Diagnostic quiz for M303: Further Pure Mathematics

Transcription

1 Diagnostic quiz for M303: Further Pure Mathematics Am I ready to start M303? M303, Further pure mathematics is a fascinating third level module, that builds on topics introduced in our second level module M208, Pure mathematics. It covers several topics that are central to much modern pure mathematics including rings and fields, and metric spaces. The better prepared you are for it the more time you will have to enjoy the mathematics, and the greater your chance of success. This diagnostic quiz is designed to help you decide whether you are ready to start studying M303, Further pure mathematics. The topics that are included in this quiz are those that we expect you to be reasonably familiar with before you start the module. We suggest that you try this quiz first without looking at any books, and only look at a book when you are stuck. You will not necessarily remember everything, and may well need to look up some things. This is perfectly all right. If you have studied the second level module M208, Pure mathematics (or its predecessor M203) then you will find all the necessary background material in this module s Handbook. It may be that even after refreshing your memory of a particular topic, you are still unable to answer a particular question. In this case it is best not to dwell too long on the question and instead to look at the solutions at the end of the quiz. You shouldn t worry if you can t do all the questions, as long as you are able to do the majority of them and are able to understand and follow the solutions of the rest. Further advice is given at the end of the quiz. About the quiz The questions are grouped into several sections, each on a specific topic. Some questions can be answered in more than one way, so do not immediately think that your solution is incorrect simply because it is different from the one that we supply! If you can complete the quiz in a reasonable time, with only the occasional need to look at other material, then you should be well prepared for M303.

2 1 Am I ready? quiz Inequalities 1 Prove the following inequalities: (a) a 2 +b 2 a+b, for any non-negative real numbers a and b; (b) a+b a + b, for any real numbers a and b. If you find these difficult then you may like to revise M208 Unit AA1. Groups 2 Consider the group G with the following Cayley table: (a) Write down the order of the group G. (b) Write down the order of each element of G. (c) Determine whether or not the group G is cyclic. (d) Find two proper subgroups of G. If you find these difficult then you may like to revise M208 Unit GTA1 and Unit GTA2. 3 Lagrange s Theorem states Let G be a finite group and let H be any subgroup of G. Then the order of H divides the order of G. A binary operation is defined on the set S = {e,a,b,c,d} by the following table: e a b c d e e a b c d a a e c d b b b d e a c c c b d e a d d c a b e Use Lagrange s theorem to prove that S does not form a group under this binary operation. If you find these difficult then you may like to revise M208 Unit GTA1 and Unit GTA4. page 2 of 13

3 Permutations 4 Two permutations f and g in S 6 are given by the following two-line symbols: ( ) ( ) f = and g = (a) Write each of f, g and the composite g f as a product of disjoint cycles. (b) Write down the order of each of the permutations f, g and g f in S 6. (c) Write down the inverse of each of the permutations f, g and g f as a product of disjoint cycles. 5 Two permutations f and g in S 5 are: f = ( 1 3 )( ) and g = ( )( 2 3 ) Write the conjugate g f g 1 as a product of disjoint cycles. If you find these difficult then you may like to revise M208 Unit GTA3. Linear algebra 6 Let 7 Let A = ( ) 1 2 3, B = and C = ( (a) Evaluate the products AB and BC. (b) Write down the associative law for three matrices A, B and C. ( ) ( ) i A =, B = and C = 0 1 i i ). ( ) 4 2i i 2. 2i 6 3 i (a) Evaluate the determinant of each of A, B and C. (b) Write down the inverse, if there is one, of each of A, B and C. If you find these difficult then you may like to revise M208 Unit I3 and Unit LA2. 8 The standard basis for R 2 is {e 1,e 2 } where e 1 = (1,0) and e 2 = (0,1). In R 2, let r be a rotation through π/2 anticlockwise about the origin and let s be reflection in the line y = x. (a) Write down the standard matrix of the linear transformations r and s. (a) Verify that sr = r 3 s. If you find these difficult then you may like to revise M208 Unit LA4. page 3 of 13

