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# Mid2Soln.pdf

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School
University of Toronto Scarborough
Department
Mathematics
Course
MATA31H3
Professor
Leo Goldmakher
Semester
Fall

Description
Carefully tear this page off the rest of your exam. U NIVERSITY OF TORONTO S CARBOROUGH MATA31H3 F : Calculus for Mathematical Sciences I R EFERENCE S HEET Properties of real numbers Throughout this course, we assume that there exists a set R endowed with two binary operations + and ▯ satisfying all the following properties: Algebraic Properties: (A1) 8a;b 2 R, we have a + b = b + a. (A2) 8a;b;c 2 R, we have a + (b + c) = (a + b) + c. (A3) There exists a number 0 2 R such that 8a 2 R, a + 0 = a. (A4) For each a 2 R there exists a number ▯a 2 R such that a + (▯a) = 0: (M1) 8a;b 2 R, we have a ▯ b = b ▯ a. (M2) 8a;b;c 2 R, we have a ▯ (b ▯ c) = (a ▯ b) ▯ c. (M3) There exists a number 1 2 R such that 8a 2 R, a ▯ 1 = a. Moreover, 1 6= 0. (M4) For each a 2 Rnf0g, there exists a numbe2 R such that a ▯ = 1: (D) 8a;b;c 2 R, we have a ▯ (b + c) = a ▯ b + a ▯ c. Order Properties: There exists a special subset P ▯ R such that (O1) 0 62 P (O2) For all a 2 Rnf0g, a 62 P iff ▯a 2 P. (O3) If a;b 2 P, then a + b 2 P and a ▯ b 2 P. Completeness Property (C) If A ▯ R is nonempty and bounded above, then there exists a least upper bound of A. N OTATION supA, the supremum of A, is the least upper bound of A (if it exists). inf A, the inﬁmum of A, is the greatest lower bound of A (if it exists). The symbol a ▯ b means a + (▯b). The symboleans a ▯ b. n+1 n 1 For all n 2 N, we deﬁne the symbol := a ▯ a, where a := a. Given a;b 2 R, the notation a > b means that a + (▯b) 2 P, and b < a is another way of writing a > b. The symbol a ▯ b means that a + (▯b) 2 P [ f0g, and b ▯ a means the same as a ▯ b. U NIVERSITY OF TORONTO SCARBOROUGH MATA31H3 : Calculus for Mathematical Sciences I MIDTERM EXAMINATION # 2 November 9, 2012 Duration – 2 hours Aids: none NAME (PRINT): Last/Surname First/Given Name (and nickname) STUDENT NO: KEY TUTORIAL: Tutorial section # Name of TA Qn. # Value Score COVER PAGE 5 1 20 2 20 3 12 4 18 5 15 6 10 Total 100 TOTAL: Please read the following statement and sign below: I understand that any breach of academic integrity is a violation of The Code of Behaviour on Academic Matters. By signing below, I pledge to abide by the Code. SIGNATURE: Midterm Exam # 2 MATA31H3 page 2 of 8 (1) Prove each of the following, explicitly justifying every step of your proof with the appro- priate property of R. (a) (8 points) 2 ▯ 2 = 4. We have 2 ▯ 2 = 2 ▯ (1 + 1) by deﬁnition of 2 = 2 ▯ 1 + 2 ▯ 1 by (D) = 2 + 2 by (M3) = (1 + 1) + (1 + 1) = 1 + 1 + 1 + 1 by (A2) = 4 by deﬁnition of 4: (b) (12 points) If a 2 (0;2), then a < 4. If a 2 (0;2), then a > 0 and a < 2, or in other words, a 2 P and 2 ▯ a 2 P: By (O3), we deduce that a + a 2 P. Note that (2 ▯ a) + (a + a) = 2 + (▯a) + a + a by (A2) = 2 + 0 + a by (A4) = 2 + a by (A3): Since 2 ▯ a 2 P and a + a 2 P, (O3) implies that 2 + a 2 P. Applying (O3) yet again to the two numbers 2 ▯ a and 2 + a, we deduce that (*) (2 ▯ a) ▯ (2 + a) 2 P: Finally, we have ▯ ▯ ▯ ▯ (2 ▯ a) ▯ (2 + a) =(2 ▯ a) ▯ 2 + (2 ▯ a) ▯ a by (D) ▯ ▯ ▯ ▯ = 2 ▯ (2 ▯ a) + a ▯ (2 ▯ a) by (M1) ▯ ▯ ▯ ▯ = 2 ▯ 2 + 2 ▯ (▯a) + a ▯ 2 + a ▯ (▯a) by (D) = 2 ▯ 2 + 2 ▯ (▯a) + a ▯ 2 + a ▯ (▯aby (A2) = 4 + 2 ▯ (▯a) + a ▯ 2 + a ▯ (▯a) by part (a) = 4 + 2 ▯ (▯a) + 2 ▯ a + a ▯ (▯a) by (M1) ▯ ▯ = 4 + 2 ▯ (▯a) + a + a ▯ (▯a) by (D) ▯ ▯ = 4 + 2 ▯ a + (▯a) + a ▯ (▯a) by (A1) = 4 + 2 ▯ 0 + a ▯ (▯a) by (A4) = 4 + 0 + a ▯ (▯a) by proof in lecture ▯ 2 2 ▯ = 4 + a + (▯a ) + a ▯ (▯a) by (A4) ▯ ▯ = 4 + (▯a ) + a 2 + a ▯ (▯a) by (A1) = 4 ▯ a + a + a ▯ (▯a) by (A2) ▯ ▯ 2 = 4 ▯ a + a ▯ a + (▯a) by (D) 2 = 4 ▯ a + a ▯ 0 by (A4) 2 = 4 ▯ a + 0 by proof from lecture = 4 ▯ a2 by (A3): Combining this with (*) above, we see that 4▯a 2 P, or equivalently, a < 4. ▯ continued on page 3 Midterm Exam # 2 MATA31H3 page 3 of 8 p (2) (20 points) Prove that 5 exists. In other words, prove that there exists a positive number 2 x 2 R satisfying x = 5. [You do not need to explicitly use the algebraic and order proper- ties in this problem, or in any of the following problems.] Let A := fx > 0 : x < 5g. Clearly A 6= ; (since 2 2 A). Also, observe that A is bounded above by 3, since 2 3 < x =) x > 9 =) x 62 A: Thus, by Completeness, supA exists. Let ▯ = supA: 2 I claim that ▯ = 5; this will prove the claim. 2 Suppose ▯ < 5. Note that in this case, ▯ 2 A. (▯ is clearly positive, since it is an upper bound of A and 2 2 A.) Since ▯ < 5, we have 5▯▯2 2 R. The Archimedean Property therefore implies the existence of a natural number n such that 2▯ + 1 n > ; 5 ▯ ▯2 from which we deduce that 2▯ 1 2 + < 5 ▯ ▯ : n n Since 2▯+ 1 ▯ 2▯+ 1 for all n 2 N, we see that n n2 n n 2▯ 1 + < 5 ▯ ▯ ; n n2 or in other words, that (▯ + 1=n) < 5: But this means that ▯ + 1=n 2 A, contradicting that ▯ is an upper bound of A. 2 2▯ Next, suppose instead that ▯ > 5. Then ▯2▯5 2 R, so by Archimedean Proper
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