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.
$\textbf{Uniqueness:}$
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.
$\textbf{Existence:}$
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!
Best Answer
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.