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 $u^3 > 3$.
Best Answer
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.