MMW - Mathematics, Patterns PDF

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GEED 10053Mathematics in the Modern WorldLesson 1: Mathematics in Our WorldAssessmentI. Patterns and Numbers in Nature(1) Give five examples each of nature having reflection symmetry and radial symmetry. Reflection Symmetry: paper airplane, mcdo logo, onion, beetle, birdsRadial Symmetry: flower peta...

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GEED 10053

Mathematics in the Modern World

Lesson 1: Mathematics in Our World Assessment I. Patterns and Numbers in Nature (1) Give five examples each of nature having reflection symmetry and radial symmetry. Reflection Symmetry: paper airplane, mcdo logo, onion, beetle, birds

Radial Symmetry: flower petals, sea urchins, leaves, jellyfish, corals

(2) Compare and contrast (a) rotation and reflection; (b) translation and rotation. a. Rotation is when a shape or pattern can be rotated or turned around a central point, and remains the same. It may be stated that a shape or pattern has a rotational symmetry of order x; this means that the shape can be turned around a central point and remains the same x times. While reflection is when a shape or pattern is reflected in a line of symmetry or a mirror line. The reflected shape or pattern will be exactly the same as original, the same distance from the mirror line and the same size. b. Translation simply moves the graph or pre-image without changing the size, shape and orientation or spinning the image, without any rotation or reflection. On the other hand, rotation spins the pre-image around a central or fixed point. Rotation is rotating an object about a fixed point without changing its size or shape

(3) Which upper case letters of the English alphabet look the same after being rotated 900? 1800? 90 degrees - O, X 180 degrees – H, I, N, O, S, X, Z

(4) Classify the following frieze patterns based on Conway’s classification.

a. Hop b. Step

c. Jump d. Step

e. Spinning Hop

II. Fibonacci Sequence (1) Enumerate the first twenty Fibonacci numbers. (starting 0) = 0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181 (starting 1) = 1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765

(2) Use F40 = 63,245,986 and F38 = 39,088,169 to find the value of F39. Show your reasoning. There are two ways to solve the value of 𝐹 . 39

First, is to simply subtract the value of𝐹40 to 𝐹38 . It is a unique property of Fibonacci numbers: each fibonacci number is a sum of two preceding fibonacci numbers. 𝐹 =𝐹 -𝐹 39

40

38

𝐹39 = 63,245,986 - 39,088,169

𝐹

39

= 24,157,817

Second, is to get the square root of the product of the two values. 𝐹 𝐹 𝐹 𝐹

39

39

39

39

= 𝐹

40

. 𝐹

38

= 63, 245, 986 . 39, 088, 169 = 2472169789339634 = 49720919.0315 ≈ 49720919

The values we came up using the two ways aren’t the same because obviously the given values for 𝐹 and 𝐹 are not Fibonacci numbers. Assuming the given values as Fibonacci numbers, 40

38

the value of 𝐹

will be either 24,157,817 or 49720919.

39

(3) Using the Binet’s formula, calculate F4

𝑛=4 𝑛

𝐹𝑛 =

(1+ 5) − (1− 5) 2

𝑛

5

4

4

(1+ 5) − (1− 5)

𝐹4 = 𝐹 4=

𝑛

4

2 5 48 5 16 5

𝐹4=3

Lesson 2: Logic and Sets Assessment 1. Write each statement in words. Let p: The plane is on time. Let q: The sky is clear. (a) p ∧ (¬ q)

The plane is on time and the sky is not clear.

(b) q → (p ∨ ¬p)

If the sky is clear then either the plane is on time or the plane is not on time.

(c) p ↔ q

The plane is on time if and only if the sky is clear.

