1. By using truth tables prove that, for all statements P and Q, the statement

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1 Author: Satiago Salazar Problems I: Mathematical Statemets ad Proofs. By usig truth tables prove that, for all statemets P ad Q, the statemet P Q ad its cotrapositive ot Q (ot P) are equivalet. I example.2.3 idetify which statemet is the cotrapositive of statemet (i) (f a = 0 a > 0). Fid aother pair of statemets i that list that are the cotrapositives of each other. Truth table P Q ~P ~Q P Q ~Q ~P T T F F T T T F F T F F F T T F T T F F T T T T Sice the last two colums are idetical, the statemets P Q ad its cotrapositive ~Q ~P are logically equivalet. The cotrapositive of statemet (i) (f a = 0 a > 0) is statemet (vii) (a 0 f(a) 0). Similarly, the cotrapositive of statemet (iii) (f a = 0 a 0) is statemet (vi) (a > 0 f(a) 0). 2. By usig truth tables prove that, for all statemets P ad Q, the three statemets (i) P Q, (ii) (P or Q) Q, ad (iii) (P ad Q) P are equivalet. Truth table P Q P or Q P ad Q P Q (P or Q) Q (P ad Q) P T T T T T T T T F T F F F F F T T F T T T F F F F T T T Sice the last three colums are idetical, the statemets P Q, (P or Q) Q, ad (P ad Q) P are logically equivalet. 3. Prove that the three basic coectives or, ad, ad ot ca all be writte i terms of the sigle coective otad where P otad Q is iterpreted as ot(p ad Q). P P ad P ~(P ad P) ~P T T F F F F T T Sice the last two colums are idetical, the statemets ~P ad ~(P ad P) are logically equivalet. Hece we ca write ~P as P otad P. P Q P ad Q ~(P ad Q) ~[~ P ad Q ] T T T F T T F F T F F T F T F F F F T F

2 Sice colums 3 ad 5 are idetical, the statemets P ad Q ad ~ ~ P ad Q are logically equivalet. But observe that, by defiitio, ~ P ad Q is writte as P otad Q. Thus ~ ~ P ad Q is logically equivalet to ~ P otad Q. The, by the first part of the problem, ~ P otad Q ca be writte as P otad Q otad P otad Q. Hece we ca write P ad Q as P otad Q otad P otad Q. P Q P or Q ~P ~Q ~P ad ~Q ~(~P ad ~Q) T T T F F F T T F T F T F T F T T T F F T F F F T T T F Sice colums 3 ad 7 are idetical, the statemets P or Q ad ~(~P ad ~Q) are logically equivalet. But observe that, by the first part of the problem, ~P ad ~Q is logically equivalet to P otad P ad (Q otad Q). Now, by defiitio, the egatio of this last statemet, amely ~(~P ad ~Q), ca be writte as P otad P otad (Q otad Q). Thus we ca write P or Q as P otad P otad (Q otad Q). 4. Prove the followig statemets cocerig the positive itegers a, b, ad c. (i) (a divides b) ad (a divides c) a divides (b + c). (ii) (a divides b) or (a divides c) a divides bc. (i) a divides b meas b = aq for some iteger q, ad a divides c meas c = ap for some iteger p. Thus b + c = aq + ap = a q + p = ak, where k = q + p is a iteger. Therefore a divides (b + c). (ii) case : a divides b meas b = aq for some iteger q. Thus bc = aq c = a qc = ar, where r = qc is a iteger. Therefore a divides bc. case 2: a divides c meas c = ap for some iteger p. Thus bc = b ap = a bp = as, where s = bp is a iteger. Therefore a divides bc. Hece, i either case, a divides bc. 5. Which of the followig coditios are ecessary for the positive iteger to be divisible by 6 (proofs ot ecessary)? (i) 3 divides. (ii) 9 divides. (iii) 2 divides. (iv) = 2. (v) 6 divides 2. (vi) 2 divides ad 3 divides. (vii) 2 divides or 3 divides. Which of these coditios are sufficiet?

