I'd like to know if the proof of the following laws of exponent is correct. First of all thanks in advance and I apologize for the extension of the following proof but it's one exercise. I hope there will not problem with this.
Definition: Let $x$ be a positive real number, and let $n\ge 1$ be a positive integer. We define $,x^{1/n}$ also known as the $n^{th}$ root of $x$ by the following formula:
$$\,x^{1/n}:=\text{ sup }(\{y\in \mathbb{R}: y\ge 0 \text{ and } y^n \le x \})$$
Lemma: Let $x,y>0$ be positive reals and let $n,m\ge 1$ be positive integers
(a) If $y = x^{1/n}$, then $y^n = x$.
(b) if $y^n = x$, then $y = x^{1/n}$
(c) $x^{1/n}$ is a positive real number
(d) We have $x<y \iff x^{1/n}<y^{1/n}$
(e) If $x>1$, then $x^{1/k}$ is an decreasing function of $k$. If $x<1$, then $x^{1/k}$ is an increasing function of $k$. If $x=1$, then $x^{1/k}=1$ for all $k$
(f) We have $(xy)^{1/n} = x^{1/n}y^{1/n} $
(g) We have $(x^{1/n})^{1/m}=x^{1/mn}$.
Proof:
(a)
Let $E=\{y\in \mathbb{R}: y\ge 0 \text{ and } y^n \le x \}$ and let $\,x^{1/n}=\text{ sup } (E)$. We have to show that either $y^n < x$ and $y^n > x$ leads to a contradiction.
Suppose $y^n < x$. If we can find an $\epsilon>0$ such that $(y+\epsilon)^n<x$ this means that $\,(y+\epsilon)\in E$ which contradicts the fact that $y$ is an upper bound for $E$.
Claim 1: If $\,0< h \le 1$ and $z>0$. Then $\,(z+h)^n \le z^n+h ((z+1)^n-z^n)$ for all $n\ge 0$.
Proof of claim 1: We use induction on $n$. Let $P(n)$ be the statement of the above claim. Clearly the base case holds. Now suppose inductively that $n$ is a natural number and $P(n)$ has already been proven. We now prove $P(n+1)$.
\begin{align} (z+h)^{n+1}=(z+h)(z+h)^{n} &\le(z+h)[z^n+h ((z+1)^n-z^n)]\\ &=(z^{n+1}+hz^n)+h (z+h)((z+1)^n-z^n)\\ &\le(z^{n+1}+hz^n)+h (z+1)((z+1)^n-z^n)\\ &=z^{n+1}+h ((z+1)^{n+1}-z^{n+1}) \end{align}
which close the induction and prove the claim. $\;\Box$
Since $y^n<x$, then $x-y^n$ is a positive real number. Let define $\epsilon< \text{min} \bigg( \frac{x-y^n}{(y+1)^n-y^n},1 \bigg)$. Thus, $0<\epsilon \le 1$ and by hypothesis $y>0$. We thus have $(y+\epsilon)^n\le x$ which is a contradiction.
Now suppose that $y^n>x$. We shall show that there is an $\epsilon>0$ such that $y-\epsilon$ is an upper bound of $E$ contradicting that $y$ is the least upper bound of $E$.
Claim 2: If $\,0< h \le 1$ and $z>h$. Then $\,(z-h)^n \ge z^n-h ((z+1)^n-z^n)$ for all $n\ge 0$.
Proof: We induct on n. The base case is when $n=0$ and clearly holds since both sides are equal. Now suppose inductively that the claim holds for some $n\ge 0$; we wish to show that also holds for $n+1$.
\begin{align} (z-h)^{n+1}=(z-h)(z-h)^{n} &\ge(z-h)[z^n-h ((z+1)^n-z^n)]\\ &=(z^{n+1}-hz^n)-h(z-h)((z+1)^n-z^n)\\ &\ge(z^{n+1}-hz^n)-h(z+1)((z+1)^n-z^n)\\ &=z^{n+1}-h((z+1)^{n+1}-z^{n+1})\\ \end{align}
which close the induction. $\; \Box$
Since $y^n>x$, then $y^n-x$ is a positive real number. Let $\epsilon < \text{min} \bigg( \frac{y^n-x}{(y+1)^n-y^n},1, y \bigg)$. Thus, $0\le \epsilon < 1$ and $y> \epsilon$. Then, we have $(y-\epsilon)^n \ge x$ contradicting that $y$ is the least upper bound since we find an upper bound which is less than $y$.
