# Show that there exists a positive real number $x$ such that $x^3 = 5$.

Here is what I've done so far:

[First, want to show $b = 5$ is an upper-bound of $S$.]

So, let: $$S = \{x \in \Bbb R : x \gt 0, x^3 \le 5\}, S \neq \emptyset$$ Assume that $b = 5$ is not an upper-bound of $S$. Then, $\exists x \in S$ s.t. $x^3 \gt 5$. But this contradicts the definition of $S$. Therefore, $b = 5$ is an upper-bound of $S$.

[Next, want to show if $c \lt b$, then $c$ is not an upper-bound.]

This is where I'm stuck... Are you supposed to pick an epsilon here? I tried this also:

Let $\epsilon \gt 0$, $\epsilon = {b - c\over 2}$. I'm not sure how to proceed from here on.

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Of course, there are numbers $c<b$ that are upper bounds, for example $c=2$. –  Hagen von Eitzen Feb 1 '13 at 17:34

What you need to do next is to establish the existence of a least upper bound. Because of the least-upper-bound property of the real numbers, this number must exist because you've shown that $5$ is an upper bound. Let this LUB of $S$ be $L$, which must be a positive real number by definition.

If $L^3 = 5$, we are done. Now, suppose $L^3 < 5$. Try and arrive at a contradiction.

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The set is non-empty because $1\in S$, so we have a non-empty subset of $\mathbb R$ that has an upper bound then $S$ has a supremum, say $\alpha$, Claim: $\alpha^{3}=5$, try to prove the claim.

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