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Discrete Mathematics Math 6A

Discrete Mathematics Math 6A. Homework 2.

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Discrete Mathematics Math 6A

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  1. Discrete MathematicsMath 6A Homework 2

  2. 1.4-16 A discrete mathematics class contains a mathematics major who is a freshman, 21 mathematics majors who are sophomores, 15 computer science majors who are sophomores, 2 mathematics majors who are juniors, 2 computer science majors who are juniors, and 1 computer science major who is a senior. Express each of these statements in terms of quantifiers and then determine its truth value. • We let P(s,c,m) be the statement that student s has class standing c and is majoring in m. The variable s ranges over students in the class, the variable c ranges over the four class standings, and the variable m ranges over all possible majors. • The proposition is smP(s, junior,m). It is true from the given information. • The proposition is scP(s,c,computer science). This is false, since there are some mathemathics majors. • The proposition is xcm(P(s,c,m)  (c≠ junior)  (m ≠mathematics)). This is true, since there is a sophomore majoring in computer science. • The proposition is s(cP(s,c,computer science)  mP(s, sophomore,m)). This is false, since ther is a freshman mathematics major. • The proposition is mcsP(s,c,m). This is false. It cannot be that m is mathematics, since there is no senior mathematics major, and it cannot be that m is computer science, since there is no freshman computer science major. Nor can m be any other major.

  3. 1.4-21 Use predicates, quantifiers, logical connectives, and mathematical operartors to express the statement that every positive integer is the sum of the squares of four integers. • xabcd ((x > 0)  x = a2 + b2 + c2 + d2), where the universe of discourse consists of all integers • 1.4-30 Rewrite each of these statements so that negations appear only within predicates (that is, so that no negation is outside a quantifier or an express involving logical connectives) • ~yxP(x,y): yx~P(x,y) • ~xyP(x,y): xy~P(x,y) • ~y(Q(y) x~R(x,y)): y(~Q(y)  xR(x,y)) • ~y(xR(x,y)  xS(x,y)): y(x~R(x,y)  x~S(x,y)) • ~y(xzT(x,y,z)  xzU(x,y,z)): y(xz~T(x,y,z)  xz~U(x,y,z))

  4. 1.4-33 Rewrite each of these statements so that negations appear only within predicates (that is, so that no negation is outside a quantifier or an expresssion involving logical connectives.) • ~xyP(x,y): xy~P(x,y) • ~yxP(x,y): yx~P(x,y) • ~yx(P(x,y)  Q(x,y)): yx~(P(x,y)  Q(x,y)) = yx(~P(x,y)  ~Q(x,y)) • ~(xy~P(x,y)  xyQ(x,y)): xyP(x,y)  (xy~Q(x,y)) • ~x(yzP(x,y,z)  zyP(x,y,z)): • x~(yzP(x,y,z)  zyP(x,y,z)) • = x(~yzP(x,y,z)  ~zyP(x,y,z)) • = x(yz~P(x,y,z)  zy~P(x,y,z))

  5. 1.6-2 Use set guilder notation to give a description of each of these sets. • {0,3,6,9,12}: {3n | n = 0,1,2,3,4} or {x | x is a multiple of 3  0  x  12} • {-3, -2, -1, 0, 1, 2, 3}: {x | -3  x  3}, where we are assuming that the universe of discourse is the set of integers. • {m, n, o, p}: {x | x is a letter of the word monopoly other than l or y} • 1.6-7 Determine whether each of these statements is true of false. • 0  : false, since the empty set has no elements. •   {0}: false: The set on the right has only one element, namely the number 0, not the empty set • {0}  : false: the empty set has no proper subsets. •   {0}: true: every element of the set on the left is an element of the set on the right; and the set on the right contains an element, namely 0, that is not the set on the left • {0}  {0}: false: the set on the right has only one element, namely the number 0, not the set containing the number 0 • {0}  {0}: false: for one set to be a proper subset of another, the two sets cannot be equal • {}  {}: true: every set is a subset of itself

  6. 1.6-12 Find two sets A and B such that A B and A  B Since the empty set is a subset of every set, we just need to take a set B that contains as an element. Thus, we can let A = and B={} as the simplest example 1.6-21 what is the Cartesian product AB C, where A is the set of all airlines and B and C are both the set of all cities in the United States? This is the set of triples (a,b,c), where a is an airline and b and c are cities. For example, (TWA, Rochester Hills Michigan, Middletown New Jersey) is an elements of this Cartesian product. A useful subbset of this set is the set of triples (a, b, c) for which a flies between b and c. For example, (Northewest, Detroit, New York) is in this subset, but the triple mentioned earliers is not.

