# Prove that there exists a number $x$ such that $x^3 = 6$

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.

## 3 Answers

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

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

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 $$\epsilon$$.

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 $$6$$.