is f(n)=O(g(n)), then g(n)= O(f(n)is true or false, give explanation?
if false give counterexample?

is f(n)=O(g(n)), then g(n)= O(f(n)is true or false, give explanation?

False:  g(n) = f(n)/O .  That is, = f(n) divided by O.

A quick two-fifty for a researcher.

The "big-O" notation is used as a convenient description of the
"growth" of a sequence or function, ie. the "behavior" as n tends to
infinity (in the present context).

Specifically, f(n) = O(g(n)) is an abbreviation for the claim that:

  as n --> +oo, the expression f(n)/g(n) is bounded by a fixed
  constant (independent of n, for all sufficiently large n)

myoarin-ga is correct in saying that the assertion:

  f(n) = O(g(n)) implies g(n) = O(f(n))

is false, but the "division" by O is either facetious or a
misunderstanding of the notation.  Left as an exercise, then, is to
find a counterexample, ie. sequences or functions such that f(n)/g(n)
remains bounded, but g(n)/f(n) doesn't.

In this respect the "big-O" notation is less precise in describing
behavior of a sequence or function than the "little-o" notation.  See
here for definition and links to the Wikipedia article and to a
contrasting definition for "little-o" notation:

[big-O notation -- NIST]
http://www.nist.gov/dads/HTML/bigOnotation.html

regards, mathtalk-ga

Yes Sir, Mathtalk,

I'll forget my highschool math and leave the answering up to you.
Thanks for the clarification of what the big O means.
Humbled, myoarin

It's false. For example log(n) is of O(n), but n is not of O(log(n)).