Unique Solution Exists to x^n=y: Real Analysis Proof

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Homework Statement


Given x > 0 and n \in N, prove that there is a unique y > 0 s.t. y^n = x exists and is unique

Homework Equations


Hint is given: consider y = 1. u.b. \{s \in R : s^n < x\}

The Attempt at a Solution


I'm not used to this style of proof (real analysis I), help would be appreciated, thanks. BTW, what does "u.b." signify?
 
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It's actually l.u.b. with a letter "L" instead of a digit "1". It means "least upper bound. Does that help?
 
oh, now its clearer. thanks
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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