Math 55 - Fall 1996 - Strain - Midterm 1 Solutions

Math 55 - Fall 1996 - Strain - Midterm 1 Solutions - TUE...

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Unformatted text preview: 09/12/2000 TUE 15:04 FAX 6434330 MOFFITT LIBRARY 001 Math 55: First Midterm r—. Solutions Problem 1: Prove that (pA q) -—+ p is a tautology. Solution: (P A q)-+p - -i(p A g) V p Definition of implication (-1]: V —-g) V p De Morgan -up V -:q V p Associativity -|p V p V —Iq Commutativity {-110 V p) V -q Associativity again T V --q Excluded middle T Domination 3333333 ' Problem 2: Let f, g and h be defined by: f:R—>R, f(z)=z3 g:Z-—>Z, g'(n)=n3 h:R-+ Z x Z, 11(3): ([ijsl) For each of the functions f, g and h, state whether the function is 1-1, whether it is onto, and whether it is invertible. Solution: Since every real number :1: has a unique cube root {/5, f is 1-] and onto and therefore invertible. Since not every integer is a cube (for example, 2 = 1:3 implies n is between 1 and 2 so 7: ¢ Z), 9 is not onto. However, 113 = 17:3 implies m = n, so 9 is 1-1. Since 9 is not onto, it is not invertible. ' - _ Since 1 = 1/2 and a: = 1/4 are both mapped to 12(3) = (0,1), 11 is not 1-1. Since (0,100) ;£ ([3], for any a: E R, h. is not onto. Since h is not 1-1 and not onto it is certainly not invertible. Problem 3: Show that Solution: 7! Zk2=1+4+9+-~+n25n2+n2+---+n2=n3=0(n3). k=l 09/12/2000 TUE 15:05 FAX 6434330 MOFFITT LIBRARY 002 Problem 4: Construct pseudocode for an algorithm which accepts input consisting of two finite sets-A = {a1,ag,...,an} and B = {b1,b2,...,bm} and a; function f : A —} B', and returns output T if f is onto and F if f is not onto. ' Solution: Boolean function onto( set A, set B, function f:A—}B): for b E B ' hi1: := F for a 6 A if(b=f(a)) hit :- 'r if(hit = F) return(onto := F) returnConto := T) Problem 5: Suppose a E b and c E (1 mod 17. Show that ac ‘5 bd mod 17. Solution: By definition of “mod,” there are integers p and 4] such that a = 13+ 1711 and c = d + 17g. Then ac — bd: (b + 17p)(d+ 17g) 4- bd L- (pq +pd+ qb)17 E 0 mod 17. By definition, ac E bd mod 17. Problem 6: Use the Euclidean algorithm to compute gcd{277, 123). Solution: :1: = 277 y = 123 r=277mod123=31 1:123 y=31 r2123 mod 31 230 32:31 31:30 r=31mod30=1 2:30 y=1 r=30mod1=0 z=1=gcd(277,123) y=0 Problem 7: Suppose a- is an integer with 0 5 a: s 1000 and zmod7=3, :mod11=5, zmod13=7. (a) Is a: uniquely determined by this information? Why or why-not? Solution: Yes, by the Chinese Remainder Theorem, because 7, 11 and 13 are pairwise relativaly prime and 7 - 11 - 13 = 1001. (b) Calcuiate 3:2 mod 7 and 1:3 mod 11. Solution: 2:2 mod 7: (.rmod 7)2=9E2m0d 7 2:3 mod 11 = (a: mod 11)3 = 125 E 4 mod 11 - ...
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This test prep was uploaded on 04/01/2008 for the course MATH 55 taught by Professor Strain during the Spring '08 term at Berkeley.

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