Homework 5 Solutions

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1 Homework 5 Solutions ECS 0 (Fall 17) Patrice Koehl koehl@cs.ucdavis.edu ovember 1, 017 Exercise 1 a) Show that the following statement is true: If there exists a real number x such that x 4 +1 = 0, then x = 1. Let x be a real number and let P be the statement considered. P is an implication of the form p q with p defined as x is a real number with x = 0 and q defined as x = 1. We note that p is false, the proposition P is therefore always true. b) If x and y are real numbers such that x < y, show that there exists a real number z with x < z < y. This is an existence proof: we only need to find one example. Let us define z = x+y. We show that x < z and z < y. z x = x+y x = y x > 0 as x < y. Similarly, y z = y x+y = y x > 0 as x < y. x < z < y: we found one real number z that satisfies x < z < y: this concludes the proof. Exercise Let x be a real number. Show that 4x = x + x x x Let us write x = n + ɛ, where n is an integer and ɛ is a real number and 0 ɛ < 1. n is the largest integer that is smaller than x; by definition, n = x. We use a proof by case (similar to the proof used in class for x ): a) : If 0 ɛ < 1/4, then 0 4ɛ < 1, 0 < ɛ + 1/4 < 1, 0 < ɛ + /4 < 1 and 0 < ɛ + 3/4 < 1., 4x = 4n + 4ɛ = 4n 1

2 and x + x + 1/4 + x + /4 + x + 3/4 = n + ɛ + n + ɛ + 1/4 + n + ɛ + /4 + n + ɛ + 3/4 = n + n + n + n = 4n b) If 1/4 ɛ < /4, then 1 4ɛ <, 0 < ɛ + 1/4 < 1, 0 ɛ + /4 < 1, and 1 ɛ + 3/4 <., and 4x = 4n + 4ɛ = 4n + 1 x + x + 1/4 + x + /4 + x + 3/4 = n + ɛ + n + ɛ + 1/4 + n + ɛ + /4 + n + ɛ + 3/4 = n + n + n + n + 1 = 4n + 1 c) If /4 ɛ < 3/4, then 4ɛ < 3, 0 < ɛ + 1/4 < 1, 1 ɛ + /4 <, and 1 ɛ + 3/4 <., 4x = 4n + 4ɛ = 4n + x + x + 1/4 + x + /4 + x + 3/4 = n + ɛ + n + ɛ + 1/4 + n + ɛ + /4 + n + ɛ + 3/4 = n + n + n n + 1 = 4n + d) If 3/4 ɛ < 1, then 3 4ɛ < 4, 1 < ɛ + 1/4 <, 1 ɛ + /4 <, and 1 ɛ + 3/4 <., 4x = 4n + 4ɛ = 4n + 3 x + x + 1/4 + x + /4 + x + 3/4 = n + ɛ + n + ɛ + 1/4 + n + ɛ + /4 + n + ɛ + 3/4 = n + n n n + 1 = 4n + 3 Based on the method of proof by case, we conclude that 4x = x + x x x is true for all x.

3 Exercise 3 This is a generalization of exercise 3: Let x be a real number and an integer greater or equal to 3. Show that x = x + x x x +. We could use a proof by case that generalizes the solution described for exercise, using case; there is however a faster and maybe more elegant solution. Let us define: f(x) = x x x + 1 x x + We show first that f(x) is periodic, with 1 being one period. For this, we need to show that: x R, f ( x + 1 ) = f(x) Let x be a real number. otice that: ( f x + 1 ) = (x + 1 ) x + 1 x +... x x = x + 1 x x + 1 x + 1 = x + 1 x x + 1 x 1 = f(x) Since this is true with no conditions on x, it is true for all x, and therefore f is periodic, with 1/ being one period. A periodic function needs to be defined only on one period, here in the interval [ 0, 1 ). Let x be in this interval. : f(x) = 0. 0 x < 1 < 1 0 x + 1 < = < x + 1 < = = 1 0 x < 1 = 1 Since f(x) = 0 on one of its period, we have f(x) = 0 x = x + x x x + x R. : Exercise 4 Let x be a real number. Show that x + x+1 = x. Let x = n, where n is an integer. By definition of floor, we have: n x < n + 1. We consider two cases: 3

4 1) n is even: there exists an integer k such that n = k. We can rewrite the inequality above as: Similarly, k x < k + 1 k x < k + 1 < k + 1 x = k. (1) k + 1 x + 1 < k + k < k + 1 x + 1 < k + 1 x + 1 = k () Combining equations (1) and (), we get x + x+1 = k = n = x ) n is odd: there exists an integer k such that n = k +1. We can rewrite the inequality above as: Similarly, k + 1 x < k + k < k + 1 < x < k + 1 x = k. (3) k + x + 1 < k + 3 k + 1 x + 1 < k + 3 < k + x + 1 = k + 1 (4) Combining equations (3) and (4), we get x + x+1 = k + k + 1 = n = x 4

5 Exercise 5 Let x be a real number. Solve the following equations: a) 3x 16 = x Let x be a real number. We notice first that 3x 16 is an integer., if x is a solution of the equation 3x 16 = x, then x has to be an integer. Since x is an integer, 3x 16 is also an integer, and therefore 3x 16 = 3x 16. The equation then becomes 3x 16 = x where x is an integer, therefore x = 8. This is the only solution. b) x 3x = 6 Let x be a real number. We notice first that x 3x + 5 is an integer, which we write as k. If there exists a solution x to the equation, then k 1 = 6, where k is the integer defined before, and therefore k = 5. However, k is even, and 5 is odd. We have reached a contradiction. the proposition, there exists a solution x to the equation, is false, and the equation does not have a solution. Extra Credit Let x and y be two real numbers such that 0 < x y. We define: a) The arithmetic mean m of x and y: m = x + y b) The geometric mean g of x and y: g = xy c) The harmonic mean h of x and y: Show that: h = ( 1 x + 1 y ) x h g m y We will proceed by steps: a) Let us show first that: i) x m y otice that: m x = x+y x = y x 0 since y x; therefore m x. Similarly, y m = y x y = y x 0; therefore y m. ii) x g y otice that g x = xy x x = x ( ) y x. Since x y and square root is an increasing function, g x 0; therefore g x. Similarly, y g = y y x y = y ( ) y x ; therefore y g 0 and y g. iii) x h y otice that 1 h is the arithmetic mean of 1 x and 1 y. From above, we can say that 1 y 1 h 1 x from which we deduce that x h y. 5

6 b) g m ote that m g = x+y x y = x+y x y = ( x y) 0; therefore m g. c) h g We note again that 1 h is the arithmetic mean of 1 x and 1 y. The geometric mean of 1 x and 1 y is 1 x 1 y, i.e. 1 g. From b) above, we have 1 g 1 h, therefore h g. From a), b), and c), we can conclude that x h g m y. 6