  7. 1.7-4 Let A={a,b,c,d,e} and B={a,b,c,d,e,f,g,h}. Find • a) A  B: {a,b,c,d,e,f,g,h} • b) A  B: {a,b,c,d,e} • c) A – B:  • d) B- A: {f,g,h} • 1.7-12 Let A and B be sets. Show that • {A  B}  A: if x is in AB, then perforce it is in A (by definition of intersection) • A  (A  B): if x is in A, then perforce it is in A  B (by definition of union) • A – B  A: if x is in A-B, then perforce it is in A (by definition of difference) • A  (B - A) = : if xA then xB-A. Therefore, there can be no elements in A  (B - A), so A  (B - A) =  • A  (B – A ) = (A  B): The left-hand side consists precisely of those thinks that are either elements of A or else elements of B but no A, in other words, thinks that are elements of either A or B (or both). This is precisely the definition of the right-hand side.

  8. 1.7-17 Let A, B, and C be sets. Show that • A  (B  C) = (A  B)  C: {x | xA  x(B  C)} = {x | xA  (xB  x C)} • = {x | (xA  xB)  x C)} = {x | (xA  B)  x C)} = (A  B)  C • b) A  (B  C) = (A  B)  C: {x | xA  x(B  C)} = {x | xA  (xB  x C)} • = {x | (xA  xB)  x C)} = {x | (xA  B)  x C)} = (A  B)  C • c) A  (B  C) = (A  B)  (A  C): • {x | xA  x(B  C)} = {x | xA  (xB  x C)} • = {x | (xA  xB)  (xA  xC))} = {x | (xA  B)  (xA  C)} • = (A  B)  (A  C) • There are many ways to prove these identities. One way is to reduce them to logical identities. Alternatively, we could argue in each case that the left-hand side is a subset of the right-hand side and vice versa. Another method would be to construct membership tables. • 1.7-20 Draw the Venn diagrams for each of these combinations of the sets A, B, and C • a) A  (B  C) b) A'  B'  C' c) (A-B)  (A-C)  (B-C)

  9. 1.7-40 Suppose that the universal set is U={1,2,3,4,5,6,7.8.9.10}. Express each of these sets with bit strings where the ith bit in the string is 1 if i is in the set and 0 otherwise. • {3,4,5}: 00 1110 0000 • {1,3,6,10}: 10 1001 0001 • {2,3,4,7,8,9}: 01 1100 1110 • 1.8-1 why is f not a function from R to R if • f(x) = 1/x?: f(0) is not defined • f(x) = x?: Things like -3 are not defined • f(x) = (x2 + 1) ?: The rule for f is ambiguous. We must have f(x) defined uniquely, but here there are two values associated with every x, the positive square root and the negative square root of (x2 + 1)

  10. 1.8-5 Find the domain and range of these functions. • the function that assigns to each bit string the difference between the number of ones and the number of zeros: domain is the set of all bit strings and the range is Z. For example, if the input is 1010000, then 3 or -3 • the function that assigns to each bit string twice the number of zeros in that string: domain is the set of all bit strings and the range is the set of even natural numbers. (the value of function can be 0,2,4,...) • the function that assigns the number of bits left over when a bit string is split into bytes (which are blocks of 8 bits): domain is the set of all bit strings and the range is {0,1,2,3,4,5,6,7} • the function that assigns to each positive integer the largest perfect square not exceeding this integer: the domain is the set of positive integers and the range is {1,4,9,16,... • 1.8-10 Determine whether each of these functions from {a,b,c,d} to itself is one-to-one • f(a)=b, f(b)=a, f(c)=c, f(d)=d: one-to-one • f(a)=b, f(b)=b, f(c)=d, f(d)=c: not one-to-one, since b is the image of both a and b • f(a)=d, f(b)=b, f(c)=c, f(d)=d: not one-to-one, since d is the image of both a and d

  11. 1.8-11 Which functions in Exercise 10 are onto? a) • 1.8-14 Determine whether f:Z  Z  Z is onto if • f(m,n) = 2m – n: onto • f(m,n) = m2 – n2: not onto, for example, 2 is not in the range. • If m2 – n2 =(m+n)(m-n), then this expression is divisible by 4 and hence cannot equal 2 • c) f(m,n) = m + n + 1: onto • d) f(m,n) = |m| – |n|: onto • e) f(m,n) = m2 – 4: not onto • onto: every element bB there is an element aA with f(a)=b • 1.8-19 Determine whether each of these functions is a bijection from R to R • f(x) = 2x + 1: bijection • f(x) = x2 + 1: not a bijection, range is [1, infinite) • f(x) = x3: bijection: • f(x) = (x2 + 1)/(x2+2): not a bijection, not injection since x and –x have the same image for all real numbers x • If a function is not a bijection, we cannot define an inverse function

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