Math2081 2010C Midterm Sol

7/30/2019 Math2081 2010C Midterm Sol

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MATHEMATICS FOR

COMPUTING SCIENCE

MID-SEMESTER TEST

Exam Date: 25th

November 2010

Reading Time: 9:00 am9:10 am

Writing Time: 9:10 am10:10 am

Examination Room: 1.4.20

Name:

ID No:

Group:

RULES

1. Calculators must not be taken into the examination room.2. There are 6 questions. You must answer only 5 of them. Total mark is 100. Each

question is worth 20 marks. Please write at the bottom of this page which question

you choose not to answer.

3. Show all your work to get full marks. Even if your answer is incorrect, you mightget partial credit for the right way of reasoning.

4. This examination contributes 20% to the total assessment of Math 2081 course.

1. There are total 8 pages (including cover) in this booklet.THE QUESTION THAT I CHOOSE NOT TO ANSWERIS:


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Question 1: (10 + 10 Marks)

a) Let A = {x N | 0 < 2x < 5} and B = {x Z | |x| < 4} be sets.Compute A, B, A \ B, A B, A B

b) Prove the following: Let A, B be sets. Then (This is one of DeMorganslaws. Prove it; do not just write follows from lecture.)

a) A = {1, 2}, B = { -3, -2, -1, 0, 1, 2, 3}, A \ B = , A B = A, A B = B (2 Marks each)

b) To show = we show both and .

: Let

.

Case 1: x A x B since x A B x by definition x for any set.Case 2: x B is symmetric.Case 3: x A or x B x or x by definition x .Since there are no more cases, we are done.

: Let x .Case 1: x , since A B A and hence Case 2: x , since A B A and hence There are no more cases and hence we are done.

From both and we get the claim.

(10 Marks. There are many ways to prove this. Partial marks will be awarded.)


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a) Let = { (x, y) | x y; x, y N}. Is reflexive/symmetric/transitive? Is an equivalencerelation? Prove your claims.

b) Give an example of a relation that is not reflexive, not symmetric, and transitive. Prove yourclaims.

a) This relation is not reflexive, since, for example, (1, 1) is not in . It is symmetric, since if x y,then y x. It is not transitive, since (4, 5) and (5, 4) , but (4, 4) . Since the relation isneither reflexive nor transitive, it is no equivalence relation. (3+3+3+1 Marks)

b) For example define = {(x, y) | x > y; x, y N}. Since never x > x, is not reflexive. Also isnot symmetric, since 5 > 4 but not 4 > 5. Finally, is transitive, since if a > b and b > c, then

obviously a > c. (4 Def + 2 + 2 + 2 Marks)


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Let f : R+ (2, ) be defined by f(x) = x22.

a) Is f one-one, onto, bijective? Prove your claims.b) Compute the inverse function of f. Prove your claims.a) f is one-one: Let x y. Then x2 y2, sine x > 0 and y > 0. Then x22 y22, and therefore f(x)

f(y). Hence f is one-one. Furthermore, f is onto: let any y (-2, ) be given. Then let x

= . Then f(x) = Since f is one-one and onto, f isbijective. (4 + 4 + 2 Marks)

b) Basically, we have done that already in the first part. Define f-1: (-2, ) R+ by

. Then f is total on domain f, since x + 2 > 0 and therefore f-1(x) is defined. We have to

show f-1

(f(x)) = x for all x domain f. So let x domain f be given. Then f-1

(f(x)) = f-1

(x2

2) = (5 marks for definition + 5 marks for proof)


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Question 4: (6 + 6 + 8 Marks)

a) Express the following series with summation and/or product symbols. Do not compute theresults.

1. 1 + 4 + 9 + + 225 = 2. (2) * (2 + 4) * (2 + 4 + 6) * * (2 + 4 + 6 + + 20) = 3. 0 + 2 + 6 + 12 + 20 + + 380 = (2+2+2 Marks)

b) Evaluate the following expressions; show your computation steps. (3+3 Marks)1.

3x2 = 63x-278x92/3x82. 3 - e = 2-33-2>3-e>3-31>e>0, so 3 - e = 1 (e = 2.7182)

c) Prove that x + yx + yx + y + 1 for all x, y R.Obviously 0 | xx| < 1for all x. Let a = | xx| + | yy|. Hence 0 a < 2. If 0 a < 1, thenx + y = x + y, by definition of the floor function. If 1 a < 2, then x + y = x + y +1,again by definition of the floor function. Since there are no more cases, we are done. (8 Marks,

marks for partial solutions)


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a) A skateboarder goes down a half-pipe. Every time he goes up the other side half of what he

travelled down on this side. If the half-pipe is 40m long in total, how much distance will he cover,

if he just keeps rolling?

b) Prove the following: Let an R and bn R for all n N. If an < bn for all n N, and furthermore then

a) He travels (20 + 10) + (10 + 5) + (5 + 2.5) + = 20 + 20 + 10 + 5 + = 20 + 20(1 + + + 1/8 +

) = 20 + 20 * meters. (10 Marks; partial marksawarded according to progress.)b) Of course, A < B iff 0 < BA. Since we know an < bn for all n N, we can always find cn > 0, cn Rsuch that bn = an + cn; (*). Hence we get:

So A < B iff 0 < BA iff 0 < C. But 0 < C is obvious, since all cn > 0 by assumption. So the claim is

proved. (10 Marks, partial marks awarded; other solutions are possible.)

First time

Second time

40m distance

etc


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a) Use the Gauss Jordan Algorithm to compute A-1 for the matrix

.

b) Let . Compute M2 and M3. What will Mn look like for n > 3? Explain.

a) (Row 22 x Row 1) { | } (Row 2 dividedby -7) { | } (Row 15 x Row 2). So the inverse matrix is [ ].(9 marks for steps, 1 mark for making inverse explicit in some way.)

b) , . From now on, all matrices M

n will be completely zero.

(4 marks each matrix, 2 marks for conjecture.)


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Math2081 2010C Midterm Sol