Help with a lemma of the nth root (without the binomial formula) I have no idea of how to solve it. I would appreciate if someone gives me a hint, please. 
Definitions Let $\,x^{1/n}:= sup\{\, y \in \mathbb{R}: y\ge0 \text{ and } y^n\le x\, \}$
Lemma: Let $x,y>0$ be positive reals, and let $n\ge 0$ be a positive integer. 
(a) If $y = x^{1/n}$ then $y^n = x$
(b) Conversely, if $y^n = x$, then $y = x^{1/n}$
The big problems is I cannot use the binomial formula to prove it. I tried to used an argument by contradiction assume $y^n < x$ and so $y^n > x$ to get contradiction. But without the binomial formula I'm not sure of what use as estimator. The hint in the book is use denseness of $\mathbb{Q}$ in $\mathbb{R}$ and the basic properties of order in $\mathbb{R}$.
I thought to use something like this: Suppose that $y^n < x$ then there is a rational number such that $y^n < q < x$ and after this create a set $E_q$ which is bounded by $q$ find its supremum and try to get a contradiction and a similar argument to $x<y^n$. But that's not work very well. Could someone give me a hint, please? 
I think I have the exercise (b) 
Proof of (b): Suppose $y^n = x$ and $y > 0$. We set $E:=\{\, z \in \mathbb{R}: z\ge0 \text{ and } z^n\le x\, \}$. It follows that $E \not= \emptyset$ since $y\in E$. Now we need to show that is bounded above but that follows because $E$ is bounded by $\text{max} \{1,x\}$.
To prove the claim is sufficient to show that $y$ is the least upper bound of $E$. 
First we have to show that $y$ is an upper bound for $E$. We may argue by contradiction, suppose that there is a $z\in E$ such that $z>y$ so $z^n>y^n=x$, i.e., $z\notin E$ a contradiction. Then for all $z\in E$ we must have $z\le y$. 
Now to conclude the proof we need to show that $y$ is the least upper bound of $E$. Let $s$ be an upper bound for $E$ and suppose $s<y$. Then by the denseness of the rational numbers we have $s<q<y$.Then $q^n<y^n=x$ and $q>0$ so $\,q\in E$ contradicting that $s$ is an upper bound. Thus, the upper bound $s$ is greater than equal to $y$, thus $y$ is the least upper bound of $E$ and by definition $y= x^{1/n}$ as desired. 
With the part (a) I'm not sure yet.
 A: For problem (a) you can use this inequality, holding for $z>0$, $0<t\le 1$ and $n$ integer:
$$
(z+t)^n \le z^n + t((z+1)^n - z^n)
$$
If $z^n<x$, then you can find $t$ such that $(z+t)^n<x$: just take $t$ such that
$$
0<t<\min\left\{\frac{x-z^n}{(z+1)^n-z^n},1\right\}
$$
which surely exists. Therefore such a $z$ can't be the supremum of the set $\{y\ge0:y^n\le x\}$ and for the supremum $\xi$ of this set it must be $\xi^n=x$.

How do you find that inequality? It doesn't matter, because you can prove it by induction on $n$. It clearly holds for $n=0$. Suppose it holds for $n$; then
\begin{align}
(z+t)^{n+1}
&=(z+t)^n(z+t)\\
&\le(z^n+t(z+1)^n-tz^n)(z+t)\\
&=z^{n+1}+tz(z+1)^n-tz^{n+1}+tz^n+t^2(z+1)^n-t^2z^n\\
&< z^{n+1}+tz(z+1)^n+t(z+1)^n-tz^{n+1}-t^2z^n\\
&=z^{n+1}+t(z+1)^{n+1}-tz^{n+1}-t^2z^n\\
&\le z^{n+1}+t(z+1)^{n+1}-tz^{n+1}
\end{align}
The hypothesis $0<t\le 1$ is used to get the second $\le$, since $t^2\le t$.
Tell your teacher an elf suggested it. ;-) Of course one can also prove it with the binomial theorem:
\begin{align}
(z+t)^n
&=z^n+\binom{n}{1}z^{n-1}t+\binom{n}{2}z^{n-2}t^2+\dots+\binom{n}{n-1}zt^{n-1}+t^n\\
&\le z^n+\binom{n}{1}z^{n-1}t+\binom{n}{2}z^{n-2}t+\dots+\binom{n}{n-1}zt+t\\
&=z^n+t((z+1)^n-z^n)
\end{align}
(which is how I first got it). The idea came from the same problem but with $n=2$, which is easier.
A: Here I provide a complete proof which was inspired by ergreg.
Proof Let $E := sup\{t \in \mathbb{R}:t \ge 0 \text{ and }t^{n}\leq x \}$, then by definition $y = sup(E)$. In order to prove $y^{n}=x$, we show that both $y^{n} <x$ and $y^{n} >x$ lead to contradictions.First suppose that  $y^{n}<x$. Let $0< \epsilon <1$ be a small real number, here we want to find an $\epsilon$ such that $(y+\epsilon)^n<x$. Since $0<\epsilon<1$, we have
$$
(y+\epsilon)^{n}-(y)^{n} \leq \epsilon((y+1)^{n}-(y)^{n})
$$
Transform it, we have
$$
(y+\epsilon)^{n} \leq (y)^{n}+\epsilon((y+1)^{n}-(y)^{n})
$$
Let the right side$$
(y)^{n}+\epsilon((y+1)^{n}-(y)^{n})\le x$$then we have $$\epsilon\le \frac {x-(y)^n}{(y+1)^{n}-(y)^{n})}$$So when 
$$
0<\epsilon<\min\left\{\frac{x-y^n}{(y+1)^n-y^n},1\right\}
$$
we have $(y+\epsilon)^n<x$, which implies that $y+\epsilon\in E$.But it contradicts that fact that $y$ is an upper bound of $E$ as $y+\epsilon >y$.
     Second suppose that  $y^{n}>x$. Let $0< \epsilon <1$ be a small real number, here we want to find an $\epsilon$ such that $(y-\epsilon)^n>x$. Since $0<\epsilon<1$, we have
$$
(y)^{n}-(y-\epsilon)^{n} \leq \epsilon((y)^{n}-(y-1)^{n})
$$
Transform it, we have
$$
(y-\epsilon)^{n} \ge (y)^{n}-\epsilon((y)^{n}-(y-1)^{n})
$$
Let the right side
$$
(y)^{n}-\epsilon((y)^{n}-(y-1)^{n})\ge x$$
then we have 
$$
\epsilon\leq \frac {(y)^n-x}{(y)^{n}-(y-1)^{n})}$$
So when 
$$
0<\epsilon<\min\left\{\frac{y^n-x}{(y)^n-(y-1)^n},1\right\}
$$
we have $(y-\epsilon)^n>x$.But this implies that $e \le y-\epsilon$ for all $e \in E$. (If $e>y-\epsilon$, then $e^n>(y-\epsilon)^n>x$, a contradiction.) Thus $y-\epsilon$ is an upper bound for, which contradicts the fact that $y$ is the least upper bound for $E$ as $y-\epsilon<y$.From these two contradictions we see that $y^n=x$, as desired.
