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chapter 1 61 - 62 CHAPTER 1 PRINCIPLES OF PROBLEM SOLVING...

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Unformatted text preview: 62 CHAPTER 1 PRINCIPLES OF PROBLEM SOLVING 14. Assume that 10g2 5 is rational. Then log2 5 = m/n for natural numbers m and 71. Changing to exponential form gives us 2”” = 5 and then raising both sides to the nth power gives 2’” : 5”. But 2’" is even and 5" is odd. We have arrived at a contradiction. so we conclude that our hypothesis. that 10g2 5 is rational. is false. Thus, log2 5 is irrational. 15. Let d be the distance traveled on each half of the trip. Let t1 and t2 be the times taken for the first and second halves of the trip. For the first half of the trip we have t1 = d/ 30 and for the second half we have 152 : d/ 60. Thus, the average speed total distance 2d 2d - 60 i 120d ( 120d totaltime _t1+t2_i+i 60—2d+dT 3d 30 60 for the entire trip is = 40. The average speed for the entire trip is 40 mi/h. 16. Let f = sin. 9 = :c. and h : 1:. Then the left—hand side of the equation is f o (g + h) = sin(33 + m) : sin 29: : 25inmc0s av: and the right—hand side is f o g + f o h = sin m + sina: : 2 sinac. The two sides are not equal, so the given statement is false. 17. Let Sn be the statement that 7" i 1 is divisible by 6. 0 5'1 is true because 71 i 1 : 6 is divisible by 6. 0 Assume Sk is true. that is. 7’“ - 1 is divisible by 6. In other words. 7’“ — 1 : 6m for some positive integer m. Then 7k+1 — 1 = 7’“ - 7 — 1 : (6m + 1) - 7 1 1 : 42m + 6 = 6(7m + 1). which is divisible by 6. s0 Sk+1 is true. 0 Therefore. by mathematical induction. 7" A 1 is divisible by 6 for every positive integer n. 18. Let Sn be the statement that 1 + 3 + 5 + - ~ - + (2n — 1): n2. 0 Si is true because [2 (1) A 1]: 1 = 12. 0 Assume Sic is true. that is, 1 + 3 + 5 + - - - —l— (2k — 1) 2 k2. Then 1+3+5+-~-+(2k—1)+[(2k+1)g1]:1+3+5+~--+(2k—1)+(2k+1) :h2+(2k+1):(k+1)2 which shows that Sk+1 is true. 0 Therefore. by mathematical induction. 1 + 3 + 5 + ~ ~ - + (2n 7 1) = n2 for every positive integer n. 19, f0(;1;) : 1:2 and fn+1(-73) : f0(fn($)) fOI‘TL Z 0. 1, 2 ..... f1(w) : f0 (fotmD : f0 ($2) : (W = 51120:): fo(f1(m)) = fo(rn4) : (2:4)2 : m . 2n+1 f3(m) : f0 (f2(x)) : f0 (x8) : (31:8)2 : 1:16 ..... Thus. ageneral formula is fn(w) : :r, ...
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