2. Construct a truth table for each proposition. (a) [(p ∧ q) ∨ r] ↔ [(p ∧ r) ∨ (q ∧ r)] p

q

r

p∧q

p∧r

q∧r

(p∧q) ∨ r

(p∧r) ∨ (q∧r)

[(p∧q) ∨ r] ↔ [(p∧r) ∨ (q∧r)]

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(b) [(p ∧ r) → (q ∧ ¬r)] → [(p ∧ q) ∨ r)] p

q

r

¬r

p ∧r

q∧¬r

p ∧q

[(p∧r) →

[(p ∧ q) ∨ r)]

[(p∧r) → (q∧ ¬r)] → [(p ∧ q) ∨ r)]

(q∧¬r)]

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3. Prove the De Morgan’s Laws by constructing truth tables. (a) ¬(p ∨ q) ⇐⇒ (¬ p) ∧ (¬ q) p

q

¬p

¬q

p∨q

(¬ p ∨ q)

(¬ p) ∧ (¬ q)

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(b) ¬(p ∧ q) ⇐⇒ (¬ p ∨ (¬ q) p

q

¬p

¬q

p∧ q

(¬ p ∧ q)

(¬ p)∨ (¬ q)

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4. Let U := Letters in the English Alphabet = {a, b, c, . . . ,x, y, z} A = {t, r, i, a, n, g, l, e, s} B = {s, q, u, a, r, e }; C = {h, e, x, a, g, o, n, s } Determine the following: (a) A ∪ (B ∩ C)

=

{t,r,i,a,n,g,l,e,s}

(b) (A ∪ B)’ ∩ C

=

{h,o,x}

(c) (A ∩ C) ∪ (B ∩ C)

=

{a,n,g,e,s}

(d) A ∩ (C ∩ U )’

=

{t,r,i,l}

(e) n[(A ∪ B) ∩ (B ∪ C)]

=

8

5. A survey of 90 customers was taken at Barnes & Noble regarding the types of books purchased. The survey found that 44 purchased mysteries, 33 purchased science fiction, 29 purchased romance novels, 13 purchased mysteries and science fiction, 5 purchased science fiction and romance novels, 11 purchased mysteries and romance novels, and 2 purchased all three types of books (mysteries, science fiction, romance novels). How many of the customers surveyed purchased

(a) mysteries only? =n(M) − [ n(SF ∩ M) +n(R ∩ M)- n(M ∩SF ∩ R)] = 44−(13+11-2) =44-22 =22 (b) mysteries and science fiction, but not romance novels? =n(SF ∩ M) - n(M ∩SF ∩ R)

=13-2 =11 (c) mysteries or science fiction? (It means the customer purchased m or sf or both) =n(M only)+n(SF only) + [n(SF ∩ M- n(M ∩SF ∩ R)] + [n(SF ∩ R-n(M ∩SF ∩ R)] + [n(M ∩ R) - M ∩SF ∩ R) +n(M ∩SF ∩ R)] =22 + 17+ 11+ 3+9+2 =64 (d) romance novels or mysteries, but not science fiction? =n(R only) + n(M only) ++ [n(M ∩ R) - M ∩SF ∩ R)] =15 +22+9 =46 (e) exactly two types (mysteries, science fiction, romance novels)? = [n(SF ∩ M) - n(M ∩SF ∩ R)] + [n(SF ∩ R) -n(M ∩SF ∩ R)] + [n(M ∩ R) - (M ∩SF ∩ R)] =11+3+9 =23

Lesson 3: Problem Solving 1. Explain why you can never be sure that a conclusion you arrived at using inductive reasoning is true. Inductive reasoning is very limited. It begins with a single observation or an inference drawn from very specific and alike situations. One weakness of inductive reasoning is also one of its most significant strengths—you are only able to establish theories based on limited evidence or knowledge. While it provides you with the opportunity to explore, it also limits the foundation available for you to use. This cannot possibly lead anyone to a fair judgment or accurate inference in a diverse world. Inductive reasoning begins with something specific and then tries to generalize, which will go wrong more often than frequent. 2. Select any two-digit number. Multiply it by 9. Then add the digits. Keep adding the digits in the answer until you get a single-digit answer. Using inductive reasoning, what

can you conjecture about any whole number multiplied by 9? Use deductive reasoning to prove that your conjecture is true. 24