3 6 divides meas = 6q for some iteger q. The followig coditios are ecessary for the positive iteger to be divisible by 6: (i) 3 divides, (v) 6 divides 2, (vi) 2 divides ad 3 divides, ad (vii) 2 divides or 3 divides. The followig coditios are sufficiet for the positive iteger to be divisible by 6: (iii) 2 divides, (iv) = 2, (v) 6 divides 2, ad (vi) 2 divides ad 3 divides. 6. Use the properties of additio ad multiplicatio of real umbers give i Properties 2.3. to deduce that, for all real umbers a ad c, (i) a 0 = 0 = 0 a, (ii) a b = ab = a b, (iii) a b = ab. (i) Prove that a 0 = 0 = 0 a = 0 0 a + (0 a) = 0 a 0 a = 0 a 0 = 0 Thus a 0 = 0 = 0 a. Additive idetity (iv) Distributive property (iii) Additive iverse (vi) Commutative property (i) (ii) Prove that a b = ab = a( b). First we show that a = ( )a by showig that a + a = 0. a + a = a + ( )a Multiplicative Idetity (v) = ( + ( ))a Distributive property (iii) = 0a Sice is the additive iverse of = 0 By part (i) of problem Thus a + a = 0 a = ( )a. The ab = ( )ab = (( )a)b = ( a)b ab = ( )ab = (( )b)a = ( b)a = a( b) Thus a b = ab = a( b). By the proof above Associative property (ii) By the proof above By the proof above Associative property (ii) By the proof above Commutative property (i) (iii) Prove that a b = ab. a + a = 0 a = ( a) ab = a b = ab = a b = b a = ( b)( a) = ( a)( b) Additive Idetity (iv) Sice a = ( a) By part (ii) of problem By part (ii) of problem Commutative property (i)

4 By part (ii) of problem Thus a b = ab. Commutative property (i) 7. Prove by cotradictio the followig statemet cocerig a iteger. 2 is eve is eve. [You may suppose that a iteger is odd if ad oly if = 2q + for some iteger q. This is proved later as Propositio.3.4.] Suppose is ot eve. The is odd, that is = 2q + for some iteger q. Thus 2 = (2q + ) 2 = 4q 2 + 4q + = 2 2q 2 + 2q + = 2p + where p = 2q 2 + 2q is a iteger. Thus 2 is odd cotradictig that 2 is eve. It follows that our iitial assumptio, that is odd, is false. Hece is eve Therefore, 2 is eve is eve. 8. Prove the followig statemets cocerig a real umber x. (i) x 2 x 2 = 0 x = or x = 2. (ii) x 2 x 2 > 0 x < or x > 2. (i) : x 2 x 2 = 0 x 2 x + = 0 (x 2) = 0 or (x + ) = 0 x = 2 or x = Thus x 2 x 2 = 0 x = or x = 2. : If x = 2, the x 2 x 2 = = 0. If x =, the x 2 x 2 = ( ) 2 2 = 0. So, i either case, x 2 x 2 = 0. Thus (x = or = 2) x 2 x 2 = 0. Hece x 2 x 2 = 0 x = or x = 2. (ii) : x 2 x 2 > 0 x 2 x + > 0 (x 2 > 0 ad x + > 0) or (x 2 < 0 ad x + < 0) (x > 2 ad x > ) or (x < 2 ad x < ) x > 2 or x < Thus x 2 x 2 > 0 x < or x > 2. : case: x > 2 2x > 4 > 2 (multiply by 2 > 0) 2x > 2 ad x > 2 x 2 > 2x > 2 (multiply by x > 0) x 2 > 2. It follows that x 2 x > 2x x = x > 2 ad so x 2 x > 2 x 2 x 2 > 0 case2: x < 0 < x 0 < 2 < x (by addig 2) 0 < x ad x < x > (multiply by < 0) ad x < x 2 > x > (multiply by x < 0) x 2 > x 2 x > x > 0 x 2 x 2 > x 2 = x > 0 x 2 x 2 > 0 Hece x 2 x 2 > 0 x < or x > Prove by cotradictio that there does ot exist a largest iteger.