4 Sets 9 For each pair of sets A and B below, find A B, A B, A B and B A. (a) A = {0,2,4,6}, B = {4,5,6,7} (b) A = [ 2,3), B = (1,7] 10 Which of the following are (i) open intervals, (ii) closed intervals, or (iii) neither open nor closed intervals? (a) [0,2] (b) (0,2] (c) R (d) (e) (0,2) [1,3] If you find these difficult then you may like to revise M208 Unit I1. 11 Which of the following sets are (i) bounded? (ii) unbounded? (iii) neither bounded nor unbounded? (a) [0,2] (b) R (c) (0,2) [1,3] (d) R [3, ) (e) A = { (x,y) R 2 : xy 1 } If you find these difficult then you may like to revise M208 Unit AA1. Relations An equivalence relation on a set X is a relation on X which satisfies the following three properties: REFLEXIVE For all x,y X, x x; SYMMETRIC For all x,y X, if x y then y x; TRANSITIVE For all x,y,z X, if x y and y z then x z. The equivalence class of x X is the set {y X:x y}. 12 Determine whether each of the following relations on a set A is an equivalence relation. If it is, find its equivalence classes. (a) A = Z, x y if x y is odd. (b) A = Z, x y if x < y. (c) A = {a/b:a,b Z {0}} a/b c/d if and only if ad = bc. 13 (a) Let A = Z and a b if and only if a b = 6n, for some n Z. Show that this is an equivalence relation and describe the equivalence class containing 5. (b) Let A = C and w z if and only if Re(w) =Re(z). Show that this is an equivalence relation and describe the equivalence class containing the element 4 2i. If you find these difficult then you may like to revise M208 Unit I3. page 4 of 13

5 Sequences 14 For each of the following sequences {a n }, determine whether it (i) converges, (ii) diverges, or (iii) neither converges nor diverges. { } { } 2n+1 n! 99 { (a) (b) (c) } 2 1/n 3n n 74 You may use the fact that the following sequences are null sequences, that is, sequences that tend to 0 as n : { } { } 1 c n n p,for p > 0;,for any real c. n! 15 The following statements are either true or false. Prove each statement that is true. For each statement that is false, give a specific example to show that it is false. (a) a n l a 2 n l2 (b) a 2 n l 2, l > 0 a n l If you find these difficult then you may like to revise M208 Unit AA2. Functions 16 For the following functions f and g, determine f f, f g and g f. f:r R x x 2 +x+1 and g:r R x x 3 x NOTE: (f g)(x) means f(g(x)). 17 For each of the following functions f, determine whether f has an inverse function f 1. If f 1 exists, find it. (a) f:r R x x 3 3 (b) f:r 2 R 2 (x,y) (x+1,y 2) (c) f:r 2 R 2 (x,y) (x 2,y 2 ) If you find these difficult then you may like to revise M208 Unit I2. Proofs 18 Use proof by contradiction to prove that, if n = a+2b, where a and b are positive real numbers, then either a 1 2 n or b 1 4 n. 19 Assume that n is a positive integer. Write down (a) the converse and (b) the contrapositive of the following implication: If 2 n 1 is prime, then n is prime. NOTE: By the contrapositive of the statement A B we mean the statement not B not A. 20 Deduce the implication in 19 by proving the contrapositive from 19(b). You should use the fact that x n 1 = (x 1)(x n 1 +x n 2 + +x+1). If you find these difficult then you may like to revise M208 Unit I2. page 5 of 13

6 2 Diagnostic Quiz Answers 1 For convenience, we shall use the symbol to denote if and only if. (a) Here a2 +b 2 a+b a 2 +b 2 (a+b) 2, for non-negative a and b = a 2 +2ab+b 2 0 2ab, on rearranging. Since a and b are both non-negative, it follows that 2ab 0. The above chain of equivalent statements shows that we must then also have a 2 +b 2 a+b. (b) Here a+b a + b a+b 2 ( a + b ) 2 (a+b) 2 ( a + b ) 2 a 2 +2ab+b 2 a 2 +2 a b +b 2 2ab 2 a b. For any real numbers a and b, we have that 2ab 2 a b. The above chain of equivalent statements shows that we must then also have a+b a + b. 2 (a) There are 6 elements in the set for G, so the order of G is 6. (b) The element 1 has order 1. The element 8 has order 2, the elements 4 and 7 have order 3, and the elements 2 and 5 have order 6. (c) The group is cyclic because either of the elements of order 6 will generate it. (d) The elements 1 and 8 form a subgroup of order 2, and the elements 1, 4 and 7 form a subgroup of order 3. There are no others. 3 If (S, ) were a group, then the table shows that {e,a} would form a subgroup of order 2. Since 2 does not divide 5, this would contradict Lagrange s theorem and so (S, ) is not a group. Note: {e,a} could be replaced by {e,b}, {e,c} or {e,d}. 4 (a) f = (1235)(46), g = (1326)(4)(5) and g f = (164)(2)(35). (b) The order of a permutation is the least common multiple of the lengths of its cycles so f has order 4, g has order 4 and g f has order 6. (c) f has inverse (1532)(46), g has inverse (1623)(4)(5) and g f has inverse (146)(2)(35). 5 (a) g f g 1 = g(13)(254)g 1 = (g(1)g(3))(g(2)g(5)g(4)) = (42)(315)= (153)(24). page 6 of 13

