Help with algorithm analysis...

Big-Oh O(f(n)) means that T(n) <= O(f(n)) (meaning that at this level the growth rate (T(n)) is bounded on top by the growth function f(n))
Big-Omega (f(n)) means that T(n) >= f(n) (meaning that at this level the growth rate is bounded on the bottom by the growth of this function)
Big-Theta (f(n)) means that T(n) = f(n) (meaning that at this level the growth rate is equal to the growth of this function)
Little-Oh (f(n)) means that T(n) < f(n).
OK, here is my question:
Suppose T1(n) = O(f(n)) and T2(n) = O(f(n)). Prove or disprove the
following. { Disprove by a counter-example. }
a) T1(n) - T2(n) = o(f(n))
b) T1(n)/T2(n) = O(1)      // Trick Question - Be careful
I think that a) is true but I just cant prove it...It's just the math that doesnt come to me...
If you guys know or can give me any clue, I would really appreciate it...TIA

Hey thx guys...nikita could you give me an example in how to do it the way u think should be done...I know that this is the way to do it but I cant figure out in my head what functions to use or how to use them. Once again thx for your time.

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    Outer Space Blake 1
    How can I automate creation of Position, based on number of authors on the fly? Let's say I have original table without Position, but I want to create a new table that will have this information.
    Regards

    If you have an existing table whose structure doesn't tell you what position the author is in, what's the algorithm you'd use to determine who was the first author, the second author, etc? If you issue a select query on a table without providing an "order by" clause, Oracle makes no guarantees about the order in which it retrieves rows.
    As an aside, why would you store position number in a varchar2 field? If it's a number, it ought to be stored as a number.
    Justin

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