given x>0 and n contained in N show show that there exists a unique positive real number $r$ such that $x = r^{n}$ I have proved half of this proof but I'm stuck on the other half because it is a little harder than the first due to negatives. 
I have considered that $(r+h)$ so that $(r+h)^n$ is less than $x$ and $h$ is between $0$ and $1$. I proved that half. However how do I prove it for $(r-h)$? 
I can take a picture of my work if you would like to see it, just let me know. I just don't have Matlab or know how to use/get it so I don't want to write it out. It would look very messy and be hard to read. Thanks for the help.
 A: I'm not quite sure what approach you were doing, but it reminds me of a proof along the lines that Rudin outline's in his Introduction to Mathematical Analysis problem.  Another, more elementary, approach would be to use the Intermediate Value Theorem to obtain existence.  Then uniqueness follows quite easily.
Note that $f(r) = r^n$ is continuous (it's a product of continuous functions), and that $f(0) = 0$ and that $f(s) > x$ if we take $s > \text{max}\{\, 1, x\, \}$.  Then it follows that there exists an $r$ such that $f(r) = x$ by the Intermediate Value Theorem.
An alternate method is to use the least upper bound property of the real numbers.  Let $S = \{ \, r : r^n < x\, \}$.  Note that $S$ is non-empty because $0 \in S$.  Also note that $S$ is bounded from above by $\max \{\, 1, x\, \}$.  So there exists a unique least upper bound $\alpha$ for $S$.  The claim is that $\alpha^n = x$.  By the least upper bound property we have
$$  (\alpha - h)^n < x $$
for any $0 < h < \alpha$.  Suppose that $\alpha^n > x$.  Let $\epsilon = \alpha^n - x > 0$.  Bernoulli's inequality says:
$$(\alpha - h)^n \geq \alpha^n\left(1 + \frac{nh}{\alpha}\right)$$
For any $0 < h < \max \{ 1, \alpha \}$.  So we can pick $h$ small enough (a formula for $h$ can be found) so that:
$$ \alpha^n - (\alpha - h)^n < \epsilon. $$
This implies that $(\alpha - h)^n > x$, which is a contradiction, so we must have $\alpha^n \leq x$.
Now we need to show that $\alpha^n \geq x$.  This follows from the same argument essentially.  Suppose that $\alpha^n < x$, and let $\epsilon = x - \alpha^n > 0$.  Using that $\alpha > 0$ and $1 > h > 0$ it's easy to show (using the binomial formula) that
$$ (\alpha + h)^n \leq \alpha^n + Ch $$
Where $C$ is some constant that depends on $\alpha$ and $n$, but not on $h$.  You can take $C = n\sum_{k = 0}^{n-1} \alpha^k$ for instance.   It follows that we can pick $h$ small enough so that
$$ (\alpha + h)^n - \alpha^n < \epsilon. $$
But this implies that $(\alpha + h)^n \leq x$ which contradicts that $\alpha$ is an upperbound for $S$.  So we must hat $\alpha^n \geq x$.  This proves that $\alpha^n = x$.
A: Having invested a little energy in the formulation of this question (see comments), guess I'll spend a little more on an answer:
I'm in the Intermediate Value Theorem school on this one. 
Before proceeding further, lauds and "+1" to breeden for an answer based on fundamental properties of the real number system $\Bbb R$.  In the following, I have presented two arguments, one using the derivative of $r^n$ and one which only uses mere continuity, in order to make clear that differentiation is not essential in order to establish the desired results, though it does make things a little easier.  Of course, the IVT is itself pretty close to the fundamentals of $\Bbb R$, depending as it does on the completeness of $\Bbb R$ (see this wikipedia page).
Having said these things, the setup is:
Let $\Bbb R_{0+}$ denote the non-negative real numbers; that is
$\Bbb R_{0+} = \{ y \in \Bbb R \mid y \ge 0 \}, \tag{1}$
and for $n \in \Bbb N$, let $f_n:\Bbb R_{0+} \to \Bbb R_{0+}$ be defined by $f_n(r) = r^n$; we have $f_n(0) = 0$ for all $n$.  Given $x > 0$, it is easy to see that there exists $y \in \Bbb R_{0+}$ with $f_n(y) = y^n > x$; if there is any doubt about this, just choose $m$ to be any integer satisfying $m > x + 1$; then $m^n \ge m > x$; set $y = m$.  Next, we note that $f_n(r)$ is continuous for all $n \in \Bbb N$; there are two ways to see this, the "with-derivatives" way and the (you guessed it!) "without-derivatives" way:
A.) with-derivatives:  note that $f_n(r) = r^n$ is differentiable, $f_n'(r) = nr^{n - 1}$; being differentiable, it is continuous;
B.) without-derivatives:  observe that for $r_0, r \in \Bbb R_{0+}$ with $r_0, r < L$,
$\vert r^n - r_0^n \vert = \vert (r - r_0)\sum_0^{n - 1}r^{n - 1 - i}r_0^i \vert$
$= \vert r - r_0 \vert \sum_0^{n - 1} r^{n - 1 - i}r_0^i < nL^{n - 1}\vert r - r_0 \vert, \tag{2}$
which shows that $r^n \to r_0^n$ as $r \to r_0$; thus, $f_n(r)$ is continuous.
We see from the above that $f_n(r)$ is continuous for each $n \in \Bbb N$, and that $f_n(0) < x < f_n(m)$; thus by the IVT there exists $r \in \Bbb R_{0+}$ with $r^n = x$; $r > 0$ since $x > 0$.
We proceed to show that $r$ is unique, in both the "with-derivatives" and the "without-derivatives" ways:
C.)  with-derivatives:  as we have seen in item (A), $f'_n(r) = nr^{n - 1}$; since $f_n'(r) > 0$ for $r > 0$, $f_n(r)$ is monotonically increasing on $\Bbb R_{0+}$; thus, given $r$ such that $f_n(r) = r^n =x > 0$, as $r$ increases (or decreases), $f_n(r)$ can never revisit $x$; to put it more formally, if $r_1^n = r_2^n = x$ with $r_1 \ne r_2$, we may take $r_2 > r_1$; but then $r_2^n > r_1^n$ by monotonicity, so if $r_1^n = x$, we must have $r_2^n > x$; likewise if $r_2 < r_1$, $f_n(r_2) < f_n(r_1) = x$; either alternative contradicts $r_1^n = r_2^n = x$; thus $r^n = x$ has a unique solution for $x \in \Bbb R_{0+}$;
D.)  without-derivatives:  the monotonicity of $f_n(r) = r^n$ can also be established by a simple demonstration based upon the binomial expansion of $(r + h)^n$ for $r, h \in \Bbb R_{0+}$; for we have
$(r + h)^n = \sum_0^n C_i^n r^{n - i}h^i = r^n + \sum_1^n C_i^n r^{n - i}h^i > r^n, \tag{3}$  
since every term in $\sum_1^n C_i^n r^{n - i}h^i$ is positive.  Given that $f_n(r) = r^n$ is monotonically increasing, we complete the argument for the uniqueness of $r$ satisfying $r^n = x$ as in (C).
Hope this helps!  Cheerio,
and as always,
Fiat Lux!!!