5 [Hit: Observe that for ay iteger there is a greater oe, say +. So begi your proof Suppose for cotradictio that there is a largest iteger. Let this larger iteger be. Suppose for cotradictio that there is a largest iteger. Observe that 0 < < +. Thus is ot the largest iteger, sice for all, + >. 0. What is wrog with the followig proof that is the largest iteger? Let be the largest iteger. The, sice is a iteger we must have. O the other had, sice 2 is also a iteger we must have 2 from which it follows that. Thus, sice ad we must have =. Thus is the largest iteger as claimed. What does this argumet prove? The proof starts with a statemet which is false (from problem 9). We also kow that the coclusio is false sice is ot the largest iteger. However all the implicatios that start with a false hypothesis are true. I fact, this argumet proves that if a largest iteger existed, it would be.. Prove by cotradictio that there does ot exist a smallest positive real umber. Suppose for cotradictio that is the smallest positive real umber. Observe that 0 < < 0 < <. Thus the umber is positive, real, ad less tha 2 2 2, cotradictig our iitial assumptio that was the smallest positive real umber. Hece there does ot exist a smallest positive real umber. 2. Prove by iductio o that, for all positive itegers, 3 divides divides meas = 3q for some iteger q. Base case: For =, = = 9 which is divisible by 3 Iductive step: Suppose ow as iductive hypothesis that, for some positive iteger k, 3 divides 4 k + 5, that is 4 k + 5 = 3q for some iteger q. We eed to show that 3 divides 4 k+ + 5, that is 4 k+ + 5 = 3q for some iteger q. The 4 k+ + 5 = 4 4 k + 5 (by iductive defiitio) = 4 3q (by iductive hypothesis) = 2q 5 = 3 4q 5 = 3p, where p = 4q 5 is a iteger. Thus 3 divides 4 k+ + 5 Coclusio: Hece, by iductio o, 3 divides for all positive itegers. 3. Prove by iductio o that! > 2 for all itegers such that 4. Base case: For = 4,! = 4! = 24 > 6 = 2 4 = 2 Iductive step: Suppose ow as iductive hypothesis that k! > 2 k for some positive iteger k 4. We eed to show that k +! > 2 k+. The 2k! >

6 2 2 k = 2 k+ (by iductive hypothesis) ad k +! = (k + )k! (by iductive defiitio) 4 + k! = 5k! (sice k 4) > 2k! > 2 2 k = 2 k+ (by iductive hypothesis). Thus k +! > 2 k+ Coclusio: Hece, by iductio o,! > 2 for all itegers Prove Beroulli s iequality ( + x) + x for all o-egative itegers ad real umbers x >. Base case: For = 0, + x 0 = = + 0 x ad so the equality holds. Iductive step: Suppose ow as iductive hypothesis that ( + x) k + kx for some o-egative iteger k ad real umbers x >. We eed to show that ( + x) k+ + (k + )x. Observe that x > + x > 0, ad that k 0 ad x 2 0 (sice for ay real umber a, a 2 0) both imply that kx 2 0. Thus ( + x) k+ = ( + x)( + x) k (by iductive defiitio). By iductive hypothesis, ( + x) k + kx + x + x k = ( + x) k+ + kx ( + x) = + x + kx + kx 2 = + k + x + kx 2 (multiply by + x > 0) + k + x + 0 = + k + x (sice kx 2 0). Therefore ( + x) k+ + (k + )x as required. Coclusio: Hece, by iductio o, ( + x) + x for all o-egative itegers ad real umbers x >. 5. For which o-egative iteger values of is! 3?! 3 is true for = 0 ad all itegers 7. For = 0,! = 0! = by iductive defiitio, ad 3 = 3 0 =, ad so the equality holds. Now we show that the iequality is also true for all itegers 7. Base case: For = 7,! = 7! = 5040 ad 3 = 3 7 = 287; ad so! 3. Iductive step: Suppose ow as iductive hypothesis that k! 3 k for some iteger k 7. We eed to show that k +! 3 k+. The 3k! 3 3 k = 3 k+ ad k +! = (k + )k! (by iductive defiitio) 7 + k! = 8k! (sice k 7) 3k! 3 3 k = 3 k+ (by iductive hypothesis). Thus k +! 3 k+ as required. Coclusio: Hece, by iductio o,! 3 for all itegers Prove by iductio o that i= i(i + ) =, + for all positive itegers. Base case: For =,

