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This is study resource 1 for Summer 2019 – 24-week cohort.

There are 12 questions in it. I have shown the work for all of them except for the questions that I had wrong. I know they are correct answers since I got feedback on Gradescope.

I hope it helps anyone and please, do not just copy and paste, you must understand the content in order to do well in the NYU Bridge To Tandon program.

Question #1

## Question #1

A. Convert the following numbers to their decimal representation. Show your work.

1. 100110112
1. 4567
1. 38A16

B. Convert the following numbers to their binary representation:

1. 6910
1. 48510
1. 6D1A16

C. Convert the following numbers to their hexadecimal representation:

1. 11010112
1. 89510

Question #2

## Question #2

Solve the following, do all calculation in the given base. Show your work.

1. 75668 + 45158 = 14303
1. 101100112 + 11012 = 11000000
1. 7A6616 + 45C516 = C02B
1. 30225 – 24335 = 34

Question #3

## Question #3

A. Convert the following numbers to their 8-bits two’s complement representation. Show your work.

1. 12410 = 01111100
1. -12410 = 10000100
1. 10910 = 01101101
1. -7910 = 10110001

B. Convert the following numbers (represented as 8-bit two’s complement) to their decimal representation. Show your work.

1. 000111108 bit 2’s comp = 30
1. 111001108 bit 2’s comp = -26
1. 001011018 bit 2’s comp = 45
1. 100111108 bit 2’s comp = -98

Question #4

## Question #4

Solve the following questions from the Discrete Math zyBook:

1. Exercise 1.2.4 – Write a truth table for each expression

B. ¬ (p ∨ q)

C. r ∨ (p ∧ ¬q)

1. Exercise 1.3.4

B. (p → q) → (q → p)

D. (p ↔ q) ⊕ (p ↔ ¬q)

Question #5

## Question #5

Solve the following questions from the Discrete Math zyBook:

Exercise 1.2.7

B. The applicant must present at least two of the following forms of identification: birth certificate, driver’s license, marriage license.

(B ∧ D) ∨ (B ∧ M) ∨ (D ∧ M)

C. Applicant must present either a birth certificate or both a driver’s license and a marriage license.

B ∨ (D ∧ M)

Exercise 1.3.7

B. A person can park in the school parking lot if they are a senior or at least seventeen years of age.

(s ∨ y) → p

C. Being 17 years of age is a necessary condition for being able to park in the school parking lot.

p → y

D. A person can park in the school parking lot if and only if the person is a senior and at least 17 years of age

p ↔ (s ∧ y)

E. Being able to park in the school parking lot implies that the person is either a senior or at least 17 years old.

p → (s ∨ y)

Exercise 1.3.9

C. The applicant can enroll in the course only if the applicant has parental permission.

c → p

D. Having parental permission is a necessary condition for enrolling in the course.

p → c

Question #6

## Question #6

Solve the following questions from the Discrete Math zyBook:

Exercise 1.3.6

B. Maintaining a B average is necessary for Joe to be eligible for the honors program.

If Joe maintains a B average, the Joe is available for the honors program.

C. Rajiv can go on the roller coaster only if he is at least four feet tall.

If Rajiv can go on the roller coaster, then he is at least four feet tall.

D. Rajiv can go on the roller coaster if he is at least four feet tall.

If he is at least four feet tall, then Rajiv can go on the roller coaster.

Exercise 1.3.10p = T, q = F, r =?

C. (p ∨ r) ↔ (q ∧ r)

When r = T:

(T ∨ T) ↔ (F ∧ T)

T ↔ F

F

When r = F:

(T ∨ F) ↔ (F ∧ F)

T ↔ F

F

Answer: False.  Whether r is true or false, the expression is always going to be false. In a bi-conditional, both the hypothesis and conclusion must be either false or true.

D. (p ∧ r) ↔ (q ∧ r)

When r = T:

(T ∧ T) ↔ (F ∧ T)

T ↔ F

F

When r = F:

(T ∧ F) ↔ (F ∧ T)

F ↔ F

T

Answer: Unknown – If r is true, the expression is false; if r is false, the expression is true.

E. p → (r ∨ q)

When r = T:

T → (T ∨ F)

T → T

T

When r = F:

T → (F ∨ F)

T → F

F

Answer: Unknown – If r is true, the expression is false; if r is false, the expression is true.

F. (p ∧ q) → r

When r = T:

(T ∧ F) → T

F → T

T

When r = F:

(T ∧ F) → F

F → F

T

Answer: True – For a conditional to be false, the hypothesis must be true and the conclusion false. Whether r is true or false, this expression will always be true.

Question #7

## Question #7

Solve Exercise 1.4.5

j: Sally got the job.

l: Sally was late for her interview

r: Sally updated her resume.

B.

If Sally did not get the job, then she was late for interview or did not update her resume.

If Sally updated her resume and was not late for her interview, then she got the job.

¬j → (l ∨ r)

(r ∧ ¬l) → j

——– NO ANSWER FOR THIS QUESTION ——

C.

If Sally got the job, then she was not late for her interview.

