Existence in real-analysis could you help me with the following problem?
I am looking at $\mathbb{R}$ construction in Cauchy sequence analysis.
Suppose the polynomial functions are continuous on $\mathbb{R}$. Let the polynomial function be such that $f (x) = x^{n}-a$, where $n\in \mathbb{N}$ and $a\in \mathbb{R^{+}}$. Prove that there exists exactly a positive real number c such that $f (c) = 0$. This number c is called the énesima root of a and is denoted $a^{1/n}$.
I have tried applying the mean value theorem, but I do not think my argument is correct since we have never used it in class
 A: You are most likely not suppose to use the Mean Value Theorem/Rolle's Theorem/Intermediate Value Theorem for this. It is highly likely that trying to prove Rolle's Theorem without assuming the property of the real numbers in question would be impossible.
All you need to do is consider $A = \{x : x \in \mathbb{R} \ | x^{n} \leq a\}$.
This set is nonempty and bounded from above, therefore the supremum of $A$ exists, call it $c$. Now you should argue that 1: $c$ is positive, 2: $c^{n} = a$, 3: prove that $c$ is unique with this property.
A: Define $f:[0,\infty)\to \mathbb{R}$ .
$f(x)=x^n-a$; $a\in \mathbb{R}^+$.
(Existance)
$f(0)=-a<0$ and by Archimedean-property $\exists M>0 $  such that $\forall x>M$ such that $x^n>a \implies f(x)=x^n-a>0,\forall x>M$ .
Then by IVT we have there exist $c>0$  such that $f(c)=0\implies c^n=a$.
(Uniqueness)
Observe that $f$ is differentiable every where on $(0,\infty)$.Also observe that $f'(x)>0$ on $(0,\infty)$.
Suppose we had $c_1,c_2\in \mathbb R$ and $c_1 \not=c_2$ such that $f(c_1)=f(c_2)$ then by Rolle's Theorem $\exists d\in (c_1,c_2)$ such that $f'(d)=0$, but this is not possible.
Hence there exist unique $c>0$ such that $f(c)=0$.
A: Lemma: If $x>0, y> 0$ then $x<y \iff x^n < y^n$ for any natural $n$.
Pf: If $x < y$ then $x^1 < y^1$.
If $x^k < y^k$ then $x^{k+1}=x^k*x < y^k*x < y^k*y < y^{k+1}$.
So by induction $x<y\implies x^n < y^n$.
And if $x=y$ then $x^n = y^n$.
And if $x>y$ then $y< x$ and $y^n < x^n$ and $x^n > y^n$.
So $x< y \iff x^n < y^n$.
....
Let $A = \{x| x>0, x^n < a\}$ and $B = \{x|x>0, x^n > a\}$.
Claim: $A$ is not empty.
Pf: If $A$ is empty then $x^n \ge a$ for all $x > 0$.  So $a> 0$ is a lower bound of $\{x^n|x>0\}$.  We also know that $a < 1$ because otherwise for all $q: 0< q < 1$ we'd have $q^n < 1 < a$ and $q \in A$.  So $a^n < a$ and $a \in A$. So we have a contradiction.
Claim: $B$ is not empty.
Pf: If $B$ is empty then $a$ is an upper bound of $\{x^n|x>\}$.  We know $a > 1$ because otherwise for all $q: q>1$ we'd have $q^n>1\ge a$. And so $a^n > a$ and $a \in B$.
Now for $x>0,y>0$ we have $x< y\iff x^n < y^n$ we have any element of $A$ is a lower bound of $B$ and every element of $B$ is an upper bound of $A$ so $\sup A$ and $\inf B$ exist in the reals and $\sup A \le \inf B$.
Now if $\sup A < \inf B$ there are $w,z: 0< \sup A < w<z < \inf B$ but $w,z\not \in A$ and $w,z\not \in B$ and $w,z >0$ so $w^2\ge a;w^2\le a; z^2 \ge a; z^2 \le a$ so $w^2 =z^2 = a$ but $w< y$ so $w^2 < z^2$ and that's a contradiction.
So $\sup A = \inf B$.
Let $\sup A = \inf B = k$.
If $k^n < a$ then 
If $k^n < a$ then.... bear with me... this is magic...
Let $0< h < \min(\frac {a-k^n}{n(k+1)^{n-1}}, 1)$.
The $(k+h)^2 - k^2 = [k+h-k][(k+h)^{n-1} + (k+h)^{n-2}k + ... + (k+h)k^{n-2}+k^{n-1}]\le$
$[k+h-k][(k+h)^{n-1} + (k+h)^{n-2}(k+h) + .... + (k+h)(k+h)^{n-2}+(k+h)^{n-1}]=$
$h[(k+h)^{n-1}+(k+h)^{n-1}+(k+h)^{n-1}+....+(k+h)^{n-1}=$
$hn(k+h)^{n-1} <$
$\frac {a-k^n}{n(k+1)^{n-1}}n(k+h)^{n-1}=$
$(a-k^n)\frac {(k+h)^{n-1}}{(k+1)^{n-1}} <$
$a-k^n$.
Therefore $(k+h)^2 < a$ and $k+h \in A$ but $k=\sup A$ so that's impossible.
So $k^n \ge a$.
If $a  < k^n$ we can do the same thing with $k = \inf B$.
Let $0 < h < {k^n- a}{nk^{n-1}}$
$k^n - (k-h)^n \le hnk^{n-1}< k^n - a$
$a < (k-h)^n$ so $k-h \in B$ which contradicts $k=\inf B$.
So $k^n = a$.
And by the Lemma at the very beginning, $k$ is unique.