7 i= i(i + ) = ( + ) = 2 ad + = + = 2 Iductive step: Suppose ow as iductive hypothesis that k i= i(i + ) = k k + for some positive iteger k. We eed to show that k+ i= i(i + ) = k + (k + ) + = k +. k + 2 But by iductive defiitio ad iductive hypothesis k+ i= i(i + ) = k i(i + ) + k + (k + 2) i= = k k + + k + (k + 2) = k2 + 2k + k + k + 2 = k + 2 k + k + 2 = k + k + 2 Coclusio: Hece, by iductio o, i(i + ) = + i= for all positive itegers. 7. For a positive iteger the umber a is defied iductively by a =, a k+ = 6a k + 5 a k + 2 for k a positive iteger. Prove by iductio o that, for all positive itegers, (i) a > 0 ad (ii) a < 5. (i) Base case: For =, a = a = > 0 Iductive step: Suppose ow as iductive hypothesis that a k > 0 for some positive iteger k. We eed to show that a k+ = 6a k + 5 a k + 2 > 0.

8 By iductive hypothesis, a k > 0 a k + 2 > 0 ad a k > 0 2a k > a k 6a k > a k 6a k + 5 > a k + 2 > 0. Thus 6a k + 5 a k + 2 > > 0 6a k + 5 a k + 2 > 0 Coclusio: Hece, by iductio o, a > 0 for all positive itegers. (ii) Base case: For =, a = a = < 5 Iductive step: Suppose ow as iductive hypothesis that a k < 5 for some positive iteger k. We eed to show that a k+ = 6a k + 5 a k + 2 < 5. By iductive hypothesis, a k < 5 6a k < 5a k + 5 6a k + 5 < 5a k + 0 = 5(a k + 2) 6a k + 5 < 5(a k + 2). Thus 6a k + 5 a k + 2 < 5 Coclusio: Hece, by iductio o, a < 5 for all positive itegers. 8. Give a sequece of umbers a, a 2,, the umber i= a(i) is defied iductively by i a i = a, ad i= k+ ii a i = a i i= Prove that i= k i= + x 2i = x2 x What happes if x =? Base case: For =, ad i= + x 2i a k + for k. for x. = + x 2 = + x 20 = + x x 2 x = x2 ( x)( + x) = = + x x x Iductive step: Suppose ow as iductive hypothesis that k + x 2i = x2k x i= for some iteger k ad for all real umbers x. We eed to show that

9 k+ i= + x 2i = x2k+. x By iductive defiitio ad iductive hypothesis, k+ i= + x 2i = + x 2i Coclusio: Hece k i= = x2k + x 2k x = x2k+ x 2 k+ + x + x 2i = x2 x i= for all itegers ad for all real umbers x. Moreover, if x =, the the formula does ot work sice x = 0 ad we caot divide by zero. However, x 2i = for all i ad so i= 9. Prove that + x 2i = 2. i 2 = + 2 i=2 for itegers 2. Base case: For = 2, 2 i=2 i 2 = 2 2 = 3 4 ad + 2 = 2 + 2(2) = 3 4 Iductive step: Suppose ow as iductive hypothesis that k i=2 i 2 = k + 2k for some iteger k 2. We eed to show that k+ i=2 (k + ) + = = k + 2. i2 2(k + ) 2k + 2 By iductive defiitio ad iductive hypothesis,

10 k+ i=2 i 2 = k i=2 i 2 = k + 2k (k + ) 2 = k + k + 2 2k k + 2 = k + 2 2k k + = k2 + 2k + 2k k + k k + 2 = 2k k + = k + 2 2k + 2 Coclusio: Hece i 2 = + 2 i=2 for all itegers 2. (k + ) Prove that, for a positive iteger, a 2 2 square grid with ay oe square removed ca be covered usig L-shaped tiles such as the oe show below. Base case: For =, a 2 2 square with oe square removed ca be covered by a sigle L-shaped tile. Iductive step: Suppose ow as iductive hypothesis that a 2 k 2 k square grid with ay oe square removed ca be covered by L-shaped tiles. We eed to deduce that a 2 k+ 2 k+ square grid with ay oe square removed ca be covered usig L-shaped tiles. If we divide the 2 k+ 2 k+ square grid i four equal square grids (as show i the figure below), we obtai four 2 k 2 k square grids (observe that 2 k+ 2 = 2 k ). Sice the 2 k+ 2 k+ square grid has oe square removed, this removed square must lie i oe of the four 2 k 2 k square grids (as show by the shaded square i the corer of the figure below). The other three 2 k 2 k square grid are complete. Now from each of the complete 2 k 2 k square grids, remove the square that touches the ceter of the origial 2 k+ 2 k+ square grid (as show i the figure below). By iductio hypothesis, all four of the 2 k 2 k square grids with oe square removed ca be covered usig L-shaped tiles. The, with oe more L-shaped tile, we ca cover the three squares touchig the ceter of the origial 2 k+ 2 k+ square grid.

