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- Prove: Let a, b, and c be integers. If (a - b) | c, then a | c. bk Suppose a, b, c are integers with (a b) | c. Then c = .). (² X . (a. b) for some integer k, so c = (alel X a, so a c.7. For n 2 1, in how many out of the n! permutations T = (T(1), 7(2),..., 7 (n)) of the numbers {1, 2, ..., n} the value of 7(i) is either i – 1, or i, or i +1 for all 1 < i < n? Example: The permutation (21354) follows the rules while the permutation (21534) does not because 7(3) = 5. Hint: Find the answer for small n by checking all the permutations and then find the recursive formula depending on the possible values for 1(n).is this functions from R to R is a bijection or not f(x)=(x2+2)/(x2+3)
- Let f (f(n) and g(n)) be asymptotically nonnegative functions. Using the basic definition of Θ notation, prove that max(f(n), g(n)) = Θ(f(n) + g(n)),Show that if f(x) and g(x) are functions from the set of real numbers to the set of real numbers, then f(x) is O(g(x)) if and only if g(x) is Ω(f(x)).Prove the following argument is valid: Væ [p(x) → (q(x) V r(x)] Væ (r(x) → q(x)] -3x q(x) 3x p(x) ..
- I. Fermat's test consists of using Fermat's Theorem (an¹=1 (mod n)) with prime n. Just pick any number a E {1, ..., n-1} and check if it satisfies the expression. If not, then it must be a composite number, that is, it is not prime. II. Although Fermat's Theorem is used to check whether a number is prime or not, we can still obtain equality (an¹-1 (mod n)) even when n is not prime. E The numbers for which this situation occurs are called Carmichael numbers or pseudoprimes. III. Both the Miller-Rabin and AKS algorithms (checking a prime number) work for any type of input and are polynomial time. However, only AKS is deterministic. IV. The Miller-Rabin algorithm guarantees when a number is prime, but when it is composite (non-prime) it can only guarantee this characteristic in probabilistic terms. Only the following items are correct: A I, II, III B I, II, IV C I, III, IV D II, III, IV E I, II, III, IVFor each set of cubes in terms of variables (a, b, c, d] obtain the minimized version for boolean fuction f(a, b, c, d) (00X1, 0XX1, 1000, 1100, 1010, 1110} {1XX0, 10XX, 11XX, 00XX} ✓ {00XX, 01XX, 0XXX, 01X1} (0100, 1010, 1XX1, XXXX, 0001} {0001, 0011, 0101, 0111, 1101, 1111, 1001, 1011} ✓ {000X, 010X, 1X00, 1X01} A f(a, b, c, d) = 1 B. f(a, b, c, d) = d C. f(a, b, c, d) = ē D. f(a, b, c, d) = a + a b Ef(a, b, c, d) = a F. f(a, b, c, d) = a dLet Z be the set of all integers. An integer a has f as a factor if a = fj for some integer j. An integer is even if it has 2 as a factor. An integer a is odd if it is not even. Prove by contradiction that an odd number cannot have an even number as a factor.