The question is as follows:

Prove the existence and uniqueness of $\sqrt[3]{x}$. More formally, prove that for every $y \in \mathbb{R}$, there exists a unique number $x$ $\in \mathbb{R}$ such that $y = x^3.$ Assume $x,y \geq 0.$

Let's prove the uniqueness first.


Suppose the solution is not unique. If $x,y \in \mathbb{R}$ and $x < y$, and $x$ and $y$ are both cube roots of some number $c \in \mathbb{R}$ where $c >0$, then we know that $$x^3 < xy < y^3.$$ But $x^3=c=y^3$, which is a contradiction $\blacksquare$.

Now, let's prove that it exists, which is a little longer.


For $x=0$ let $y=0$, so then $y^3=0$. Assume $x>0.$

Let $S$ be the set defined by $S=\left \{y \in \mathbb{R} : y \geq 0, y^3 <x \right\}.$ We will use the completeness axiom to prove this. Let's check two important conditions. Firstly, $S$ is nonempty because $0 \in S.$ Secondly, if $y=x+1$, then $$y^3=x^3+3x^2+3x+1>x.$$ Therefore, $S$ is bounded above. Hence $ \beta =\sup S$ exists.

My claim is that $\beta^3 = x.$ Let's prove this by contradiction.

First suppose that $\beta^3 < x.$ Then $\exists z$ s.t. $z>\beta$ and $z^3 <x.$ But then $z \in S$ and $z>\beta.$ This is a contradiction.

Now let's suppose that $\beta^3 > x.$ Then $\exists z$ s.t. $0 \leq z < \beta$ and $x <z^3.$ So, $z$ is not an upper bound of $S$.

Thus we have proven that $\beta^3 = x$, which means that $\sup S = \sqrt[3]{x}$, and finally, that $y^3 = x$ $\blacksquare$.

Thank you so much for taking the time to read this. I would like to know, would you give full credit for this proof? Am I missing important information? Does the proof make sense and is it coherent?

Once again, thanks in advance!

  • $\begingroup$ The $xy$ in the middle of the chain of inequalities needs one of the variables squared. $\endgroup$ Commented Sep 5, 2018 at 1:50
  • $\begingroup$ How do you know the existence of $z$? Maybe you need to write such an $z$ explicitly. $\endgroup$ Commented Sep 5, 2018 at 1:50
  • $\begingroup$ @RossMillikan Can you explain further? $\endgroup$
    – user545426
    Commented Sep 5, 2018 at 2:08
  • $\begingroup$ @EclipseSun I tried to do that, but the algebra got a bit messy and nothing canceled out to give me a nice solution, so I assumed I did it wrong. $\endgroup$
    – user545426
    Commented Sep 5, 2018 at 2:09
  • 1
    $\begingroup$ You want the three quantities to all be third powers. If $x=4,y=5$ you are claiming $64 \lt 20 \lt 125$. If you put $x^2y$ in the middle it is $80$ and the inequalities are true. $\endgroup$ Commented Sep 5, 2018 at 3:18

1 Answer 1


Your proof is nearly perfect, just notation errors.
There are similar ideas as your previous question: Proving the supremum of a set in the general case

Existence: After fixing some $y>0$, your set $S$ should be $S=\{x\in\mathbb{R}:x^3<y\}$, you do not need $x\ge0$. But your ideas are the same. Using your set $S$:
For $\beta^3<x$, its more clear to write 'there exists $z\in\mathbb{R}$ s.t. $\beta^3<z^3<x$' instead of $\beta<z$ and $z^3<x$, and likewise for the other case.
For $\beta^3>x$, $z$ is an upper bound of $S$! The contradiction should be 'So $\sup S\neq\beta$.'
Using my set $S$, you should conclude $\beta^3=y$.
Uniqueness: See Ross Millikan's comment.


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