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# Mid1Soln.pdf

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

Description
U NIVERSITY OF TORONTO SCARBOROUGH MATA31H3 : Calculus for Mathematical Sciences I MIDTERM EXAMINATION # 1 October 3, 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 15 4 15 5 10 6 15 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 # 1 MATA31H3 page 2 of 8 (1) (20 points) Use induction to prove that Xn 4 = 4 (4 ▯ 1) 8n 2 N: 3 k=1 Xn k 1 2 n [Recall that4 is a fancy way of writing 4 + 4 + ▯▯▯ + 4 .] k=1 Proof: By induction. Let n n o X k 4 n A = n 2 N : 4 = (4 ▯ 1) : k=1 3 4 1 2 A, since 4 3 (4 ▯ 1). Now, suppose n 2 A. Then from the deﬁnition of A, Xn k 4 n 4 = 3 (4 ▯ 1): k=1 n+1 Adding 4 to both sides yields n+1 X k 4 n n+1 4 = (4 ▯ 1) + 4 k=1 3 n+1 n+1 4 ▯ 4 3 ▯ 4 = + 3 3 4 ▯ 41▯ 4 = 3 4n+2▯ 4 = 3 4 n+1 = (4 ▯ 1): 3 Thus, we see that whenever n 2 A, we must have n + 1 2 A as well. By induction, we conclude that A = N, which is precisely what is claimed. ▯ continued on page 3 Midterm Exam # 1 MATA31H3 page 3 of 8 p (2) (20 points) Prove tha5 62 Q. There are (at least!) two different approaches: p Proof 1: Suppose 5 2 Q. Then 9a 2 Z and 9b 2 N such that a is fully reduced, b and p a 5 = : b Squaring both sides and simplifying shows that 2 2 (y) a = 5b : 2 In particular, a is a multiple of 5. It follows (see Lemma below) that a is a multiple of 5. So, we can write a = 5k for some k 2 Z. Plugging this back into (y) and simplifying yields 2 2 b = 5k : Exactly as above, this implies that b is a multiple of 5. Thus, both a and b must be a multiples of 5. But this contradicts our assumptionbtis fully reduced! ▯ 2 Lemma: Suppose a 2 Z, and a is a multiple of 5. Then a is also a multiple of 5. Proof. Let k be the smallest integer such that 5k > a. (Such an integer exists by the Well-Ordering Property of N.) This implies that 5(k ▯ 1) ▯ a, since otherwise, k wouldn’t be minimal. So, we have 5k ▯ 5 ▯ a < 5k: This means that a = 5k ▯ r; where r = 1;2;3;4; or 5. Moving things around and squaring both sides, we see that 2 2 2 r = a ▯ 10kr + 25k : Note that the entire right hand side is a multiple of 5, since a is a multiple of 5 by 2 2 hypothesis, and 10kr and 25k are clearly multiples of 5. We thus conclude that r must be a multiple of 5. There are only ﬁve possible values of r, and it’s easy to check that the only one of these for which r is a multiple of 5 is r = 5. Thus, we conclude that a = 5k ▯ 5, which is a multiple of 5 as claimed. ▯ —————————- —————————- p Proof 2: Let A = fn 2 N : n 5 2 Zg. If A = ;, then we’re done. Suppose instead that A 6= ;. By the Well-Ordering Property of N, A must have some least element; call it a. I now claim that p (*) a( 5 ▯ 2) 2 A: p Since a( 5 ▯ 2) < a, this would contradict the minimality of a. So, why is (*) true? First, note that a( 5 ▯ 2) 2 Z, since a 5 2 Z and 2a 2 Z. Next, p since a( 5 ▯ 2) > 0, we deduce that it is actually a natural number. Finally, we have p p p a( 5
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