How to Think Like a Mathematician Solutions to Exercises

Transcription

1 How to Think Like a Mathematician Solutions to Execises Septembe 17, 2009 The following ae solutions to execises in my book How to Think Like a Mathematician. Chapte 1 Execises 1.10 (i) 5 (ii) 3 (iii) (iv) 0 (v) Infinite (vi) 2 (vii) 2 (viii) 3. Execises 1.20 (i) { 1, 1, 2, 3,, 5, 7} (ii) Z. Execises 1.23 (i)(a) {0, 1, 2, 3,, 5}, (b), (c) {0, 1, 5}. (ii) Z,, Z. Execises 1.33 My intention was that the domain would be the lagest in R. (i) R\{(5 13)/2, (5+ 13)/2}, (ii) R, (iii) Many answes ae possible. Hint: We can always find a line between two points in the plane so we could use a linea function such as f(x) = ax+b whee a and b ae detemined by using the equations f( 1) = 5 and f(3) = 2. One you have masteed the idea behind this ty finding a quadatic whose gaph passes though the points. Then move on to functions involving highe odes o sines, cosines and the exponential function. Execises 1.3 (i) A B = {2}, A B = {0, 2, 3,, 5, 6, 7, 8, 10} (o X\{1, 9} if you pefe), A\B = {0,, 6, 8, 10}, B\A = {3, 5, 7}, A B has 2 elements (which is a lot to type), X A has 66 elements, A c = {1, 3, 5, 7, 9}, B c = {0, 1,, 6, 8, 9, 10}. (ii) The quadatic has oots 2 and 7 so the union is {2, 3,, 5, 6, 7, 8, 9, 10} and the intesection is {7}. (iii) The pecise answes will depend on the sets you take. You should always get equalities of sets except in (d) and (f). (This does not exclude the situation whee you examples give equalities in (d) and (f). Howeve, thee exist examples whee they don t have equality.) (iv)(a) A and B disjoint. 1

2 (b) A B A C (A B) c (c) An example of fom (i): (v) Not applicable. Chapte 2 (A B) (A B) o A (B C). Solutions not applicable to this chapte. Chaptes 3 and The solutions to these chaptes will depend on you pesonal answes and witing style. Execises 3.2 (i) Don t foget to explain what a, b, c, α, β, γ and h ae and explain what ae assumptions and what ae conclusions. (ii) Thee ae some mathematical mistakes. The second line is actually f (x) yet the student has equated it to f. Also they say (taking into account the pevious comment) that f (x) = 6x 2 2x + 18 implies that x = 1 and 3. This is not tue, it is the equation f (x) = 0 that gives us the values fo x. Equality signs ae missing in the calculation of x. Some notation is missing. When finding maxima and minima we put the values of x into the expession fo the second deivative and so we should eally have d 2 y dx 2 etc, athe than just d2 y x=1 dx 2. Chapte 5 The point of this chapte is to get you thinking about solving poblems and so giving the answes would defeat the pupose of the chapte. Keep coming back to them! Howeve, I should say that (v) and (vi) ae classics and so you should be able to find some answes with a web seach. 2

3 Chapte 6 Execises 6.10 Both tuth tables should have the following fom A B F F T F T F T F F T T F Execises 6.11 (i) Statements: (a), (d) and (e). Statement (d) is tue but the tuth o othewise cannot be known fo (a) and (e). (Well, (e) will be known if you ae eading this afte 2089.) (ii) The tuth table should be A B (a) (b) (c) (d) (e) (f) F F T T T T T F F T T F T F T F T F T F T T T F T T F F F T T F (iii)(a) A is false and B is tue. (b) A is tue and B is false. (c) A is false and B is false. (d) A is false o B is false. (iv) The tuth tables look like the following A B C (a) (b) (c) F F F F F F F F T F T T F T F F F F F T T F T F T F F F F F T F T T T F T T F T T F T T T T T F (v)(a) The tuth table can be constucted as follows: A not A A o (not A) F T T T F T (b) not(a and not A) (c) Yes, it is but maybe I should have put this question in one of the chaptes on implications. (d) The tuth table looks like A not A A and (not A) F T F T F F 3

4 Chapte 2 Execises 2.10 (i) Simila to Example 2.. (ii) Simila to Example 2.7. (iii) Simila to Example 2.6. (iv) We ll pove that 17 divides 3 n + 3n+2 fo all n N as a woked example. (If this wasn t an execise in a chapte on induction, then, because of the n N pat, alam bells should ing in you head and you should automatically think I ll use induction.) Initial case: We want to show that 17 divides 3 n + 3n+2 when n = 1. We have that 3 n + 3n+2 = = = = 1105 = So 17 divides 3 n + 3n+2 when n = 1. Induction step: Now suppose that 17 divides 3 n + 3n+2 when n = k fo some paticula k, i.e., 3 k + 3k+2 = 17m fo some m Z. Let s conside the expession fo n = k + 1: 3 (k+1) + 3(k+1)+2 = 3 ( k+ + 3k+5 = 3 k) 3 + 3k+5 ( = 17m 3k+2) 3 + 3k+5, by the inductive hypothesis, = 17m 3 3k k+5 = 17m 3 3k k+2 3 = 17m 3 3k+2 ( 3 3) = 17m 3 3k+2 (81 6) = 17m ( 3 3k+2 17 = 17 m 3 3k+2). So 17 divides 3 (k+1) + 3(k+1)+2 as equied. We have shown that the statement fo n = k implies that the statement fo n = k + 1 is tue. Theefoe by the Pinciple of Mathematical Induction the statement is tue fo all n N. That is, 17 divides 3 n + 3n+2 fo all n N. (v) Hint: sin(n + 1)x can be taken to be the complicated side as we can expand it to something else. Also 1 cos θ 1.

