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 February 2nd, 2018, 07:32 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, 08:42 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,553 Thanks: 1403 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 $t_1 = t_2$ Thanks from greg1313, Maschke, Country Boy and 1 others Tags cubes, positive, proof, uniqueness Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Tau Math 40 December 31st, 2016 06:23 PM annakar Number Theory 1 December 4th, 2012 11:06 AM aeromantang Applied Math 3 December 28th, 2011 10:34 AM jstarks4444 Number Theory 2 November 16th, 2010 12:19 PM johnmath Advanced Statistics 1 July 24th, 2010 06:56 PM

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