11 Thus we ca cover the origial 2 k+ 2 k+ square grid with oe square removed usig L-shaped tiles Coclusio: Hece, for a positive iteger, a 2 2 square grid with ay oe square removed ca be covered usig L-shaped tiles. 2 k 2 k+ 2 k 2 k+ 2 k 2 k 2. Suppose that s a real umber such that x + s a iteger. Prove by iductio o that x + x is a iteger for all positive itegers. [For the iductive step cosider x k + x k x + x.] Strog iductio is used. Base case: For =, x + x = x + s a iteger Now x + x 2 = x 2 + x is a iteger sice the square of a iteger is a iteger ad thus x 2 + x 2 is a iteger (sice for ay iteger a, a 2 is a iteger) provig the result for = 2. Iductive step: Suppose ow as iductive hypothesis that x k + x k is a iteger for all positive itegers k for some iteger k 2. We eed to show that x k+ + x k+ is a iteger. By iductive hypothesis, x + x x k + x k is a iteger sice the product of two itegers is a iteger. But x + x k + x x k = xk+ + x k+ + xk + x k ad, by iductive hypothesis, x k + x k is a iteger. Thus x k+ + x k+ must be a iteger Coclusio: Hece, by iductio o, x + x is a iteger for all positive itegers. 22. Prove that i= i= / for positive itegers ad positive real umbers. [it does ot seem to be possible to give a direct proof of this result usig iductio o. However it ca be proved for = 2 m for m 0 by iductio o m. The geeral result ow follows by provig the coverse of the usual iductive step: if the result holds for = k +, where k is a positive iteger, the it holds for = k.]

12 case : If all the terms of the sequece are equal, that is x = x 2 = = x, the i= which implies that i= = x = x = x = i= / case 2: If ot all the terms of the sequece are equal. Clearly this case is oly possible whe >, ad it is proved by iductio. First we prove the iequality whe = 2 m for m ad the, usig this result, we deduce that the iequality is true for all positive itegers. Base case: For m =, = 2 m = 2 = 2. So we have two terms, x ad x 2, ad sice they are ot equal, we have x x 2 x x 2 0 x x 2 2 > 0 x 2 2x x 2 + x 2 2 > 0 x 2 + 2x x 2 + x 2 2 > 4x x 2 x + x 2 2 > 4x x 2 2 x + x 2 > x 2 x 2 x + x 2 > x 2 x 2 ad so 2 2 i= 2 i= /2 Iductive step: Suppose ow as iductive hypothesis that i= i= / where = 2 m, for some positive iteger m. We eed to show that 2m + 2 m+ i= 2m + i= The by iductive hypothesis 2m + 2 m+ i= /2 m +. = x + x x 2 m + 2 m+ i= /

13 = x + x x 2 m m = x + x x 2 m 2 2 m + x 2 m + + x 2 m x 2 m m = x + x x 2 m 2 2 m + x2m + + x 2m x 2 m + 2 m 2 m x x 2 x 2 m + x 2 m + x 2 m +2 x 2 m + 2 m 2 2 m 2 x x 2 x 2 m m x 2 m + x 2 m +2 x 2 m + 2 = m x x 2 x 2 m + 2m + = x x 2 x 2 m + 2m + = i= /2m + Hece, by iductio o m, the result is true for all positive itegers m. Thus the iequality is true for the atural powers of 2, that is for = 2,4,8,6, Now we proceed to prove the iequality for all positive itegers. If is ot equal to some atural power of 2, the it is certaily less tha some atural power of 2, sice the sequece 2,4,8,6,, 2 m, is ubouded above. Therefore let m be some atural power of 2 that is greater tha. Also let i= = α ad expad our list of terms such that x + = x +2 = = x m = α. The α = x + x x m = x + x x m m = + x + x x m = x + x x + m x + x x m = x + x x + m α m ad so = x + x x + x x m m m x x 2 x x + x m m = x x 2 x α m