5 (vi) Hint: In the induction step we can show that (x + y)(x + y) k = (x + y) x k y =0 = x k +1 y + x k y +1 =0 At this point a numbe of students stat witing out the summations in full so that they eaange and gathe them togethe. I.e., they would like to gathe the tems x a y b fom the two summations. We can do this by changing the vaiable in one of the summations. Let s = + 1, so when = 0, s = 1 and when = k, s = k + 1. We also have = s 1. Then the second summation can be changed: =0 x k y +1 = = =0 x k (s 1) y (s 1)+1 s 1 s=1 x k s+1 y s. s 1 s=1 Now hee s the pat that upsets some people. The summation does not depend on the name of the vaiable used in the summing and so I can change the s to. But s = 1 they say, we can t take s =. The answe is that this is diffeent to the pevious. Sounds confusing but mathematicians do this type of change all the time! We have x k s+1 y s = s 1 s=1 =1 In conclusion this means we have shown that =0 Thus we get that (x + y)(x + y) k = x k y +1 = =0 =1 x k +1 y + x k +1 y. 1 x k +1 y. 1 =1 x k +1 y. 1 Fom hee you can sepaate out the tem = 0 in the fist summation and = k + 1 in the second so that you can add the two emaining pats of the summations togethe. In my oiginal notes fo the solution to this poblem I added togethe ( ) k and ( k 1) by using the definition in tems of factoials. Instead you can use the identity fom the statement of execise (viii), which amounts to the same thing. (vii) Change n 2 1 to (2n 1) 2 1 as descibed on page 172. In this execise we can also pove the statement by the diect method. It is easy to calculate that (2n 1) 2 1 = n(n 1). Now fo any n eithe n o n 1 is even and hence so is thei poduct. This means that n(n 1) = 2m fo some m Z. 5

6 Theefoe, (2n 1) 2 1 = 2m = 8m and we can conclude that the numbe is divisible by 8. (viii) Simila to (vi). (ix) Simila to Example 2. and execise (i) but with moe complicated algebaic manipulation. A supising esult, don t you think? That the squae of the sum of the fist n numbes is equal to the sum of the cubes of the fist n numbes is supising to me. (x) This execise is incoect. The statement is not even complete. See the coections on the website. (xi) This question is about counting subsets and to some extent about how you oganize you witing of mathematics. In the inductive step we conside that X has k + 1 elements and so it contains a subset Y of k elements so that X = Y {x} fo some x X. By the inductive hypothesis Y has 2 k subsets. As Y is a subset of X, it must be tue that X has these as subsets too. Next we can see that if Z is a subset of Y, then Z {x} will be a subset of X that we haven t aleady counted. Thee should theefoe be 2 k of these. Thus X should have 2 2 k subsets. The aim of the execise is to igoously count these subsets and to wite you answe so that is compehensible. This is the oganizational pat, you need to give good notation so that the counting pocess is easy/clea. (xii)(a) The inductive hypothesis is x k 1 = m(x 1) fo some m Z so x k = m(x 1) + 1. The est is just algeba. (b) The fomula is x n 1 x 1 = xn + x n x + 1 and the latte can be witten succinctly as n i=0 xi. (xiii) We have 2 n < 2 n n 1 < 2 n 1 2 < 1. So the statement is false. It is impotant that we use, see Chapte 21 on common mistakes. The induction step should be staightfowad. (xiv) The induction step involves dealing with the two cases, and, sepaately. (xv) This one should be staightfowad as I think that neithe side of the inequality is moe complicated than the othe so it doesn t matte which you pick to wok on. (xvi) The algeba fo this is the hadest pat so don t woy if you answe is long. You may need to use the fomula in (xii)(b). The poblem is about compaing the aithmetic and geometic means of a collection of numbes. Chapte 25 Execises 25.2 (i) One answe is that the poof goes wong fo k < 7 simply because those cases ae not tue. I.e., the initial case is missing. One could also ague that sometimes the induction step is false when k < 7. That is, in 6