43

99

24× 9=216

43× 9=387

99× 9=891

2+ 1+ 6=9

3+ 8+ 7=18 ∼ 1+ 8=9

8+ 9+1=18 ∼1+8=9

Using inductive reasoning, we could say that the single-digit sum of any two-digit number multiplied by 9 is 9. While using deductive reasoning, there is a well-known divisibility rule that a number is divisible by 9 if and only if 9 divides the sum of its digits. By multiplying a number with 9, you are making it a multiple of 9, and hence the sum of its digits must be divisible by 9. Since you can iterate the digit adding process until you get down to a single digit number, the result must be equal to 9. 𝑘 due to the fact that 10 − 1 , where k≥1, thus proving the divisibility test of 9. 3. Use Polya’s Four Steps to solve the following problems. (a) Susie’s age this year is a multiple of 5. Next year, her age is a multiple of 7. What is her present age? Susie’s age now: 5k Susie’s age next year: 5k+1 = 7m --------Find an integer solution for “k” and “m”. Sequence of 5k+ 1= 6,11,16,21 Sequence of 7m= 7,14,21 --------Equation: 5k+1=21 5k=21-1 5k=20

(b) Consider a square whose side is 1 unit. If the measure of its side is doubled, what will be its new area as compared to the smaller square? How about if the side of the smaller square was tripled, what will be its new area? Area is the space covered by an object. While square is a polygon with four sides with the same length. In order to get the area of a square, we have to multiply side by side. Given that the side of a square is 1 unit and if we doubled the length of the side, we will have,

𝐴𝑠𝑞=1 unit × 1 unit ∼ 1 unit sq. 𝐴𝑠𝑞=2 units × 2 units 𝐴𝑠𝑞=4 units sq.

Therefore, doubling the side length multiplies the area by 4 or it gets 4 times bigger than the original. If the side of the smaller square was tripled, then new area will be,

𝐴𝑠𝑞=3 units × 3 units 𝐴𝑠𝑞= 9 units sq. Therefore, tripling the side lengths multiplies the area by 9 or it gets 9 times bigger than the original. (c) How many perfect squares are there between 1,000,000 and 9,000,000? Square root of 1,000,000 is 1000. (smallest perfect square in this range) Square root of 9,000,000 is 3000. (largest perfect square in this range) Subtract the smallest from the largest perfect square then add 1, 3000-1000= 2000 2000+ 1= 2001 , This equation tells us there are 2001 perfect squares in between 1,000,000 and 9,000,000 inclusive, and if we are going to exclude 1000 and 3000, then there are 1999 perfect squares. (d) Determine the number of different triangles that can be drawn given eight noncollinear points? Given that there are 8 non-collinear points, we have to choose three points to be used as the vertices out of them in order to form a single triangle. Therefore, we will use the formula:

8

8! 3!*5!

8

8!7!6!5!4!3!2!1! 3!2!1!*5!4!3!2!1!

8

8!7!6! 3!2!1!

8

336 6

𝐶3 = 𝐶3 = 𝐶3 = 𝐶3 =

8

𝐶3 = 56

Or let us name the 8 non-collinear points as 1,2,3,4,5,6,7,8. We can choose three out of them, 1,2,3

1,2,4

1,2,4

1,2,5

1,2,6

1,2,7 and so on.

We will exclude the kind of sets of points like (1,2,3), ( 2,1,3), (3,2,1) which are actually just the same. All in all, we will have 56 distinct triangles.

(e) There are 25 students asked by their literature instructor regarding the type of literary works they prefer to read. He found out that 10 prefer to read novels, 11 prefer to read short stories, 15 prefer to read poems, 5 for both novels and short stories, 4 both short stories and poems, 7 for both novels and poems, and 3 prefer all. How many students prefer none of the given types of literary works?

If 3 students prefer reading all the three types of literary works, this means 1 student prefers reading novels only, 5 students prefer short stories only, and 7 students prefer poems only as shown in the computations above. Thus, 1 + 5 + 7 + 3 students who prefer all the three types of

literary works equals 16. In order to get the number of students who prefer none of the given literary works, we should subtract the total number of students to the obtained value. Therefore 25-16=9. There are 9 students who prefer none of the literary works....