Proof of uniqueness of positive cubes

Question

Prove that there exists exactly one positive $t\in \mathbb{R}$ with $t^3=c$ for $c>0$.

I assume that most of my derivation needs to be algebraic and using some simple axioms. So for example, if I define the set:

$$T:=\lbrace t\in \mathbb{R}: t> 0, t^3 \leq c \rbrace$$

I must show that the cubed root of $c$ is the supremum of $T$ and then prove its uniqueness. But I'm not sure how to go about this. How can I concisely prove this?

Answer 1

"At most one": if $t^3 = c = s^3$, then $t^3-s^3 = 0$ so $(t-s)(t^2+ts+s^2)=0$. Hence either $t=s$ or $t^2+ts+s^2 = 0$. But the latter is to say that $(t+s)^2 = ts$, or equivalently $t+s = \sqrt{ts}$ (this is valid since we are assuming $t, s$ are both positive), and this violates AM/GM because $t, s > 0$.

"At least one": this is just from the intermediate value theorem, taking $x \mapsto x^3$ for $x$ between $0$ and $c$ (if $c \geq 1$), or taking $x \mapsto x^3$ for $x$ between $0$ and $1$ (if $c < 1$).

Answer 2

If $f(t) = t^3$ and $f(t_1) = f(t_2) = c$ with $c > 0$, then, by the mean value theorem, there is a $t_0$ with $0 < t_1 \le t_0 \le t_2$ with $f'(t_0) = 0$.

But $f'(t) = 3t^2$ so that $f'(t_0) = 0$ implies that $t_0 = 0$ which contradicts $0 < t_1 \le t_0 \le t_2$.

Answer 3

Uniqueness is pretty easy: if $0 < t < s$, then $t^3 < s^3$, so you cannot have two different $t, s > 0$ such that $t^3 = s^3 = c$.

For existence, let $s = \sup T$.

Assume by contradiction that $s^3 < c$. We will show that there is $\epsilon > 0$ such that $(s+\epsilon)^3 < c$, which contradicts the definition of $\sup T$. Let $\Delta = c - s^3$. We have $\Delta > 0$ by assumption. We have

$$ (s+\epsilon)^3 = s^3 + 3 \epsilon s^2 + 3 \epsilon^2 s + \epsilon^3 = s^3 + \epsilon(3 s^2 + 3 \epsilon s + \epsilon^2) $$

For $\epsilon < s$, $3s^2 + 3 \epsilon s + \epsilon^2 < 7 s^2$, so $(s+\epsilon)^3 = s^3 + \epsilon(3 s^2 + 3 \epsilon s + \epsilon^2) < s^3 + \epsilon \cdot 7 s^2$, so it's enough to pick $\epsilon$ such that $\epsilon < s$ and $\epsilon \cdot 7s^2 < \Delta$, that is, $\epsilon < \frac{\Delta}{7 s^2}$.

Thus, we obtained a contradiction, and therefore $s^3 \geq c$. We now need to prove that $s^3 \leq c$ to obtain $s^3 = c$. For this, we can use the same method as above: assume by contradiction that $s^3 > c$, and find $\epsilon > 0$ such that $(s-\epsilon)^3 > c$. This will show that $s - \epsilon$ is an upper bound of elements of $T$, and so $s - \epsilon \geq \sup T = s$, a contradiction. The argument is the same as above, with only minor changes.

Note that the argument above is basically about proving that $x \mapsto x^3$ is a continuous function.