7 6 The (i,j)-entry in a matrix product XY is the dot product of the ith row of X with the jth column of Y. So ( ) AB = 2 1 (a) ( ) 1 0 = and 0 1 BC = 2 1 ) 2 4 ( = ; (b) The associative law for three matrices is (AB)C = A(BC). 7 (a) A = (1 1) 0 = 1, B = (0 i) ( i i) = (i 2 ) = ( 1) = 1 and C = (4 2i) 3 i) ((2i 6) (i 2) = 12 10i+2i 2 (2i 2 10i+12) = 0. (b) A 1 = 1 ( ) ( ) =, B 1 = 1 ( ) ( ) i i i i = 1 i 0 i 0 C has no inverse. 8 (a) The linear transformation r maps e 1 to e 2 = 0.e 1 +1.e 2 and maps ( e 2 ) to e 1 = 1.e 1 +0.e 2. So the standard matrix of r is The linear transformation s maps e 1 to e 2 = 0.e 1 +1.e( 2 and maps ) 0 1 e 2 to e 1 = 1.e 1 +0.e 2. So the standard matrix of s is. 1 0 Alternatively, you use the fact that the matrix of a rotation anticlockwise ( through ) an angle θ about the origin is cosθ sinθ, and that the matrix of a reflection in the sinθ cosθ straight line ( through the origin ) that makes an angle φ with the cos2φ sin2φ x-axis is. sin2φ cos2φ ( )( ) ( ) (b) The matrix of sr is the matrix = The ( matrix ) for r 3 is the cube of the matrix of r, which is 0 1 so the matrix of r s is the matrix ( )( ) ( ) = We see that the matrix of sr is the same as the matrix of r 3 s, verifying that sr = r 3 s. page 7 of 13

8 9 (a) A B = {0,2,4,5,6,7}, A B = {4,6}, A B = {0,2} and B A = {5,7}. (b) A B = [ 2,7], A B = (1,3), A B = [ 2,1] and B A = [3,7]. 10 (a) [0,2] is a closed interval. (b) (0,2] is an interval, but is neither open nor closed. (c) R is an interval. By convention, it is both open and closed. (d) is the empty set. It is not an interval. (e) (0,2) [1,3] is the interval [1,2), which is neither open nor closed. 11 Note that no set can be neither bounded nor unbounded! (a) [0,2] is a bounded set. (b) R is an unbounded set. (c) (0,2) [1,3] = [1,2) is a bounded set. (d) R [3, ) = (,3) is an unbounded set. (e) A = { (x,y) R 2 : xy 1 } is the region containing the origin bounded by the rectangular hyperbola with equation xy = 1 and its reflection in the x-axis (viz. the curve with equation xy = 1, together with the hyperbolas themselves). It is an unbounded set. y x 1 A 2 12 To show that something is not an equivalence relation, you only need to show that one of the conditions fails. In these solutions, we have checked them all for completeness, but you don t need to! (a) This relation is NOT reflexive. For example, 2 2 since 2 2 = 0, which is not odd. The relation is symmetric. For, if x y, then x y is odd, so that y x = (x y) is also odd and therefore y x. The relation is NOT transitive. For example, 5 2 since 5 2 = 3, which is odd, and 2 1 since 2 1 = 1, which is odd however 5 1 since 5 1 = 4, which is not odd. Thus the relation is NOT an equivalence relation. (b) This relation is NOT reflexive. For example, 2 2 since 2 is not less than 2. The relation is NOT symmetric. For example, 2 3, but 3 2 since 3 is not less than 2. The relation is transitive. If x y and y z then x < y < z so x z. Thus the relation is NOT an equivalence relation. page 8 of 13

