Math 55 - Fall 1996 - Strain - Midterm 1 Solutions

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

• Test Prep
• 2

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

This is the end of the preview. Sign up to access the rest of the document.

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 - ...
View Full Document

• Spring '08
• STRAIN

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern