Real Analysis : $x^n = a$ has only one solution. I have a question regarding the following exercises.
(1)
Let $n \in \mathbb {N}$. For all $a \in \mathbb {R}$ with $a \geq 0$ there exists $x \in \mathbb {R}$  with $x \geq 0$ such that $x^n = a$.
(2) The equation $x^n = a$ has just one solution for $x \geq 0$. That solution is called $n$-th root of $a$.
These two problems seem intuitive. Especially the second one since to me it seems that there is only some small step required to get me the right result / proof. However, Taking the $n$-th root obviously is not the solution.
 A: *

*Use continuity of the function $f(x):=x^n$.


 $f(0)\le a$ and $f(\max(a,1))\ge a$ so by the intermediate value theorem there must be an $x$ such that $f(x)=a$.



*$f(x)$ is a strictly growing function.


 $x>y\implies x^n>y^n$. So $x^n=b^n=a\implies x=y$.


Alternatively, for 1. we can show that the subset $$A:=\{x\in\mathbb R:x^n<a\},B=\mathbb R\setminus A$$ defines a Dedekind cut.

*

*$A\ne \emptyset$ and $A,B$ partition $\mathbb R$.


*$x^n<a\land y^n\ge a\implies x^n<y^n\implies x<y$, by monotonicity.


*if $y^n\ge a$, $x:=\min_{y\in\mathbb R}(y)\in B$ because $x^n\ge a$. (By monotonicity, infimum and power commute.)
A: An alternative approach for (2). Suppose $x^n=y^n=a$ where $x,y\geq 0$ and $x\neq y$. Since
$$0=x^n-y^n=(x-y)\times\sum_{k=0}^nx^ky^{n-k},$$
and $x-y\neq0$, we must have $\sum_{k=0}^nx^ky^{n-k}=0$. But every term of this sum is nonnegative, and $x,y$ are different so the first and last terms can't both be $0$, contradiction.
A: Too trivial?
Uniqueness of solution.
1)$a>0;$
Assume that $x^n =y^n=a.$ Then $x, y >0.$
$(x/y)^n=1$ $\Rightarrow x/y=1;$ $x=y;$
2)$a=0;$
Implies $x=y=0.$
Appended:
$ b:=x/y =1$ is the only real number $b>0$ that satisfies
$b^n=1;$
1)Assume $b>1$.
Then $b^2>b>1$. Continuing inductively
we get  $b^n>1;$
2)Assume $b<1;$
Similarly we get  $b^n<1;$
3)$b=1$ satisfies $b^n=1,$ and
we are done.
