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Modify the Theorem that states There exists a positive real number x such that $x^2 = 2$.

Show that there exists a positive real number $u$ such that $u^3 = 3$.

So far, I have come up with the following but I am getting stuck:

Let $S = \{ x \geq 0 | x^3 \leq 3\}$. Note that $1 \geq 0$ and $1^3 = 1 \leq 3$. So $ 1 \in S$. So $S \neq \emptyset$. Therefore, $\forall x \in S, x < 3$. So $S$ is bounded above. So, by the completeness axiom, $S$ has a supremum. Say $u = \sup S$. Suppose by way of contradiction $u^3 \neq 3$. Then either $u^3 < 3$ or $u^3 > 3$.

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  • $\begingroup$ You want to prove that there exists $u\in\mathbb{R}$ such that $u^3=2$ or $u^3=3$? $\endgroup$ – Larara Oct 24 '14 at 4:02
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Now if $u^3 \lt 3$, find a real $v \gt u$ such that $v^3 \lt 3$ Then $u$ is not the sup, contradiction. The other way is similar.

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