb 24 at 15:09
    
am I using the symbol wrongly? I have to create a set $H=\lbrace r_1(r_2r_3+r_2r_4): r_2,r_3\in \mathbb{R},r_1\in\mathbb{Q}\rbrace$ is it? –  nanopotato Feb 24 at 15:14
1  
You are using the set element $\in$ correctly, but drawing a conclusion from it wrongly. For example, the set of algebraic numbers includes all the rationals and the set of irrationals that are not transcendental, and this set is countable. –  abiessu Feb 24 at 15:23
    
You might consider reviewing HenningMakholm's answer prior to considering this question closed. –  abiessu Feb 24 at 16:07

3 Answers 3

up vote 2 down vote accepted

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.

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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.


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

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I'm not certain that I'm understanding this problem correctly. In particular, you state "since there is an injection from $F$ to $\Bbb Q,F$ is (at most) countable." I'm having trouble getting this to match up with the answer I posted. Can you help me sort out what I'm missing? –  abiessu Feb 24 at 15:49
    
I think I need more help with "one easily sees that no nonzero $x$ can be a root for more than one $a$..." –  abiessu Feb 24 at 15:52
    
@abiessu: For each (nonzero) choice of $x$, consider $a$ to be the unknown in the equation $x^4+ax-5=0$. Then the equation is of first degree and therefore can have only one solution for $a$. –  Henning Makholm Feb 24 at 15:54
    
Thank you, that's what I was missing. I let the degree of the function in $x$ distract me... –  abiessu Feb 24 at 16:05

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.

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so my second expression is right? And the first expression is only consider the rational part so it's wrong? –  nanopotato Feb 24 at 15:16
    
The second is confuzing, but I think it's OK. –  5xum Feb 24 at 15:25

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