Practice Problem Midterm 2

1. Let f (x) : R ! R be a function and L a real number. L is the limit of f (x) as x approaches to infinity, (written L = lim f (x)), if for any ✏ > 0, x!+1 there is a positive constant M so that |f (x) L| < ✏ when x > M (a) Use the bi-conditional sentence and symbols to rewrite the definition (Do not use words). L = lim f (x) () x!+1 Solution: L = lim f (x) () 8✏ > 0, 9M > 0, 8x 2 R, x > M =) |f (x) x!+1 L| < ✏ (b) Write down a useful denial; In other word explain what it means that L 6= lim f (x). x!+1 L 6= lim f (x) () x!+1 Solution: L 6= lim f (x) () 9✏ > 0, 8M > 0, 9x 2 R, x > M ^ |f (x) x!+1 L| ✏. (c) Use the definition to prove that lim (1 + x1 ) = 1 x!+1 Solution: For any ✏ > 0, we choose M = 1✏ . Hence, we have that x>M = 1 1 1 =) 0 < < ✏ =) | + 1 ✏ x x 1| < ✏. (d) Use (b) to prove that lim ( x1 ) 6= 2 x!+1 Solution: We take ✏ = 1. For any M > 0, we choose that x = M + 1. Hence, we have that x>M ^| 1 x 2| = 2 1 >2 M +1 1 = 1 = ✏. 2. Let a be a positive integer. We say a is called perfect if there is a positive integer n so that a = n2 . (a) Give a symbolic form of the definition (do not use words) a is perfect () Solution: a is perfect () 9n 2 N, a = n2 (b) Write down a useful denial; In other word explain what it means that a is not perfect. a is not prefect () Solution: a is not prefect () 8n 2 N, a 6= n2 (c) Use the direct proof to show that if both a and b are perfect, then ab is also perfect. Solution: Proof. Since a is perfect, there is N1 2 N so that a = N12 . Similarly, since b is perfect, there is N2 2 N such that b = N22 . Hence, we have that ab = N12 N22 = (N1 N2 )2 . Because N1 N2 is an integer, then ab is perfect. 3. Let m and n be two real numbers. If m > 1 and n > 1, then mn + 1 > m + n (a) Give a symbolic form of the statement. (Do not use words). Hint: Do not forget the universal quantifiers. Solution: 8m 2 R8n 2 R, n > 1 ^ m > 1 =) mn + 1 > m + n (b) Use the direct argument to prove the statement. Solution: Proof. Because n > 1 and m > 1, then we have that m 1 > 0 and n > 1. Hence, we have that (n 1)(m 1) > 0 =) mn m n + 1 > 0 =) mn + 1 > m + n. (c) Prove that if x = m or x = n, then x3 (m + n + 1)x2 + (n + m + mn)x mn = 0 Solution: Proof. When x = n, we have that x3 (m + n + 1)x2 + (n + m + mn)x mn = n3 (m + n + 1)n2 + (n + m + mn)n mn = 0. mn = m3 (m + n + 1)m2 + (n + m + mn)m mn = 0. When x = m, we have that x3 (m + n + 1)x2 + (n + m + mn)x Page 2 (d) Recall that (Vieta’s formulas) Let x1 , x2 and x3 be three roots of ax3 + bx2 + cx + d = 0, where a 6= 0. Then, we have x1 + x2 + x3 = ab . Use the Vieta’s formula to find the third root of the equation in (c). (You do not need to prove the Vieta’s formula.) Solution: Let x3 be the third root. From Vieta’s formula, we have that m + n + x3 = (m + n + 1) = m + n + 1. 1 Then, we have that x3 = 1. 4. Let a and b be two positive integers (a) use the proof by contradiction to show that if a3 is divisible by 3, then a is divisible by 3. (Hint: (1)If x is divisible by 3, then x = 3k + 1 for some k 2 Z or x = 3k + 2 for some k 2 Z. (2) We have the following equation (a + b)3 = a3 + 3a2 b + 3ab2 + b3 Solution: Proof. Suppose that a is not divisible by 3 and a3 is divisible by 3. Then a = 3k + 1 for some k 2 Z or a = 3k + 2 for some k 2 Z. We have two cases: Case I: If a = 3k + 1 for some k 2 Z, then we have that a3 = (3k + 1)3 = 1 + 9k + 27k 2 + 27k 3 = 1 + 3(3k + 9k 2 + 9k 3 ). That is to say, a3 is not divisible by 3. It is a contradiction with the assumption 3 | a3 . Case II: If a = 3k + 2 for some k 2 Z, then we have that a2 = (3k + 2)3 = 8 + 36k + 54k 2 + 27k 3 = 2 + 3(2 + 12k + 18k 2 + 9k 3 ). That is to say, a3 is not divisible by 3. It is a contradiction with the assumption 3 | a3 . (b) Use (a) to prove that if a3 = 3b3 , then both a and b are divisible by 3. Solution: Proof. Since a3 = 3b3 , then 3 | a3 . We use (a) to get that 3 | a. Then, there is a k 2 Z such that a = 3k. Hence, we have that 3b3 = a3 = (3k)3 = 27k 2 =) b3 = 9k 3 . We get that 3 | b3 . Then, we use (a) again to get that 3 | b. Page 3 p (c) Use (b) to prove by contradiction that 3 3 is not rational. (Hint: For any positive rational number x, x can be written as follows, p x= , q where p and q are positive integers and they have a unique common divisor, 1.) Solution: Proof. Suppose p • 3 3 = pq p 3 3 is rational. Then, there is two positive integers p and q satisfying that • p and q has a unique common divisor, 1. Then, we have that p3 = 3q 3 . We use (b)to get 3 | p and 3 | q. 3 is a common divisor of p and q, which leads to a contradiction with the fact that p and q has a unique common divisor, 1. 5. Let {an }1 n=1 be a sequence of real numbers: a1 = 2; an = 12 an 1 + 1 2 for n > 1 (a) Use the inductive argument to prove that an = 1 2n 1 + 1 for n 1. Solution: Proof. When n = 1, a1 = 2 = 1 + 211 1 . Suppose that an = 2n1 1 + 1 is true. We have that 1 1 an+1 = an + 2 2 1 1 1 = ( n 1 + 1) + 2 2 2 1 = n + 1. 2 (b) Use the inductive argument to prove that k P n=1 Page 4 an = k + 2 1 2k 1 . Solution: P Proof. When k = 1, then 1n=1 an = a1 = 2 = 2 + 1 211 1 . k P Suppose that an = k + 2 2k1 1 is true. We have that n=1 k+1 X n=1 an = k X an + ak+1 n=1 =k+2 =k+2 =k+3 1 2k 1 1 2k 1 1 . 2k + ak+1 + 1 +1 2k (c) Let {xn }n=1 be a sequence of real numbers and L a real number. L is called the limit of the sequence {xn }n if 8✏ > 0, 9N 2 N, 8n 2 N n N =) |xn L|  ✏ Notations: L = lim xn . n!1 Use the definition to prove that lim an = 1 n!1 Solution: Proof. For any ✏ > 0, we choose that N = [log2 ( 1✏ )] + 1 + 1. Then we have that for any n 2 N 1 n > N > log2 ( ) + 1 =) n ✏ 1 1 1 > log2 ( ) =) n 1 < ✏ =) |an ✏ 2 1| = |1 + 1 2n 1 We can conclude that lim an = 1. n!1 6. Let A be a finite subset of Z. We define the maximum of A as follows: max A = {x 2 A | 8a 2 A, a  x}. Example: If A = {1, 2, 3, 5}, then max A = 5. If A = {2, 5, 10}, then max A = 10. (a) Let a and b two real integers. If A = {a, a + b}, find max A. Page 5 1| < ✏. Solution: If b 0, then max A = a + b. If b < 0, then max A = a. (b) Let C be non-empty subset of Z. C has no maximum if and only if for any a 2 C, there is a1 2 C so that a1 > a. Rewrite the right-hand side statement in the symbolic form. C has no maximum () Solution: C has no maximum () 8a 2 C, 9a1 2 C, a1 > a. (c) (Bonus) Use (b) and inductive argument to prove that if C has no maximum, there is an increasing sequence {an }n in C (equivalently, {an }1 n=1 satisfies an+1 > an ). (You don’t prove the statement (b) and You can directly use it. ) Solution: Proof. We use the induction to construct the increasing sequence {an }n=1 When n = 1, choose any element a 2 C and define a1 = a. Suppose that we find a finite sequence of {ak }nk=1 in C with ak < ak+1 for k  n Since C has no maximum, there is an+1 2 C with an < an+1 . 1. (d) (Bonus) Let B be a finite subset of Z. Use (c) to prove by contradiction that B has a maximum. Solution: Proof. Suppose that a finite set B has no maximum. We use (c) to find an increasing sequence of {an }1 n=1 in B. Then, B has infinitely many elements which leads to contradiction with the fact that B is a finite set. (e) (Bonus) Let A be a finite subset of Z. Modify the definition of maximum to give the definition of the minimum. Solution: min A = {x 2 A | 8b 2 C x Page 6 b}