14 α m x x 2 x α m α x x 2 x α x x 2 x i= i= / Hece, by iductio o, the iequality is true for all positive itegers ad all positive real umbers. 23. For o-zero real umbers x we may exted Defiitio to a defiitio of powers x for all itegers by defiig x m = x m for itegers m > 0. With these defiitios prove the laws of expoets for ay o-zero real umbers x ad y ad itegers m ad : (i) x y = xy ; (ii) x m+ = x m x ; (iii) x m = x m. [Hit: Start from exercise 5.7.] (i) First, we prove the result for the o-egative itegers. Base case: For = 0, x y = = xy Iductive step: Suppose ow as iductive hypothesis that x k y k = xy k for some o-egative iteger k. We eed to show that x k+ y k+ = xy k+. The, by iductive defiitio ad iductive hypothesis, x k+ y k+ = x x k y y k = xy xy k = xy k+ as required to prove the result for = k +. Coclusio: Hece, by iductio o, x y = xy for ay o-zero real umbers x ad y ad o-egative itegers. Now we prove the result for the o-positive itegers by provig that x y = xy for all o-egative itegers. The, by the defiitio of x m, we ca coclude that the result is true for all the o-positive itegers. Base case: For = 0, x y = = xy Iductive step: Suppose ow as iductive hypothesis that x k y k = xy k for some o-egative iteger k. We eed to show that x k+ y k+ = xy k+. The, by iductive defiitio ad iductive hypothesis, x k+ y k+ = x x k y y k = xy xy k = xy k+

15 as required to prove the result for = k +. Coclusio: Hece, by iductio o, x y = xy for ay o-zero real umbers x ad y ad o-egative itegers. Therefore x y = xy for ay o-zero real umbers x ad y ad ay iteger. (ii) First, we prove the result for the o-egative itegers. Base case: For = 0, x m+ = x m = x m x 0 = x m x Iductive step: Suppose ow as iductive hypothesis that x m+k = x m x k for some o-egative iteger k. We eed to show that x m+k+ = x m x k+. The, by iductive defiitio ad iductive hypothesis, x m+k+ = xx m+k = xx m x k = x m xx k = x m x k+ as required to prove the result for = k +. Coclusio: Hece, by iductio o, x m+ = x m x for ay o-zero real umber x ad o-egative itegers m ad. Now we prove the result for the o-positive itegers by provig that x m+ = x m x for all o-egative itegers m ad. The, by the defiitio of x m, we ca coclude that the result is true for all the o-positive itegers. Base case: For = 0, x m+ = x m = x m x 0 = x m x Iductive step: Suppose ow as iductive hypothesis that x m+k = x m x k for some o-egative iteger k. We eed to show that x m+k+ = x m x k+. The, by iductive defiitio ad iductive hypothesis, x m+k+ = xx m+k = xx m x k = x m xx k = x m x k+ as required to prove the result for = k +. Coclusio: Hece, by iductio o, x m+ = x m x for ay o-zero real umber x ad o-egative itegers m ad. Therefore x m + = x m x for ay o-zero real umber x ad ay itegers m ad. (iii) First, we prove the result for the o-egative itegers. Base case: For = 0, x m = = x 0 = x m

16 Iductive step: Suppose ow as iductive hypothesis that x m k = x mk for some o-egative iteger k. We eed to show that x m k+ = x m k+. The, by iductive defiitio ad iductive hypothesis, x m k+ = x m x m k = x m x mk = x m+mk (by part (ii) of the problem) = x m k+ as required to prove the result for = k +. Coclusio: Hece, by iductio o, x m = x m for ay o-zero real umber x ad o-egative itegers m ad. Now we prove the result for the o-positive itegers by provig that = xm x m for all o-egative itegers. The, by the defiitio of x m, we ca coclude that the result is true for all the o-positive itegers. Base case: For = 0, x m = = x 0 = x m Iductive step: Suppose ow as iductive hypothesis that = xm k x mk for some o-egative iteger k. We eed to show that x m k+ =. xm k+ The, by iductive defiitio, iductive hypothesis, ad part (ii) of the problem x m k+ = x m x m k = x m x mk = x m +mk = xm k+ as required to prove the result for = k +. Coclusio: Hece, by iductio o, = xm x m for ay o-zero real umber x ad o-egative itegers m ad. Therefore x m = x m for ay o-zero real umber x ad ay itegers m ad. 24. Fiboacci s rabbit problem may be stated as follows: How may pairs of rabbits will be produced i a year, begiig with a sigle pair, if i every moth each pair bears a ew pair which become productive from the secod moth o? Assumig that o rabbits die, express the umber after moths as a Fiboacci umber ad hece aswer the problem. Usig a calculator ad the Biet formula (Propositio 5.4.3) fid the umber after three years. The th Fiboacci umber is give by the followig formula: u = α β 5

