Download MATH104 Exam May 2007 Solutions: Mathematics Problems and Their Solutions and more Exams Mathematics in PDF only on Docsity!
MATH104 Exam May 2007, Solutions
All questions except q.1 are standard homework examples
- tau, omega, θ, ρ. (1 mark each.)
a) f (n) = −18 + 10n. b) f (n) = 4(n + 1) c) f (n) = n if n is even and −n − 1 if n is odd. (2 marks for a), 4 marks for b) and 6 marks for c).)
a) a 6 = 1 and a < 2. (2 marks) b) x ≤ y or y ≤ z (3 marks) c) There exist x and y with f (x) < f (y) and x ≥ y (3 marks) d) For all N ∈ N there exists x ∈ R such that f (x) ≥ N or g(x) ≥ N. (4 marks)
a) f is not injective if there exist x, y ∈ R such that f (x) = f (y) but x 6 = y. (2 marks) b) Let x, y ∈ R and suppose f (x) = f (y). Then 10 − 3 x = 10 − 3 y and rearranging gives 3x = 3y so x = y. (3 marks) c) Let x = 0, y = −1. Then f (x) = f (y) = 0 but x 6 = y. Hence f is not injective. (4 marks)
d) Let x, y ∈ R, x ≥ 0 , y ≥ 0, Suppose that f (x) = f (y). Then x^2 + x = y^2 + y, which gives x^2 − y^2 + x − y = 0, that is (x − y)(x + y) + (x − y) = 0, that is (x − y)(x + y + 1) = 0. But x > 0 and y > 0 so that the second factor cannot be 0. Hence the first factor is 0, giving x = y. (4 marks)
For integers m and n the statement m|n means that there exists an integer k with n = km. (1 mark)
a) R is not an equivalence relation. Property (ii) fails, since for example 1 R 2 is true but 2 R 1 is false (since 2 = 1 + 1). (3 marks) b) R is an equivalence relation. For let x, y, and z be any integers. Then i) x − x = 0 and 10|0 since 0 = 0 × 10, so x R x. ii) If x R y then x − y = 10k, for some k ∈ Z so y − x = 10(−k), and −k ∈ Z so y R x. iii) If x R y and y R z then x − y = 10k and y − z = 10l for some integers k, l so x − z = (x − y) + (y − z) = 10(k + l), i.e. x R z. (4 marks)
1
c) R is not an equivalence relation. In this case only property (iii) fails. For example, 2 R 6 since 2 divides 2 and 6, and 6 R 9 since 3 divides 6 and 9, but 2 R 9 is false since there is no prime dividing both 2 and 9. (6 marks)
a) Let m and n be integers and suppose that m and n are both odd. Then m = 2k+1, n = 2l+ for some integers k, l. Hence mn = (2k + 1)(2l + 1) = 2(2kl + k + l) + 1, which is of the form 2 k′^ + 1 for an integer k′^ and hence odd. The converse states: Let m, n ∈ Z. If mn is odd then m and n are odd. This is true: Suppose that mn is odd and that m or n is even, say m. Then m = 2k for an integer k so that mn = 2kn is even, giving a contradiction. [Or we could prove the contrapositive: if m is even or n is even, then mn is even, in the same way.] (6 marks) b) Suppose, for a contradiction, that m and n are integers and that 6m + 9n = 22. Then 3(2m + 3n) = 22 and it follows that 3|22. But this is false since 22 = 3 × 7 + 1. This contradiction proves the result. (3 marks) c) The contrapositive is: Let a, b ∈ R. If (a + b)^2 ≤ 4 ab then a = b. Proof: From (a + b)^2 ≤ 4 ab we have a^2 + 2ab + b^2 ≤ 4 ab, so (a − b)^2 ≤ 0. This is possible only if a − b = 0, that is a = b. (6 marks)
Context X is a subset of euclidean n-space. (1 mark) Hypothesis X is closed and X is bounded. (1 mark) Conclusion X is compact.(1 mark) Contrapositive Let X be a subset of euclidean n-space. If X is not compact then X is not closed or X is not bounded. (2 marks) a) Nothing: hypothesis of theorem not fully satisfied. (2 marks) b) X is not closed or X is not bounded. From contrapositive.(2 marks) c) X is not bounded. From (b) and ruling out ‘not closed’. (2 marks) d) Nothing: not hypothesis of theorem or contrapositive. (2 marks) However if the converse of the theorem is true then X compact would imply both X closed and X bounded. Since neither is the case we can deduce X is not compact. (We could deduce this merely from one of ‘X not closed’, ‘X not bounded’.) (2 marks)
a) False. (x = 0 gives cos x = 1.) (2 marks) b) True. (Take n = −3.) (3 marks) c) False. (Take x = 2. Then there is no y with cos y = x.) (3 marks) d) True. (Take m = 1.) (3 marks) e) False. (Take y = 12. Then x < y − 1 = − 12 cannot hold for x ∈ R+.) (4 marks.)
2
