Discrete Mathematics with Graph Theory (3rd Edition) 227

Discrete - Section 8.2 16(a No B y its very definition 225-< is not reflexive(b Yes(A j is a partially ordered set Reflexivity Since f = f we have

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Section 8.2 225 16. (a) No. By its very definition, -< is not reflexive. (b) Yes, (A, j) is a partially ordered set. Reflexivity: Since f = f, we have f j f· Antisymmetric: If f j 9 but f #- g, then f -< g. This means f = O(g) but 9 #- O(f), so 9 -f, It follows that if f j 9 and 9 j f, then f = g. Transitivity: See the solution to Exercise 10. 17. [BB] Reflexivity: For all n ~ 1, If(n)1 ::; If(n)l, so with e = 1, no = 1 in Definition 8.2.1 we see that f is O(f) and hence f ::=:: Symmetry: If f ::=:: g, then f = O(g) and 9 = O(f). Hence, 9 = O(f) and f = O(g), so 9 ::=:: Transitivity: If f ::=:: 9 and 9 ::=:: h, then f = O(g) and 9 = O(h). Therefore, f = O(h) by Exercise lO(a). Since h = O(g) and 9 = O(f), an argument identical to the one just given yields that h = O(f) as well. Thus, f::=:: h. 18. [BB] Since 10g2 n = O(n) by Proposition 8.2.9 and n = O(n), we have nlog2 n = 0(n 2 ) by Proposition 8.2.3. Thus, it remains only to show that n 2 #- O(n 10g2 n). Assume to the contrary that
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.

Ask a homework question - tutors are online