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Prove that there is a positive number $x$ such that $ x^3= 5$


Define $S=\{x\in\mathbb{R}\mid x>0,x^3<5\}$.

Set $S$ is not empty because $1\in S$ and is bounded above because $2^3=8>5>x^3,\forall x\in S$, by the completeness axiom, $S$ has a least upper bound, $b$.

Consider that if $b^3>5$, we can pick a small $\epsilon=(b^3-5)/(3b^2+3))$ such that $b-\epsilon<b$, then we can have $$(b-\epsilon)^3=b^3-3b^2\epsilon+3b\epsilon^2-\epsilon^3>b^3-3b^2\epsilon-3\epsilon^3>b^3-3b^2\epsilon-3\epsilon=5\tag 1$$

So $(b-\epsilon)^3>5>x^3,\forall x\in S$ which contradicts $b$ is the least upper bound.

Now consider that $b^3<5$, then so we can pick a $\epsilon=(5-b^3)/(3b^2+3b+1)$ such that $b+r>b$. Then we have $$(b+r)^3=b^3+3b^2\epsilon+3b\epsilon^2+\epsilon^3<b^3+3b^2\epsilon+3b\epsilon+\epsilon=5 \tag 2$$

This shows $b$ is in $S$ which contradicts $b$ is an upper bound. Thus, by the positivity axiom, $b^3=5$.


I am following the direction provided from the book, but I am not really good at inequity, so I am not sure $(1)$ and $(2)$ correct or not. If not, can someone give me a hit or suggestion to make $(1)$ and $(2)$ correct? Thanks in advanced.

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  • $\begingroup$ @yurnero @ graydad Thanks. $\endgroup$
    – Simple
    Oct 9, 2015 at 2:54
  • $\begingroup$ The asker of the linked question (link provided below) noticed a minor mistake in your first part proof that shows if $b^3>5$ then there is an $\epsilon>0$ such that $(b-\epsilon)^3>5$. You can easily fix by redefining $\epsilon = \frac{b^3-5}{3(b^2+1)}$: math.stackexchange.com/questions/2914671/… $\endgroup$
    – Michael
    Sep 12, 2018 at 19:02
  • $\begingroup$ I see that Clayton has edited the question itself in accordance with my comment above. $\endgroup$
    – Michael
    Sep 12, 2018 at 19:27
  • $\begingroup$ It should also be noted that there is another minor inaccuracy (I think?) where we should take the first epsilon as the smaller of 1 and the one given, just so that $\epsilon^3 \leqq \epsilon$. $\endgroup$
    – anak
    Jan 29, 2019 at 19:13

1 Answer 1

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The first part of your proof looks good! Nice choice of $\varepsilon.$ That establishes that $S$ has a least upper bound $b$ and $(1)$ shows that $b \leq 5$.

There are some bumps on the second half but you are definitely on the right track. First, to be pedantic, looks like you used $r$ on two occasions instead of $\varepsilon$. No biggie. The real problem is with your reasoning about $b$. Note that $b$ is still acting as the least upper bound for $S$, and it is perfectly fine for a least upper bound of a set to be an element of the set itself. For example, consider the closed unit interval $[0,1]$. It's least upper bound is $1$, a member of $[0,1]$.

That being said, $(2)$ was executed excellently and still completes the proof. The reasoning is that $(2)$ reveals a number $b+\varepsilon \in \Bbb{R}$ such that $b^3<(b+\varepsilon)^3<5$. Hence $b+\varepsilon \in S$, so the contradiction is that you've found an element in $S$ larger than the least upper bound of $S$.

P.S. You ask great questions. Complete with $\LaTeX$, a clear summary of what you've tried so far and you respond to input from other MSE users. Keep it up!

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  • $\begingroup$ To the proposer: You cannot "pick" $\epsilon.$ It is completely determined by $ b$. $\endgroup$ Sep 13, 2018 at 2:16

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