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I want to show that there exists a real number $x$ such that $x^3 = 6$. Here is what I have so far. $\\$


Let $S = \{x \mid x \in \mathbb{R}, x \geq 0, x^3 < 6\}$. By this definition, $S$ is nonempty since $0 \in S$, and also $S$ is bounded above since $2^3 = 8 > 6$. Thus, by the Completeness Axiom, $S$ has a least upper bound; call it $b$. We will show that $b^3 = 6$ (and hence, there exists a real number such that $x^3 = 6$) by showing that we cannot have $b^3 > 6$ or $b^3 < 6$.

First, for the sake of contradiction, suppose we had $b^3 > 6$. Then, we will show that we can choose a suitably small positive number $\epsilon$ such that $b - \epsilon$ is also an upper bound for $S$, which contradicts $b$ being the least upper bound. But, I'm not sure about how to find $\epsilon$. I tried expanding:

$$(b - \epsilon)^3 = b^3 - 3b^2\epsilon + 3b\epsilon^2 - \epsilon^3,$$

and from here, I think I'm supposed to use greater-than equalities to try and come up with $\epsilon$, but I'm not really sure how to do that. Any help is appreciated.

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Note that$$b^3 - 3b^2\varepsilon + 3b\varepsilon^2 - \varepsilon^3>6\iff b^3-6>3b^2\varepsilon-3b\varepsilon^2+\varepsilon^3.$$Now, take $\varepsilon\in\left(0,1\right)$ such that$$\varepsilon<\dfrac{b^3-6}{6b^2\varepsilon}\tag1$$and that$$\varepsilon<\dfrac{b^3-6}2.\tag2$$Then\begin{align}3b^2\varepsilon+\varepsilon^3&<3b^2\varepsilon+\varepsilon\text{ (because $\varepsilon<1$)}\\&<\frac{b^3-6}2+\frac{b^3-6}2\text{ (by $(1)$ and $(2)$)}\\&=b^3-6.\end{align}Therefore$$3b^2\varepsilon-3b\varepsilon^2+\varepsilon^3<3b^2\varepsilon+\varepsilon^3<b^3-6.$$

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  • $\begingroup$ Apologies Jose, perhaps I'm a little slow today; I cannot follow the logic to arrive at $(2)$. Could you expand? $\endgroup$ – Kevin Sep 25 '18 at 8:22
  • $\begingroup$ @Kevin I don't understand your question. I chose a $\varepsilon\in(0,1)$ which satisfies conditions $(1)$ and $(2)$ so that $3b^2\varepsilon+\varepsilon^3<b^3-6$. $\endgroup$ – José Carlos Santos Sep 25 '18 at 8:24
  • $\begingroup$ Ah yes, I understand now, sorry I did say I was being slow today. My question was junk! $\endgroup$ – Kevin Sep 25 '18 at 8:26
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If you write it like so:

$$(b-\epsilon)^3 = b^3 - \epsilon(3b^2 - 3b\epsilon + \epsilon^2)$$

you can see that (if $\epsilon < \frac b2$, which can safely be assumed), you have $$(b-\epsilon)^3 = b^3 - \epsilon\cdot M$$

where $M>b$.

Now, look at the values of $b^3$ and $6$. There is a "gap" assumed to be between them, call it $\delta = b^3-6$. Now, clearly, $6=b^3-\delta$, but your number $b^3-\epsilon\cdot M$ can be made greater than $b^3-\delta$ if you pick a small enough $\epsilon$.

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First you can discard the positive term $3b\epsilon^2$ and focus on the rest factoring out $\epsilon$. $$ (b-\epsilon)^3>b^3-(3b^2+\epsilon^2)\epsilon $$ Since for small positive values of $\epsilon$ the expression $(3b^2+\epsilon^2)$ is bounded, the last term can be made as small as you like as to eventually show that the whole thing is greater than $6$.

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