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 11, 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 11. 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 11. Since
(0,100) ;£ ([3], for any a: E R, h. is not onto. Since h is not 11 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 setsA = {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 whynot? 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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 Spring '08
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