9 (c) This relation is reflexive because a/b a/b since ab = ba. The relation is symmetric. If a/b c/d then ad = bc, but then cb = da so c/d a/b. The relation is transitive. If a/b c/d and c/d f/g then ad = bc and cg = df. This implies that a = bc/d and g = df/c, so ag = bf and a/b f/g. Thus the relation is an equivalence relation. The equivalence classes correspond to Q, the set of rational numbers. 13 (a) This relation is reflexive because a a since a a = 0 = 6 0. The relation is symmetric. If a b then a b = 6n, but then b a = 6n = 6 ( n) so b a. The relation is transitive. If a b and b c then a b = 6n 1 and b c = 6n 2. This implies that a b+(b c) = 6n 1 +6n 2 = 6(n 1 +n 2 ), but a b+(b c) = a c, so a c = 6n 3 and a c. Thus the relation is an equivalence relation. The equivalence class containing 5 comprises all a Z with a 5 = 6n, for some n Z. That is all a Z satisfying a = 6n+5, or {..., 7, 1,5,11,17,...}. (b) This relation is reflexive because w w since Re(w) =Re(w). The relation is symmetric. If w z then Re(w) =Re(z), so z w. The relation is transitive. If w z and z t then Re(w) =Re(z) and Re(z) =Re(t) so Re(w) =Re(t) and w t. Thus the relation is an equivalence relation. The equivalence class containing 4 2i comprises all z C with Re(w) = 4. That is 4+ai for any a R. 14 (a) Let a n = 2n+1 3n 2 4, n 1. The dominant term here is the n2 on the denominator, so we divide both numerator and denominator by n 2, so that a n = 2n+1 2 3n 2 4 = ( ) 1 n n ( ) 2 1 n , as n = 0 3 = 0. Hence the sequence {a n } converges to 0. page 9 of 13

10 (b) Let a n = n! 99 2 n, n 1. The dominant term here is the n! on 74 the numerator, so we look at the reciprocal sequence {1/a n } and divide both numerator and denominator by n!. Thus 1 = 2n 74 a n n! 99 = 2 n n! 74 1 n! n! , as n = 0 1 = 0. { } 1 Hence the sequence converges to 0. It follows that, since a n a n > 0 for all n 1, the original sequence {a n } tends to as n. In particular, {a n } is divergent. (c) As n, 1 0 so that n 2 1 n 2 0 = 1. (Here we have used the fact that the function f(x) = 2 x is continuous at 0.) In particular, the sequence {a n } is convergent (to 1). 15 (a) This statement is true. It is an immediate consequence of the Product Rule for convergent sequences. (Recall that a sequence {a n } is convergent (with limit l) if, for each positive number ε, there is a number X such that a n l < ε, for all n > X.) Alternatively, we can prove it directly, as follows. Since the sequence {a n } is convergent, then: (i) there is some positive number K such that a n < K for all n 1, and (ii) for any positive number ε, there is a positive integer N such that ε a n l < for all n > N. K + l So, let ε be any positive number and N a positive integer chosen so that (b) above holds. It then follows from (a) and (b) above that, for n > N, we have a 2 n l 2 = (an l) (a n +l) = a n l a n +l ε < K + l ( a n + l ) ε < (K + l ) = ε. K + l This completes the proof that the sequence {a 2 n } is convergent to l 2. page 10 of 13