Hence, we must have $y^n=x$.
(b)
Suppose $y^n = x$, we need to show that $y = x^{1/n}$. We argue for contradiction, suppose $y \not= x^{1/n}$. Then by the trichotomy of the real numbers either $x^{1/n}<y$ or $x^{1/n}<y$.
If $\,x^{1/n}<y$ and since $x^{1/n}$ is the lub of $E$ so that it is non-negative (otherwise we have a contradiction, because $0 \in E$) and it is not zero, because by (a) if were zero that implies $x=0$ but contradicts our hypothesis, then $x^{1/n}$ is positive. This means $0<x^{1/n}<y$, which implies $0<x<y^n=x$ a contradiction
If $\,y<x^{1/n}$ and since $y>0$ using a similar argument as above we have $0<y^n=x<x$ which is a contradiction. Therefore $y = x^{1/n}$.
(c)
We can use the same argument as above when shows that $x^{1/n}$ is non-negative. But also cannot be zero, since $x>0$.
(d)
($\Rightarrow$) For the sake of contradiction suppose $x^{1/n}\ge y^{1/n}$ by (c) we already know that $0<y^{1/n}$. Moreover by (a) we have $y=(y^{1/n})^{n}$ and $x=(x^{1/n})^{n}$. Thus, $0<y^{1/n}\le x^{1/n}$ implies $0<y \le x$ which is a contradiction.
($\Leftarrow$) Suppose $x^{1/n}<y^{1/n}$ then $x<y$
(e) When $x>1$ we have to show that if $k,\ell \in \mathbb{N}-\{0\}$ and $\mathcal k<\ell$, then $x^{1/k}\ge x^{1/\ell}$. We argue this for contradiction.
Suppose $x^{1/k}< x^{1/\ell}$. Then using (d) and the properties of the exponentiation with natural numbers we have $x^{\ell}< x^{k}$. Let $m=\ell-k$. Thus, $x^{k+m}=x^k x^m< x^{k}$.
It is easy to show that $x^n>0$ for each positive integer so its inverse is also positive. Thus $x^m = x^{\ell-k}<1$. We claim that this is a contradiction.
Claim 3: If $x>1$ and $n$ is a positive integer, then $x^n>1$. If we can prove this, then since $m=\ell-k$ is a positive integer we have the desired contradiction.
Proof: The base case is when $n=1$ and is trivial. Now suppose the claim holds for $n$. We shall show that also holds for $n+1$ which close the induction. $x^{n+1}= x x^n> x$ and since $x>1$ the results follows. $\Box$
So, by the claim 3 and the fact $k-\ell\ge 1$, we have that $x^m = x^{\ell-k}<1$ is a contradiction.
If $x<1$, so $1<1/x$ and by what we have said above, if $k<\ell$ we thus have $(1/x)^{1/k}\ge (1/x)^{1/\ell}$.
We claim that $1/x^{1/k} =(1/x)^{1/k} $. Let $a= x^{1/k}$ and $b = (1/x)^{1/k}$. Thus, $a^k = x$ and $b^k= 1/x=1/a^k=(1/a)^k$. Then by b) we have $b=1/a$. In other words we have $(1/x)^{1/k} = 1/x^{1/k}$ as desired.
Then $1/x^{1/k}\ge 1/x^{1/\ell}$ and thus $x^{1/\ell} \ge x^{1/k}$ which prove that is increasing.
The case when $x=1$ is trivial.
(f)
We have to show $(xy)^{1/n} = x^{1/n}y^{1/n} $. Let $a = x^{1/n}$, $b= y^{1/n}$ and $c=(xy)^{1/n}$. Then, $a^n=x,\,b^n=y,\,c^n =xy$. Thus, $c^n= xy =a^nb^n =(ab)^n$, which implies that $c=ab$, i.e., $(xy)^{1/n}=x^{1/n}y^{1/n}$ as desired.
(g)
We have to show $(x^{1/n})^{1/m}=x^{1/mn}$. Let $a =x^{1/n}$, $b= a^{1/m}$. Then $a^n = x$, $b^m = a$ so $x=a^n =(b^m)^n= b^{mn}$. Thus, $b=x^{1/mn}$, which means $(x^{1/n})^{1/m}=x^{1/mn}$ and we're done.