Let α and β be real numbers such that 0 < α < β. Show that n^a is in O(n^b) but n^b is not in O(n^a)

Question

This is big O notation.

Answer 1

Are you talking ring theory or are you talking about big O notation?

If it is ring theory, what is O?

I haven't seen big O notation since undergrad, sorry.

Answer 2

(n^a)/(n^b) = n^(a-b) which converges to 0 since a-b is < 0. This makes n^a bigO(n^b). Actually it is littleO(n^b), which means not only is it bounded but it actually goes to 0.

The reverse, (n^b)/(n^a) = n^(b-a) which goes to infinity, so is not bounded.