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February 2nd, 2018, 07:32 PM   #1
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Exclamation Proof of uniqueness of positive cubes

Prove that there exists exactly one positive $t\in \mathbb{R}$ with $t^3=c$ for $c>0$.

I assume that most of my derivation needs to be algebraic and using some simple axioms. So for example, if I define the set:

$$T:=\lbrace t\in \mathbb{R}: t> 0, t^3 \leq c \rbrace$$

I must show that the cubed root of $c$ is the supremum of $T$ and then prove its uniqueness. But I'm not sure how to go about this. How can I concisely prove this?
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February 2nd, 2018, 08:42 PM   #2
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suppose there are $t_1\neq t_2$ such that $t_1^3=t_2^3=c$


$t_1^3 - t_2^3 = 0$

$(t_1 - t_2)(t_1^2 + t_1 t_2 + t_2^2) = 0$

well we've assumed that $t_1 \neq t_2 $ so it must be that

$t_1^2 + t_1 t_2 + t_2 ^2 = 0$

$\left(t_1 - \dfrac{t_2}{2}\right)^2 + \dfrac{3 t_2^2}{4} = 0$

It should be pretty clear that this has no real solutions.

Thus the original assumption is incorrect and $t_1 = t_2$
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