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So I'm reading a book which requires no prior knowledge other than highschool algebra. I'm required to prove that the function from $\mathbb{R}$ to $\mathbb{R}$ $f(x) = x^3-x $ is a surjection, basically just using the definition of surjection and I'm guessing some clever manipulation that I haven't been able to come up with. So using calculus knowledge like in this question is not allowed. Any ideas on how to proceed?

Sine of the Time
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zlaaemi
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    What happens if $x$ is very big negative number? What happens if $x$ is a very big positive number? – lulu Dec 11 '22 at 00:35
  • @lulu The functions yields a really big positive or negative number respectively. Since x^3 sort of grows faster than x. Tho I don't know how to formalise this idea without calculus. – zlaaemi Dec 11 '22 at 00:39
  • Well, there is a formula for solving cubic equations. Not a very simple one though. – Mark Dec 11 '22 at 00:40
  • Even a limit? Pretty weak notion. try this: Show that $f(10^n)>10^n$. for large positive $n$. And similarly for $f(-10^n)$. – lulu Dec 11 '22 at 00:44
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    @Mark I was thinking using the cubic formula to somehow prove that the polynomial x^3-x-a, where a is any real number, always have at least one real solution. I think this should prove that the function is a surjection, would it not? – zlaaemi Dec 11 '22 at 00:45
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    @lulu That will not help if OP is not allowed to use the intermediate value theorem anyway. – Mark Dec 11 '22 at 00:45
  • @Mark But surely that's allowed. Invoking Cardano is so much deeper than the intermediate value theorem. – lulu Dec 11 '22 at 00:46
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    @zlaaemi Yes, if you are familiar with the cubic formula then it will work. – Mark Dec 11 '22 at 00:46
  • Just nitpicking: the fact that the cubic function is well defined as a function from $\mathbb R$ into $\mathbb R$ probably requires the completeness of $\mathbb R$, which is an advanced and subtle notion. – Taladris Dec 11 '22 at 00:50
  • @Taladris I don't understand how $\mathbb{R}$ being a field requires it to be complete? – donaastor Dec 11 '22 at 00:57
  • you are reading a book. can you tell more about that detail? does the book state your question as a theorem or an exercise? what high school is it? not all high schools are the same. what knowledge exactly are we allowed to use? does it have to be derived from even more elementary knowledge because it is also not always the case in high schools? – donaastor Dec 11 '22 at 00:59
  • @donaastor: replace $\mathbb R$ by $\mathbb Q$. Is $f(x)=x^3-x$ seen as a function from $\mathbb Q$ into $\mathbb Q$ surjective? – Taladris Dec 11 '22 at 01:12
  • @donaastor Here is what the full statement looks like https://imgur.com/a/Uqk963W. There is no specific high school, I'm in college but I grabbed the book and I'm trying to figure out the author's intended solution to the problem, since the book requires no knowlege of calculus or the fundamental theorem of algebra, I assume the intended solution shouldn't use any of that and should instead use some clever manipulation like in the other similar exercises. – zlaaemi Dec 11 '22 at 01:14
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    @donaastor Note that $f:\Bbb{Q}\to\Bbb{Q}$, $f(x)=x^3-x$ is not surjective because $x^3-x-1=0$ does not have rational roots (by the rational root theorem, the only candidates are $-1,1$, but neither is a root). So, proving surjectivity definitely requires the completeness of the reals. The most intuitive form of which is the intermediate value theorem. Otherwise, one would have to use the explicit Cardano formula (which of course uses square and cube roots...completeness again... but let's say we allow this. Even then, Cardano's formula is pretty complicated compared to quadratic formula). – peek-a-boo Dec 11 '22 at 01:15
  • Okay, so I think at least some form of intermediate value theorem should be allowed because I remember we were using it in my high school. Something like the fact that continuous functions map compacts to compacts. On the other hand, I think the high schoolers should be familiar with the Newton's method of approximating the solutions to non-linear equations, even though it's more conplex than intermediate value theorem. Maybe using that you could prove it "easily"? – donaastor Dec 11 '22 at 01:41
  • @Taladris I know that it's not surjective on Q but you were talking about well-defined-"ness", so that got me confused... – donaastor Dec 11 '22 at 01:43
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    Well the polynomial $x^3-x=a$ is actually solvable quite easily in $\cos$ or $\cosh$ (see this https://math.stackexchange.com/q/2934230/399263) but hyperbolic functions are out of scope. On the other hand continuity neither. So I don't see how you can possibly solve this in elementary manner. – zwim Dec 11 '22 at 01:53
  • Just came up with a nice pure high school algebra proof that this function is injective, but rechecked the question before writing it up. Not sure whether this would help for surjectivity though. – quarague Dec 11 '22 at 07:10

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By definition of a surjective function the question statement is equivalent to:

Let $P(x)=x^3-x+c$, where $c\in\mathbb R$, then prove that the polynomial $P(x)$ always has at least one real root.

