Is $g(x)=x^3$ surjective on $\Bbb R$?

Question

Definition of surjective:

Let $X$ and $Y$ be sets and let $f:X\to Y$ be a function. $f$ is surjective if for each $y\in Y$ there is some $x\in X$ such that $f(x)=y$

Solution attempt:

$$g(x) = x^3$$ $$f: \Bbb R \to \Bbb R$$

Take any $y \in \Bbb R$. Then there exists $x\in X$ s.t x is $\sqrt[3]{y}$ (cube root of $y$). Since cube root is a function on $\Bbb R \to \Bbb R$, $g$ is surjective.

Is this proof correct? Otherwise can you please comment on what's wrong or any improvements for making it more rigurous?

Answer 1

Let me summarize some of the comments as an answer.

The logic in your proof is correct. As an answer to an exercise in an elementary set theory class I would accept it.

I would change some wording: what makes this work is not that "cube root is a function" but that every real number has a real cube root. In a calculus class or a class on real analysis you would have to explain why that is true.

Answer 2

Simply note that $f(x)=x^3$ is a countinuos function and

• $\lim_{x\to +\infty} x^3 = +\infty$
• $\lim_{x\to -\infty} x^3 = -\infty$

thus by IVT $f(x)=x^3$ assume all values $y\in \mathbb{R}$.

Answer 3

Take any y∈R. Then there exists x∈X s.t x is y√3 (cube root of y).

Is this proof correct?

That depends.

It is true that for any $y$ if $\sqrt[3]y$ exists then $f(\sqrt[3]y) = y$. And therefore if for all $y$ and real $\sqrt[3]y$ does exist, this does indeed prove $f$ is surjective.

But the question is, do we know that for any $y$ there exists and value $x = \sqrt[3] y$ so that $x^3 = y$? If so, how do we know it? And is it acceptable for us to state that and with what degree of reference? And finally, have we ever actually seen a proof of that?

And all of those depend on what level of class we are in and how much detail is expected.

tl;dr; This proof is valid but somewhere along the line it needs to be proven that: Proposition: For all $y\in \mathbb R$ the exists an $x \in \mathbb R$ so that $x^3 = y$.

I suppose what makes me uneasy is that the question of "Prove $f(x) = x^3$ is surjective" and "Prove for any $y \in \mathbb R$ then there is an $x$ so that $x^3 = y$" are absolutely identical and equivalent! So this proof is nothing other than restating the statement in other words. The proof is a complete failure!

However that's too harsh on a student, I think. It's quite likely the purpose of the exercise was to make sure the student understand definitions in meaning. In which case, understanding that the two statements are equivalent IS the proof! In which case: The proof is complete and perfect and excellently done!

So ... which is it? I don't know. I'm not in your class. Can you prove Proposition: For all $y\in \mathbb R$ the exists an $x \in \mathbb R$ so that $x^3 = y$? Or can you cite it as axiom?

(It is true, of course, and can be proven either by concepts of continuity, the least upper bound principal, and/or the archimedean principal/completeness of the reals. But then again it is equally likely that for your class this can be a given axiom.)