This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 55: Sample Final Exam, 9 December 2009 3: (a) B is countable: in fact, a 11 correspondence between B and the natural numbers is provided by the function f ( b b 1 b 2 ...b n ) = b + 2 1 b 1 + ... + 2 n b n . In other words, B is the set of binary representations of the natural numbers. (b) F is uncountable, by the diagonal process. Suppose F were countable or finite, and arrange the elements of F into a sequence f ,f 1 ,f 2 ,... . Define g : N { , 1 } by g ( n ) = 1 f n ( n ). Then g ( n ) negationslash = f n ( n ) for any n , so g negationslash F . This contradicts the assumption that we have listed all the elements of F , so F cannot be countable or finite. (c) Since a program is a finite bit string, the number of C programs is countable by (a). But functions are uncountable by (b), so C programs cannot evaluate all of them. 4: (a) Subtracting the number of edges of C n from the number of edges of the complete graph K n gives  E  = n ( n 1) / 2 n = n ( n 3) / 2. (b) Since n ( n 3) / 2 negationslash = n for n negationslash = 5, C n and C n have different numbers of edges and hence cannot be isomorphic. For n = 5, they are isomorphic: the map 1 1, 2 4, 3 2, 4 5, 5 3 preserves edges. (c) By symmetry and the handshaking theorem, all vertices of C n have the same degree deg( v ) = 2 n ( n 3) / 2 n = ( n 3), which is even iff n is odd. By Eulers theorem, C n has an Euler circuit iff n is odd. 5: (a) A partial order is a relation which is reflexive, antisymmetric and transitive. R is a partial order because for any a , b and c Z n , a  a , a  b b  a a = b and a  b b  c a  c . (b) The matrix M has M ij = 1 whenever i  j , so summing the entries in column j counts exactly d ( j ) divisors i of j Z n ....
View Full
Document
 Fall '08
 STRAIN
 Math, Natural Numbers

Click to edit the document details