Suppose that $x_1,x_2,x_3$ are roots of the polynomial $P(x)$, such that $x_1,x_2,x_3\in\mathbb C\setminus \mathbb R$.

Let $x_1=a+bi$, where $a\in\mathbb R, b\in \mathbb R\setminus \{0\}$.

We have,

$$P(a{\color{#c00}{+}}bi)={\color{#c00}{ib}}(3a^2-b^2-1)\\ +\left(a^3-a(3b^2+1)+c\right)=0$$

This implies that, $3a^2-b^2-1=0$ and $a^3-a(3b^2+1)+c=0$.

This leads to,

$$P(a{\color{#c00}{-}}bi)={\color{#c00}{-ib}}(3a^2-b^2-1)\\ +\left(a^3-a(3b^2+1)+c\right)=0.$$

Thus, we have shown that, if $x_1=a+bi$ is one of the roots of the polynomial $P(x)=x^3-x+c$, then $x_2=a-bi$ is also one of the roots of $P(x)$.

Finally, using Vieta's formulas we observe that:

$$ \begin{align}&x_1+x_2+x_3=0\\ \implies &x_1+x_2=-x_3\\ \implies &x_3=-2a\in\mathbb R.\end{align} $$

which contradicts the assumption $x_1,x_2,x_3\not\in\mathbb R$. Therefore, there exist at least one $x_i$, such that $x_i\in\mathbb R.$

This completes the proof.

lone student
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    @nonuser Thank you very much. The criticism might be: How do you know that the cubic equation has 3 roots? The Fundamental theorem of algebra is in all high school textbooks (without proof). – lone student Dec 11 '22 at 09:27
  • "The question is equivalent... " no justification given for the most important part of the answer... You are basically hand-waving. Add the relevant details, and in this case the details you omitted are very important. – Bertrand Wittgenstein's Ghost Dec 11 '22 at 09:27
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    @BertrandWittgenstein'sGhost I dont agree. OP already knows that, proving surjectivity $\mathbb R\to\mathbb R$ is equivalent to shows that the cubic with real coefficients has always one real root. See comments. See the link, OP provided. – lone student Dec 11 '22 at 09:31
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    What are you talking about? It doesn't matter if you agree or not. You deserve the downvote for 2 reasons. 1: you posted an incomplete answer (ommited the relevant details). 2: Even if you were to provide the additional details it would clearly require mathematics beyond "simple algebra," so it won't be an answer to the question anyways. For these two reasons I downvoted your answer, and I will keep the downvote unless you fix the answer. I don't even think you fully read the question... what does the portion after the link says? – Bertrand Wittgenstein's Ghost Dec 11 '22 at 09:37
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    Conditioned on the fundamental theorem of algebra being high-school level (which I'm not really sure about but willing to accept) I think your answer works. Imagine if it weren't allowed : then contradiction wouldn't even be an option. Indeed, the FTA tells us that all the "roots" are complex, but failing that we don't even know what "roots" mean. I'll give you +1, because I cannot, for the life of me, see how one would do it without even that. – Sarvesh Ravichandran Iyer Dec 11 '22 at 11:45
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    @BertrandWittgenstein'sGhost I will acept this answer because honestly I think it's as good as it's going to get. I would have prefer avoiding the fundamental theorem of algebra since its proof is outside of the scope of what you see in highschool, but I think it might be impossible without it. – zlaaemi Dec 11 '22 at 12:20
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    +1 There is no elementary proof, because it must rely on completeness; but this is a neat proof relying on the single big hammer of the FTA (really a theorem in analysis), and the good-to-know (and elementary) algebraic facts that for polynomials with real coefficients $z$ is a root if and only if $\overline{z}$ is, and also piecing it together with Vieta's formula is cute. Of course, this isn't my preferred approach, since I could convince most people of the IVT, but not the FTA (though I've "known" it since middle school), and also because (the common) proofs of FTA require complex analysis. – peek-a-boo Dec 11 '22 at 12:25