Homework 12; Due Monday, 4/9/2018

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Cuthbert Gardner
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1 Homework 2; Due Monday, 4/9/28 Answer-Only Questions Credit based solely on the answer Question For each of the following relations, determine whether each one is symmetric, transitive, or refleive If the relation is an equivalence relation, give the equivalence classes a) Define on the set P 3 by p q if p) q) for all R Solution Not Refleive, Symmetric, Not Transitive b) For the set N, define a b if a b and b a Solution Refleive, Symmetric, Transitive The equivalence class of n N is given by [n] = {n} c) For the set Z, define a b if a b and b a Solution Refleive, Symmetric, Transitive The equivalence class of n Z is given by [n] = {n, n} d) On the set P 5, define p q if p3) q3) Solution Refleive, Not Symmetric, Transitive e) For a set A, define a relation on its power set PA) by B C if C B Solution Refleive, Not Symmetric, Transitive f) For a finite set A, define a relation on its power set PA) by B C if B = C Solution Refleive, Symmetric, Transitive The equivalence class of B PA) is given by [B] = {C PA) : C = B } g) For the set of Michigan State students, say that two students are related if they have taken a class together Solution Refleive, Symmetric, Not Transitive Question 2 a) Given the assignment fn) = 4n+2 4, find an appropriate choice of domain so that the codomain of f is N What is the range of this function? Recall the range of a function f : X Y is the set {y Y X, f) = y}) MSU Due: 4/9/28

2 Solution Let the domain of f be D = { 2 } Then f 2) = N The range of f is the set {} b) Repeat part a) for gn) = n 4 + n 8 Solution Let the domain of f be D = {8} Then f8) = 3 N The range of f is the set {3} c) Give an eample of a rule defining a function with domain A = N {} and codomain B = {, 2,, 598} Solution Let f : A {, 2,, 598} be defined via the rule f) = for all A Full Justification Questions Provide complete justifications for your responses Question 3 Define a relation on L = {f P f) = a for some a R} by f g if and only if f) d = g) d a) Prove that is an equivalence relation on L Solution Let f, g, h L be arbitrary Observe that a) f) d = f) d = f f This shows refleivity of b) f) d = g) d = g) d = f) d Thus f g = g f This shows symmetry of c) f) d = g) d and g) d = h) d = f) d = h) d Thus f g and g h) = f h This shows transitivity of We conclude that as defined above is indeed an equivalence relation on L b) Find and give a complete list of all the distinct equivalence classes of Prove that your list is accurate and complete Recall, this means you must show that your equivalence classes are pairwise disjoint and that their union is L) Solution Let f, g L be arbitrary Then f) = a and g) = b from some a, b R Let us assume that f g Then f) d = g) d = a = b 2 2 = b = ±a From this, we see that if f g, then g = ±f We then claim that the equivalence class of f L, where f) = a for some a, is MSU 2 Due: 4/9/28

3 [f] = {g L : g = ±f} We now have to prove that the collection {[a] : a } is a partition of L To that end, first note that a [a] for any a by refleivity of Therefore, none of the equivalence classes is empty Second, we show that the equivalence classes, as defined above, are pairwise disjoint So, let a, b with a b Assume by way of contradiction that there eists c such that c [a] [b] Then a d = c d = b d, from which we get that a = c = b, which in turn implies that a = b since a, b This is a contradiction to a b Finally, we have to show that [a] = L a To that end, observe that if f a [a], then f) = b for some b R, and so f L This shows that a [a] L Now, let f L be arbitrary Then f) = c for some c R, and c [ c ] a [a] This shows that L a [a] Therefore, a [a] = L as needed Question 4 Prove that for all n N and for every polynomial, p, of degree n, d n+ p) = dn+ for all R Hint: Use induction on the degree of p; you may assume without proof) the power rule for differentiation Solution Let the conditional statement be, for all n N, Q n) : deg p) = n = dn+ p ) = R dn+ The base case is for n = Note that deg p) = means that p ) = a + a for some a, a R We have d 2 d2 p ) = d2 d a 2 + a ) = d d a ) = for all R This shows that Q ) is true Assume now that Q k) is true for a generic and fied k N; that is deg p) = k = dk+ p ) = R dk+ MSU 3 Due: 4/9/28

