Math2081 2012B Midterm Sol

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MATH2081

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MATHEMATICS FORCOMPUTING

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MID-SEMESTER TEST

Exam Date: 23rd July 2012

Reading Time: 9.009.10

Writing Time: 9.1010.10

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. There is a total of 100

marks. Each question is worth 20 marks. Please write at the bottom of this pagewhich 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 the Math2081 course.5. 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) Write down the following sets and then draw them into a Venn diagram.

i) A = { x | x N, x 12, x is even }ii) B = { 3*y | y N, 1 y 7 }iii) C = { z

2| z Z, -5 < z < 5 }

b) Prove the following. Let A, B be finite sets. Then | A | + | B | = | A B | iff A B =.a) A = {0, 2, 4, 6, 8, 10, 12}; B = {3, 6, 9, 12, 15, 18, 21}; C = {0, 1, 4, 9, 16}

b) Proof: Let | A | = n and | B | = m, and A = {a1, a2, ..., an} and B = {b1, b2, ..., bm}.

Assume A B , A B = {b1, ,bk} and let | A B | = k, k 1. We have | A | + | B | = n +m. But then | A B | = n + mk, since A B = {a1, , am, bk+1, , bm}. Since k 1, n + m n + m

k, and hence | A | + | B | | A B |. Let A B =. Then we get A B = {a1, a2, ..., an, b1, b2, ..., bm}, by definition of andhence | A B | = n + m = | A | + | B |.

A B

C

0, 4 9

6, 122, 8, 10 3, 15, 18,

21

1, 16


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a) Let A x A be a relation. Define the inverse relation -1 = { (b, a) | (a, b) }. Prove thefollowing. If is symmetric, then -1 is symmetric.

b) Let N x N by defined by = { (x, y) | x y + 4 }. Is reflexive, symmetric, and transitive?Is an equivalence relation? Prove your claims.

a) So let be symmetric. Now let (a, b) -1. By definition of-1 we get that (b, a) . Since issymmetric, we have (a, b) . Again by definition of-1 we get (b, a) -1. Hence -1 is symmetric.

b) Reflexive: since, for all x N we have x + 4, we get (x, x) .Not symmetric: we have (10, 15) but (15, 10) , since not 15 10 + 4.Not transitive: We have (10, 7) and (7, 4) , but obviously (10, 4) , since not 10 4 + 4.

So is no equivalence relation.


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Let a R, a 0, and let f : R0 R0 be defined by .a) Is f one-one, onto, bijective? Prove your claims.b) Does f have an inverse function? If so, find it and prove your claims. If not, prove that f can

not have an inverse function.

a) One-one: Let x y. Then x/a y/a and . So f(x) f(y).Onto: Let y R0 be given. Define x = a * y2. Then Since f is one-one and onto, f is bijective.

b) Define f-1(x) = a * x2. Let x R0 be given. Then() .


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Question 4: (5 + 5 + 5 + 5 Marks)

a) Simplify | 3| x + 7 | | for 7 < x 0 for -7 < x < -4) = |4x | =4x since4x > 0 for

7 < x


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a) Express using summation and/or production symbols; do not compute.

1) 13 + 12 + 11 + ... + 1 + 0 + (-1) + (-2) =2) 2 * 6 * 12 * 20 * ... =3) -3 + 69 + 1215 + ... =4) 1 + 22 + 333 + 4444 + ... + 999999999 =

b) Prove or disprove the following. Let n N, n 2, x1, ..., xn N. Then .

a)

1) 2) 3) 4)

b) This is not true. For example, let n = 2, x1 = 1 and x2 = 3. Then . (Most people actually used , which is also not true.)


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

b)

Prove that for all 2x2 matrices A and B we have that (A * B)

T

= B

T

* A

T

.

a) (Row 12*Row 2) { | } (Row 2 /(-3)) { |

} (Row 12*Row 2)

So the inverse matrix is [ ].

b) Just compute both sides. Let and [ ]. Then (A * B)T =[ ]

[ ] . Now compute B

T * AT = [ ]

[

]. Since both sides are equal, the claim is proven.


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Math2081 2012B Midterm Sol