17 where α = ad β = 5 2 ad is the umber of moths. Thus after oe year (2 moths) there are u 2 = α2 β 2 = 44 rabbits 5 ad after three years (36 moths) there are u 36 = α36 β 36 = rabbits Let u be the th Fiboacci umber (Defiitio 5.4.2). Prove, by iductio o (Without usig the Biet formula Propositio 5.4.3), that u m+ = u m u + u m u + for all positive itegers m ad. Deduce, agai usig iductio o, that u m divides u m. Strog iductio is used. Base case: For =, u m+ = u m+ = u m u + u m u 2 = u m + u m (sice, by the iductive defiitio of the Fiboacci sequece, u = u 2 = ). For = 2, u m+ = u m+2 = u m u 2 + u m u 3 = u m + 2u m (sice u 3 = u + u 2 = + = 2). Note that by the first case u m = u m+ u m. The u m+2 = u m+ u m + 2u m = u m+ + u m Iductive step: Suppose ow as iductive hypothesis that u m+k = u m u k + u m u k+ for all positive itegers k for some positive iteger k 2. We eed to show that u m+k+ = u m u k+ + u m u k+2. The, by the iductive defiitio of the Fiboacci sequece, u m+k+ = u m+k + u m+k ad by iductive hypothesis u m+k+ = u m u k + u m u k+ + u m u k + u m u k = u m u k + u k + u m u k+ + u k = u m u k+ + u m u k+2 = u m u k+ + u m u k+2 Coclusio: Hece, by iductio o, u m+ = u m u + u m u + for all positive itegers m ad. u m divides u m meas that u m = u m q for some iteger q. Base case: For =, u m = u m ad so u m divides u m Iductive step: Suppose ow as iductive hypothesis that there exists some iteger q such that u mk = u m q for some positive itegers m ad k. We eed to show that u m k+ = u m p for some iteger p. Observe that all the Fiboacci umbers are itegers sice every umber is the sum of the previous two umbers, which i tur are itegers. The, by the first part of the problem ad by iductive hypothesis, u m k+ = u mk +m = u mk u m + u mk u m+ = u mk u m + u m qu m+

18 = u m u mk + qu m+ = u m p where p = u mk + qu m+ is a iteger ad so u m divides u m k+ Coclusio: Hece, by iductio o, u m divides u m for all positive itegers m ad. 26. Suppose that poits o a circle are all joied i pairs. The poits are positioed so that o three joiig lies are cocurret i the iterior of the circle. Let a be the umber of regios ito which the iterior of the circle is divided. Draw diagrams to fid a for 6. Prove that a is give by the followig formula. a = + C,2 + C,3 + C,4 = The followig are the drawigs correspodig to a, a 2, a 3, a 4, a 5, ad a 6. For a 6, it is ot feasible to show that the ceter is ot the itersectio of three lies sice we are workig with small diagrams. However ote that the poits o the circle do ot form a regular hexago circumscribed about a circle for otherwise we would obtai three lies cocurret at the ceter of the circle. Thus there is aother regio (ot visible i such diagram) at the ceter of the figure. Base case: For =, we ca see that the iterior of the circle is divided ito oe regio ad a = + C,2 + C,3 + C,4 = = Iductive step: Suppose ow as iductive hypothesis that a k = + k k k 2 5k for some positive iteger k. We eed to show that a k+ = + k k + k k The by defiitio, a k+ = k + + C k, 2 + C k, 3 + C k, 4 k! = k + + 2! k 2! + k! 3! k 3! + k! 4! k 4! = k k k + 6 k k k 2 + k k k 2 k 3 24 = + 7 k k2 2 k k4 = + k 4 + k 2k2 + k 3 24 = + k k + k2 3k = + k k + k k as required to prove the result for = k +.

19 Coclusio: Hece, by iductio o, a = for all positive itegers.