Existence of $n^{th}$ root Disclaimer : This is homework (Tagged)
I am trying to prove that Real numbers have nth roots. Most references do the following :
Let $a \in \mathbb{R}, n \in \mathbb{N}, n > 0$ Then we consider a set  X = {$t | t^n< a$}. Then we prove that X is nonempty (since it contains 0) and that it is bounded above ($a+1$ is an upper bound). Then through the property of Real numbers, this set will have a supremum. Let that supremum be f. Most books now deal with 2 cases : $f^n<a , f^n>a$ and show a contradiction thus proving $f^n = a$ and I am comfortable with this way of doing it. I have finished my homework. Books closed. EOF.
However, is there any way to prove this without the use of contradiction? 
 A: The idea: Show the Newton-Raphson interation scheme to find the positive root of $f(x)= x^n-a $ converges, where we fix $n\geq 2, a>0.$  This is sufficient to prove existence, and uniqueness follows from $0<x<y \implies 0< x^n < y^n.$ 
Newton-Raphson: $$x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)} = \frac{1}{n} \left( (n-1)x_k+ \frac{a}{x_k^{n-1}} \right) .$$
Pick $x_0=a.$ If $x_k\to x$ then taking limits of the recurrence and rearranging gives $x^n=a$ as we hope. 
We show $(x_k)$ is positive and monotonically decreasing for $k\geq 1$, and thus convergent: By the Arithmetic-Geometric mean inequality, 
$$ x^n_{k+1} = \frac{1}{n^n} \left( (n-1)x_k+ \frac{a}{x_k^{n-1}} \right)^n = \left( \frac{ x_k + x_k + \cdots x_k + \frac{a}{x_k^{n-1}} }{n} \right)^n \geq x_k^{n-1} \frac{a}{x_k^{n-1}} = a$$ for all $k\geq 0.$ Hence $a/x_k^n\leq 1$ for all $k\geq 1$ so using the recurrence gives: $$nx_{k+1}=(n-1)x_k + \frac{a}{x_k^{n-1}} = x_k\left( n-1 + \frac{a}{x_k^n} \right) \leq x_k(n-1+1) = nx_k$$
so the sequence is decreasing as required. 

Simple introduction to Newton-Raphson: The aim is to approximate the root of a function. To do so, take a decently close estimate of the root $x_0.$ Draw the graph of $y=f(x)$ with some root, and mark the point $(x_0, f(x_0))$ which is somewhat near-by. Draw the tangent to the curve at that point, notice that the $x$-intercept of this tangent is a better approximation to the root than $x_0$ was, call it $x_1.$ Repeat this, and in the limit (in "ideal" situations) $x_n$ tends to the root. Simple analytic geometry yields $x_{k+1}$ in terms of $x_k, f(x_k), f'(x_k)$ and we get the Newton-Raphson recurrence. 
This is often a very quickly converging method to approximate roots of functions (each iteration doubles the number of accurate decimal places), but it can fail if your initial approximation is too far away, or too close to another root, or if $f'(x_k)=0$ (so that the tangent is horizontal so never intercepts the $x$-axis). For this problem I showed that by choosing $x_0=a$ this method ends up converging (which is something you should strongly suspect to happen by sketching the graph and a few of the tangents), to something that must be a root of $x^n-a.$
