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February 2nd, 2018, 08:32 PM  #1 
Senior Member Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0  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? 
February 2nd, 2018, 09:42 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198 
suppose there are $t_1\neq t_2$ such that $t_1^3=t_2^3=c$ then $t_1^3  t_2^3 = 0$ $(t_1  t_2)(t_1^2 + t_1 t_2 + t_2^2) = 0$ 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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cubes, positive, proof, uniqueness 
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