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MATH 240; EXAM # 1, 100 points, October 18, 2005 (R.A.Brualdi)
TOTAL SCORE (11 problems; 100 points possible):
Name: These R. Solutions
TA: Anders Hendrickson (circle time) Mon 12:05 Mon 1:20 Wed 12:05 Wed. 1:
- [8 points] Let P (x) and Q(x) be predicates where the universe of discourse for x is some set U. Let A = {x : P (x) is true} and let B = {x : Q(x) is true} be the truth sets of P (x) and Q(x), respectively. Circle all the predicates below that have truth set equal to A ∩ B?
(a) YES P (x) ∧ ¬Q(x)
(b) YES ¬(P (x) → Q(x))
(c) YES ¬(¬P (x) ∨ Q(x))
(d) ¬(Q(x) ∨ P (x))
(e) ¬(P (x) ∧ Q(x))
- [8 points] Let f : A → B and g : B → C be functions where A, B, C are finite sets. Circle all the CORRECT statements below.
(a) If f is surjective, then |A| ≤ |B|.
(b) If |A| ≤ |B|, then f is injective.
(c) YES If f is surjective and |A| = |B|, then f is injective
(d) YES If f and g are both injective, then g ◦ f is injective.
(e) If g is surjective, then g ◦ f is surjective.
- [8 points] Circle all the CORRECT statements below, or circle (e):
(a) YES d−ne = −bnc.
(b) d− 2. 999999999 e = − 3
(c) dx + ye = dxe + dye
(d) YES dx − 0. 5 e is the closest integer to x, rounding down in the case of ties.
(e) None are correct.
- [6 points] The number of different functions f : A → B from a set A of m elements to a set B of n elements equals:
(a) mn
(b) YES nm
(c) mn
(d) m + n
(e) None of the above
- [10 points] For each of the functions f (n) below, give the simplest function g(n) such that f (n) = Θ(g(n)).
(a). 01 n^3 − 1000 n^2 + 5n + 35: Θ(n^3 )
(b) 4 n
(^5) +3n (^4) log n− 3 n+ 2 n^3 +5n^2 − 6 n+8 : Θ(n
(c) f (n) = bncn: Θ(n^2 )
(d) f (n) = sin n: Θ(1)
(b) Alice wants to send Bob the important message 5. Compute the encrypted message (an integer between 1 and 32) that she sends?
c = 5^3 modulo 33 and this is 26.
(c) April, a different friend of Bob, has sent Bob a message which he received as 2. What message (an integer between 1 and 32) did April send Bob? This is m = 2^7 modulo 33 which equals 29.
- [10 points]
(a) If it exists, use the Euclidean Algorithm to find the multiplicative inverse of 14 modulo 45 (trial and error not acceptable).
We get
45 = 3 · 14 + 3 14 = 4 · 3 + 2 3 = 1 · 2 + 1 2 = 2 · 1 + 0
So GCD is 1 and an inverse exists. Using the equations in the reverse order we get that 1 = 5 · 45 − 16 · 14 Since −16 is 29 modulo 45, we get the inverse to be 29.
(b) Use your answer above — calculator answer not acceptable — to find a solution to 14x ≡ 47 mod 45 where x is between 0 and 44.
x = 29 · 47 or 29 · 2 or 58 which is 13 modulo 45.
- [10 points] Give a proof by contradiction that if a and b are nonnegative numbers, then a + b 2
ab. Suppose not. Then a + b 2
ab
Squaring we get a^2 + 2ab + b^2 4
< ab
This gives a^2 + 2ab + b^2 < 4 ab or a^2 − 2 ab + b^2 < 0 or (a − b)^2 < 0, a contradiction since a square is always at least zero.
