$\forall x \in \mathbb{R}^{+}, n \in \mathbb{N}$, prove that $\exists y \in \mathbb{R}^{+}$ such that $y^n = x$. Q: Let $n$ be a natural number and $x$ a positive real number. Prove that there is a positive real number $y$ such that $y^n = x$. Is $y$ unique?
\begin{align*}
  y^n &= x \\
  y &= x^{1/n} \\
  y &\in \mathbb{R}^{+} \\
\end{align*}
So, yes, I can see that there is a positive real number $y$ for any natural $n$ and for any positive real $x$. And I can see that $y$ is unique. How can I prove this?
 A: For the purposes of this answer, I take
$\Bbb R_+ = \{ r \in \Bbb R \mid r \ge 0 \}, \tag 1$
that is, $\Bbb R_+$ will be the set of non-negative reals.
For $n \in \Bbb N$ we consider the function
$f_n: \Bbb R_+ \to \Bbb R_+, \; f_n(y) = y^n; \tag 2$
clearly,
$f_n(0) = 0^n = 0; \tag 3$
next we choose $M \in \Bbb N$ such that
$M > x; \tag 4$
then
$f_n(M) = M^n \ge M> x; \tag 5$
we note that $f_n$ is continuous for every $n \in \Bbb N$; we now have
$f_n:[0, M] \to \Bbb R, \; f_n(0) = 0, \; f_n(M) = M^n; \tag 6$
since $x > 0$ we also have
$x \in (0, M^n); \tag 7$
it now follows from the intermediate value theorem that
$\exists \Upsilon \in (0, M), \; f_n(\Upsilon) = \Upsilon^n = x; \tag 8$
as for the uniqueness of $\Upsilon$, we note that $f_n$ is strictly monotonically increasing on $\Bbb R_+$; that is, 
$0 < y_1 < y_2 \Longrightarrow f_n(y_1) < f_n(y_2); \tag 9$
we can see this by looking at the well-known identity
$y_2^n - y_1^n = (y_2 - y_1) (y_2^{n - 1} + y_2^{n - 2}y_1 + y_2^{n - 3} y_1^2 + \ldots + y_2y_1^{n - 2} + y_1^n); \tag{10}$ 
clearly every term in the sum $y_2^{n - 1}y_1 + y_2^{n - 2}y_1 + \ldots + y_1^{n -1}$ is positive, hence as is the sum itself; since $y_2 - y_1 > 0$, the right-hand side of (10) is positive, so that
$f_n(y_2) = y_2^n > y_1^n = f_n(y_1); \tag{11}$
now using (9) we see that
$\Upsilon' < \Upsilon \Longrightarrow f_n(\Upsilon') < f_n(\Upsilon) = x, \tag{12}$
$\Upsilon' > \Upsilon \Longrightarrow f_n(\Upsilon') > f_n(\Upsilon) = x, \tag{13}$
which shows that $\Upsilon$ must be unique.
A: Hint: $y = \sup \{ a \in \mathbb{R}^{+} : a^n \le x \}$ gives existence after some work.
Uniqueness is much easier: $y_1^n = x = y_2^n$ implies $0=y_1^n - y_2^n=(y_1 -y_2)z$, where $z >0$ if $x>0$.
