This problem shows the existence of an nth root.
Let a > 0 and n >= 3 be an integer. Define the set S = {x > 0 : x^n <= a}.
1. Show that S is bounded above and thus that b = sup S exists.
2. Show that b^n = a. To do this, show that it cannot be true that b^n < a or b^n > a.
Let's do it the most elementary way since I presume this is the very beginning of your course.
1. For the existence of the supremum, it would be easier to define , so that it is obvious that is non-empty (it contains 0). Otherwise, this should be proven.
Anyway, to show that is bounded from above, you can notice that, if , then, because , hence , and is an upper bound for .
2. Suppose by contradiction that . We must show that there exists such that . To do this, I'll use the following computation:
if , then , hence
This is because of the usual formula with .
Then, for , .
Let's choose . Then and thus , so that . However, . This contradicts the fact that .
For the other part, this is the same idea, using with an small enough so that . You can use a similar estimate: for , we have
, hence .