7 the poof, k 2 + 3k k 2 + k k is tue fo k 7 but in fact it is also tue fo k 3 while not tue fo k = 1 and 2. (ii) Simila to (i). The inductive step is tue fo all k 1. (iii) Hint: Execises 25. (i) The two initial cases ae easy. The induction step is moe complicated. Hee s how I would attempt it. I am tying to show you that, sometimes, solving a poblem involves dead ends and efinement and that it may not be obvious, afte it is witten up, whee the answe came fom. Let s begin the poof of the induction step: Assume that A(k) and A(k 1) ae tue, i.e., x k < ( 7 ) k and xk 1 < ( 7 ) k 1. Then x k+1 = x k + x k 1, by definition, ( ) 7 k ( ) 7 k 1 < +, by the inductive hypothesis, ( ) 7 k ( ) 7 k < +, an obvious step towads gouping tems, ( ) 7 k = 2 > 7 ( ) 7 k. So unfotunately, at the end I don t get the less than ( 7 k+1 ) I would like. The poblem aises in the obvious step towads gouping tems whee I used ( 7 ) k 1 ( < 7 ) k. Taking the RHS tem has led to something bigge than I would like, so I should take something smalle on the RHS. 7

8 So let s ty again: x k+1 = x k + x k 1, by definition, ( ) 7 k ( ) 7 k 1 < +, by the inductive hypothesis, ( ) 7 k ( ) 7 k < + a, when 1 < 7 a, ( ) 7 k = (1 + a). So if I took a = 3/, (and this is OK as 1 < 7 3 ), then I would get ( ) 7 k ( (1 + a) = ) ( ) 7 k ( ) ( ) 7 7 k ( ) 7 k+1 = =. And this is what I wanted! Now, obviously this is just ough wok and I wouldn t hand in all this woking out if I wee submitting it as pat of an assignment. Instead I would just do the following: x k+1 = x k + x k 1, by definition, ( ) 7 k ( ) 7 k 1 < +, by the inductive hypothesis, ( ) 7 k < + 3 ( ) 7 k ( = ) ( ) 7 k ( ) ( ) 7 7 k = ( ) 7 k+1 =. Hence A(k + 1) is tue. Notice that in the above, in the polished final answe, I didn t explain whee the 3/ came fom I just used it and the eade is left to veify the faily simple obsevation that ( ) 7 k 1 ( < 3 7 ) k. This is anothe example whee we cove ou tacks. (ii) We need A(2) tue as well o else we can t use the inductive step to pove that A(3) is tue. In othe wods we can t get stated without it. Execises 25.6 (i) Not applicable. (ii) The poof doesn t involve induction. Ty contadiction instead. Execises 25.7(i)(a) {1, 1, 2, 3, 5, 8, 13}, (b) The algebaic manipulation afte adding the foms fo x k and x k 1 gets a bit messy but stick with it, tems magically disappea. (c) We have x 3(k+1) = x 3k+3 = x 3k+2 + x 3k+1, by definition, = (x 3k+1 + x 3k ) + x 3k+1, again by definition, = 2x 3k+1 + 2m, by the inductive hypothesis. 8

9 The othe details ae faily staightfowad. In fact, it is possible to pove that x 3n = n ( n i=1 i) 2 i x i. You may like to attempt the poof of this. (d) The fomula is n i=1 x 2i 1 = x 2n. (e) The fomula is n i=1 x 2i = x 2n (ii) It is easie if you simplify the k 1 type tems fist but this is not stictly 3 2 2k 2 necessay. Also, don t foget that ( 1) k+1 = ( 1) 2 ( 1) k 1 = 1.( 1) k 1 = ( 1) k 1. (iii) The factoization can be done in a numbe of ways (we don t need induction). I can see that is divisible by 10 and so 2 and 5 ae factos. So I need to facto To do this I find the squae oot as it gives me the highest numbe that I need to check to find a facto. If thee is a facto geate than the squae oot it must be balanced by one smalle than the squae oot. (Think about it.) My calculato gives 1287 = 3 13 fom which I deduce that 3 2 is a facto of I now need to facto 1287/9 = 13. Again I use my calculato, this time it says that so I need only to check divisibility by the pimes fom 2 to 11. Howeve, when thinking about 11 I notice that the numbe 13 is of the fom abc whee a + c = b and so I know that 13 is divisible by 11. I can wok out that 13/11 = 13. This is pime so I can now stop my factoization. Putting this all togethe we get = The numbe is done in a simila manne. I can see that the numbe is even so I can divide by 2. The esulting numbe is even also and so I can divide by 2 again. Altenatively, I can see that the last two digits ae divisible by and so the numbe itself is divisible by. (A poof of this is equied in Execises 27.23(ix)(b)). We have 17836/ = 59. This is not even so is not divisible by 2. Again I find the squae oot so that I have a bound on the factos I should look fo, hee 59 = Thus, 59 = Assuming I don t know the divisos of 91 I each fo my calculato again and get so we only need check the pimes 2, 3, 5, and 7 fo divisibility. Obviously 2 and 5 can be ignoed as 91 is not even and does not end in 0 o 5. Also, 3 can be discaded as 90 is obviously a multiple of 3. This leaves 7, a quick calculation shows that 91/7 = 13. As this is pime we stop. Putting it all togethe we get: = = Kevin Houston School of Mathematics Univesity of Leeds Leeds UK LS2 9JT k.houston@leeds.ac.uk khouston 9