4 For the inductive step, we will show that Q k + ) is true; that is Observe that deg p) = k + = dk+2 p ) = R dk+2 for some b,, b k+ R Now, deg p) = k + = p ) = b + b + + b k k + b k+ k+ d k+2 dk+ p ) = dk+2 d k+ ) d d p ) = dk+ d k+ + b + + kb k k + k + ) b k+ k) = for all R The last equality holds since b + + kb k k + k + ) b k+ k is a polynomial of degree k, whence its k + ) st derivative is zero by the inductive assumption Finally, since k N was arbitrary, we conclude, by mathematical induction, that Q n) is true for all n N Question 5 Consider the set of functions Prove or disprove the following two statements: A = {f : R R f) = sinn) for some n Z} a) For all f A, there eists an a N {} such that f ) + af) = for all R Solution The statement above is true Proof Let f A be arbitrary then f) = sinn) for some n Z and all R Observe that f ) = d d n cosn) = n2 sinn) = n 2 f) Let a = n 2 N {} Then for all R, as needed f ) + af) = n 2 f) + n 2 f) = b) There eists a Z such that for all f A, f ) + af) = for all R MSU 4 Due: 4/9/28

5 Solution The statement above is false Proof Assume by way of contradiciton that a Z, f A, R, f ) + af) = Let f, f 2 A be given by f ) = sin) and f 2 ) = sin2) Then and f ) + af ) = R sin) + a sin) = R a = f 2 ) + af 2 ) = R 4 sin2) + a sin2) = R a = 4 By assuming the quantified statement is true, we arrived at = a = 4, a contradiction This means that the original statement is false Question 6 Consider the function F : P 2 P 2 given by Show that F is a bijection F p)) = 2 p ) Solution We will first show that F, as defined above, is injective To that end, let p, q P 2 be arbitrary Then p) = a 2 + b + c and q) = d 2 + e + f for some real numbers a, b, c, d, e, f Assume that F p)) = F q)) for all R; we will show that p = q follows Observe that as needed ) ) F p)) = F q)) = 2 p = 2 q = 2 a + b ) + c = 2 d + e ) + f 2 2 = a b c 2 = d e f 2 = a = d, b = e, and c = f = p = q, Net, we show that F is surjective To that end, let q P 2 be arbitrary Then q) = d 2 + e + f for some real numbers d, e, f We need to find p P 2 such that F p) = q Let p) = f 2 e d Then p P 2, and F p)) = 2 f 2 e d ) = f + e + d 2 = q), MSU 5 Due: 4/9/28

6 as needed Finally, since F is both injective and surjective, it is indeed bijective 2 Show that F = F Solution We need to show that F is its own inverse To do so, it suffices to show that F F p)) = p for all p P 2 So, let p P 2 be arbitrary Then p) = a 2 + b + c for some real numbers a, b, c We calculate as needed )) F F p))) = F 2 p = F 2 a + b )) + c 2 = F c 2 b a ) = 2 c b ) 2 a = a 2 + b + c = p), Question 7 For each n N, n >, define the interval, I n, as [ n, + ] n Define X = is equal to X I n Find a simpler epression for X and provide a proof that this simpler epression n=2 Solution We claim that X =, 2] 3 To prove this, first we will show that X, 3 2] So, let a X be arbitrary Then a [, + n n] for some n 2 Thus, n a + n = < n a + n + 2 = a, 3 ] 2 Since a was arbitrary, we conclude that X, 3 2] Net, we show that, 3 2] X To that end, let b, 3 2] be arbitrary We have the following two cases: a) If 2 b 3 2, then b X since b I MSU 6 Due: 4/9/28

7 b) If < b < 2, then by the Archimedian Property, there eists an m N such that < m < b Moreover, b < < + m This implies that b I m, whence b X Since b was arbitrary, we conclude that, 3 2] X, and consequently, X =, 3 2] 2 Define Y = is equal to Y I n Find a simpler epression for Y and provide a proof that this simpler epression n=2 Solution We claim that Y = [, ] To prove this, first we will show that Y [, ] So, let a Y 2 2 be arbitrary Then a [, + ] n n for all n 2 It is clear that a ; we claim that a 2 To prove this, assume by way of contradicition that a > Then a >, and by the Archimedian Property, there eists m N such m > Then a > and so a > +, which means that a m m a I m, a contradiciton Therefore, a, and a [, ], as needed 2 Net, we show that [, ] Y To that end, let b [, ] be arbitrary Then 2 2 n 2 b < + n for any n 2, which implies that b Y, as needed Finally, we conclude that Y = [ 2, ] MSU 7 Due: 4/9/28

MATH 220 (all sections) Homework #12 not to be turned in posted Friday, November 24, 2017

MATH 220 (all sections) Homework #12 not to be turned in posted Friday, November 24, 2017 Definition: A set A is finite if there exists a nonnegative integer c such that there exists a bijection from A