Prove there exists a real number x so that $x^n=y$. Let y be a positive real number and let n be an element of natural number. Prove there exists a real number $x$ so that $x^n=y$. 
I'm not sure how to go about solving this problem.
 A: $$\{x\in \mathbb{R}^{\ge 0}:x^n <y \}$$
why the set is non-empty and bounded? what properties have its l.u.b.?
A: We can show the existence of $n^{\text{th}}$ roots using only basic properties of real numbers. 
Take any real number $a \gt 0$.
The following two exercises can be solved using the triangle inequality, the reverse triangle inequality, and the binomial expansion formula:
Exercise 1: If $x \gt 0$ and $x^n \lt a$, there exist a $y$ such that $x \lt y$ and $y^n \lt a$. 
Exercise 2: If $x \gt 0$ and $x^n \gt a$, there exist a $y$ such that $y \lt x$ and $y^n \gt a$. 
Let $\mathbb{R}^{\gt 0} = (0,+\infty)$ be the set of positive real numbers.
Let $I = \{x\in \mathbb{R}^{\gt 0}:x^n \lt a \}$ and $R = \{x\in \mathbb{R}^{\gt 0}:x^n = a\}$ and $F = \{x\in \mathbb{R}^{\gt 0}:x^n \gt a \}$.
Exercise 3: $I$ and $F$ are nonempty sets and $I$ has no maximal element and $F$ has no minimal element.
Exercise 4: $R$ contains at most one element and $\mathbb{R}^{\gt 0} = I \cup R \cup F$.
Exercise 5: $I \, \cap \, R \, \cap F = \emptyset$.
We need to show that $R$ is not the empty set.
Exercise 6: Prove that every number in $F$ is an upper bound of the set $I$.
Exercise 7: Prove that every number in $I$ is a lower bound of the set $F$.
By exercise 3 and exercise 6, we can write $\alpha = \text{lub of } I$.
Now since $\alpha \in \mathbb{R}^{\gt 0}$, by exercise 4 it must belong to $I$, $R$, or $F$. By exercise 3 $I$ has no maximal element and so $\alpha \notin I$. Again by exercise 3, $F$ has no minimal element so if if $\alpha \in F$ we could find a $\tau \in F$ with $\tau \lt \alpha$. But by exercise 6, $\tau$ would be another upper bound of $I$, but one that is smaller than $\alpha$, which is absurd.
We must conclude that $\alpha \in R$.

I set this up so that the exercises are built up on a symmetric foundation - a verbose and leisurely analysis. But the  final argument does not use all the pieces.
Exercise 8: Show that  $n^{\text{th}}$ roots exists using a more direct argument.