11 (b) This statement is false. For example, let a n = ( 1) n, for n 1, and l = 1. Then (a n ) 2 = 1 for n 1, so that a 2 n 1 = l 2 as n. However the subsequence a 2k+1 = ( 1) 2k+1 = 1, for k 1, converges to the limit 1, which is different from 1. It follows that, although a 2 n 12 as n, nevertheless a n 1 (indeed {a n } is not even convergent). 16 All the functions f f, f g and g f have domain R and codomain R. Then: f f:x f(f(x)) = f(x) 2 +f(x)+1 = ( x 2 +x+1 ) 2 + ( x 2 +x+1 ) +1 = ( x 4 +2x 3 +3x 2 +2x+1 ) + ( x 2 +x+1 ) +1 = x 4 +2x 3 +4x 2 +3x+3; f g:x f(g(x)) = g(x) 2 +g(x)+1 = ( x 3 x ) 2 + ( x 3 x ) +1 = ( x 6 2x 4 +x 2) + ( x 3 x ) +1 = x 6 2x 4 +x 3 +x 2 x+1; and g f:x g(f(x)) = f(x) 3 f(x) = ( x 2 +x+1 ) 3 ( x 2 +x+1 ) = ( x 6 +3x 5 +6x 4 +7x 3 +6x 2 +3x+1 ) ( x 2 +x+1 ) = x 6 +3x 5 +6x 4 +7x 3 +5x 2 +2x. 17 (a) The function f is one-one. For, if f(x 1 ) = f(x 2 ), then x = x3 2 3 so that x3 1 = x3 2, and so x 1 = x 2. It follows that f has an inverse whose domain is f(r). Then, either by sketching the graph of f or using the facts that f(x) as x and f(x) as x, we see that f(r) = R. Next, if y = f(x) = x 3 3, we have x 3 = y +3 and so x = 3 y +3. It follows that f 1 is the function: f 1 :R R x 3 x+3 (b) The function f is one-one. For, if f(x 1,y 1 ) = f(x 2,y 2 ), then so that (x 1 +1,y 1 2) = (x 2 +1,y 2 2) (x 1,y 1 )+(1, 2) = (x 2,y 2 )+(1, 2), and so x 1 = x 2 and y 1 = y 2. It follows that f has an inverse whose domain is f(r 2 ). But f(x,y) is simply (x,y) translated 1 to the right and 2 down, so that clearly f(r 2 ) is just R 2 itself. page 11 of 13

12 Next, if f(x,y) = (x+1,y 2) = (X,Y), we have f 1 (X,Y) = (x,y) = (X 1,Y +2). It follows that f 1 is the function: f 1 :R 2 R 2 (x,y) (x 1,y +2) (c) The function f is not one-one, since f(2,3) = (4,9) and f( 2, 3) = (4,9), although (2,3) ( 2, 3). It follows that f has no inverse function. 18 Suppose that n = a+2b, where a and b are positive real numbers. Suppose also that a < 1 2 n and b < 1 n. Then 4 n = a+2b < 1 2 n+2( 1 4 n) = n. This contradiction shows that the supposition that a < 1 2 n and b < 1 4 n must be false. That is, we must have either a 1 2 n or b 1 4 n. 19 (a) The converse of the implication is the following implication: If n is prime, then 2 n 1 is prime. (b) The contrapositive of the implication is the following implication: If n is not prime, then 2 n 1 is not prime. 20 We now prove the contrapositive of the original implication in 19. Suppose that n is a positive integer that is not prime. If n = 1, then 2 n 1 = 2 1 = 1, which is not prime. If n > 1, then n = ab where 1 < a, b < n. It follows that on factoring. 2 n 1 = 2 ab 1 = (2 a ) b 1 = (2 a 1) ( (2 a ) b 1 +(2 a ) b a +1 ), Now 2 a 1 > 1, since a > 1; and similarly (2 a ) b 1 +(2 a ) b a +1 > 1, since both a and b are greater than 1. Hence 2 n 1 is not prime. We have thus proved the required contrapositive implication in both the cases n = 1 and n > 1. It follows that the original implication is also true, for any positive integer n. page 12 of 13

13 3 How did you get on? Now that you have finished the quiz, how did you get on? The information below should help you to decide what, if anything, you should do next. Most of this quiz was unfamiliar to me. I don t think I have met any of the topics before. Perhaps you should consider studying a level 1 mathematics module such as MST124, Essential mathematics 1, or MST125, Essential mathematics 2. The MathsChoices and Study at the OU websites may help you decide. Some of these topics were familiar to me, but not all of them. Perhaps you should consider studying a level 2 mathematics module such as M208, Pure mathematics. The Study at the OU website may help you decide. I have met all these topics before, but I am a bit rusty on some, and had to look up how to do a few of them. You will benefit from doing some revision before starting M303. See below for some possible sources of help. I could do all of these questions. You are well prepared for M303. You can study it with confidence. Register now. What resources are there to help me to revise and prepare for M303? If you have studied the Open University course M208, Pure Mathematics (or its predecessor M203), then you could use parts of this to revise for M303, Further pure mathematics. Each group of diagnostic quiz questions has references to M208, so you should concentrate on the highlighted M208 material for those areas you found difficult. You will find the M208 Handbook here. There are many good mathematical resources on the internet. In general, Wikipedia is reliable for mathematics and the Wolfram Mathworld website is quite comprehensive. Do contact your Student Support Team via StudentHome if you have any queries about M303. page 13 of 13