If Sally did not get the job, then she was late for her interview.

j → ¬l

¬j → l

D.

If Sally updated her resume or she was not late for her interview, then she got the job.

If Sally got the job, then she updated her resume and was not late for her interview.

(r ∨ ¬l) → j

j → (r ∧ ¬l)

Question #8

## Question #8

Solve the following questions from the Discrete Math zyBook:

Exercise 1.5.2

Using the following table as reference

C. (p → q) ∧ (p → r) ≡ p → (q ∧ r)

F. ¬ (p ∨ (¬p ∧ q)) ≡ ¬p ∧ ¬q

I. (p ∧ q) → r ≡ (p ∧ ¬r) → ¬q

Exercise 1.5.3

C. ¬r ∨ (¬r → p)

D. ¬ (p → q) → ¬q

Question #9

## Question #9

Solve the following questions from the Discrete Math zyBook:

Exercise 1.6.3

c: There is a number that is equal to its square.

∃x (x2= x)

d: Every number is less than or equal to its square.

∀x (x ≤ x2)

Exercise 1.7.4

b: Everyone was well and went to work yesterday.

∀x (¬S(x) ∧W(x))

c: Everyone who was sick yesterday did not go to work.

∀x (S(x) → ¬W(x))

d: Yesterday someone was sick and went to work.

∃x (S(x) ∧W(x))

Question #10

## Question #10

Solve the following questions from the Discrete Math zyBook:

Exercise 1.7.9

c: ∃x ((x = c) → P(x))

TRUE Elements {a,b,d,e} render the statement true ie. When x=c is false (when x is not c) then the values for the predicates P,Q,R are true.

d: ∃x (Q(x) ∧ R(x))

When x = e, Q(e) ∧ R(e) is True.

e: Q(a) ∧ P(d)

T ∧ T

T

Expression is True.

f: ∀x ((x ≠ b) → Q(x))

Q(a) = T

Q(c) = T

Q(d) = T

Q(e) = T

Expression is true.

g: ∀x (P(x) ∨ R(x))

When x = c, expression is false, therefore, not all values are true.

Expression is false.

h: ∀x (R(x) → P(x))

when x = a; F → T = T

when x = b; F → T = T

when x = c; F → F = T

when x = d; F → T = T

when x = e; T → T = T

For all, expression is true.

i: ∃x (Q(x) ∨ R(x))

when x = a; T ∨ F = T

when x = b; F ∨ F = F

when x = c; T ∨ F = T

when x = d; T ∨ F = T

when x = e; T ∨ T = T

For at least one, expression is true.

Exercise 1.9.2

b: ∃x ∀y Q(x, y)

True – there exists a row in Q where all the values are true (2nd row).

c: ∃x ∀y P(y, x)

TRUE. For x = 1, P(1, x), P(2, x) and P(3, x) are all true.

d: ∃x ∃y S(x, y)

False – every value is false.

e: ∀x ∃y Q(x, y)

False – for every x (1, 2, 3), exists at least one y such that has true but when x = 1, all y’s are false.

f: ∀x ∃y P(x, y)

when x=1, y=1: true

when x=2, y=1: true

when x=3, y=1: true

Hence, true.

g: ∀x ∀y P(x, y)

False – Not all x and all y are true.

h: ∃x ∃y Q(x, y)

True – exist at least one x and one y for which it is true.

i: ∀x ∀y ¬S(x, y)

True – All x & y’s are false, and they become true by the negation.

Question #11

## Question #11

Solve the following questions from the Discrete Math zyBook:

Exercise 1.10.4

c: There are two numbers whose sum is equal to their product.

∃x ∃y (x+y = xy)

d: The ratio of every two positive numbers is also positive.

∀x ∀y ((x > 0 ∧ y > 0) → x/y > 0)

e: The reciprocal of every positive number less than one is greater than one.

∀x ((x > 0 ∧ x < 1) → 1/x > 1)

f: There is no smallest number.

¬ ∃x ∀y (x <= y)

g: Every number besides 0 has a multiplicative inverse.

∀x ∃y (x ≠ 0 → xy = 1)

Exercise 1.10.7

c: There is at least one new employee who missed the deadline.

∃x (N(x) → D(x))

d: Sam knows the phone number of everyone who missed the deadline.

∃x ∀y D(y) -> P(Sam, y)

e: There is a new employee who knows everyone’s phone number.

∃x ∀y N(x) → P(x,y)

f: Exactly one new employee missed the deadline.

I do not have the answer for this one but one of the TAs gave me this feedback: “For this one, you should review zyBooks Participation Activity 1.10.3 on the exclusion of exactly one. “

Exercise 1.10.10

c: Every student has taken at least one class besides Math 101.

∀ x ∃ y ((y ≠ Math 101) ∧ T(x, y))

d: There is a student who has taken every math class besides Math 101.

∃ x ∀ y ((y ≠ Math 101) → T(x, y))

e: Everyone besides Sam has taken at least two different math classes.

∀ x ∃ y ∃ z ((x ≠ Sam) → ((y ≠ z) ∧ T(x, y) ∧ T(x, z)))

f: Sam has taken exactly two math classes.

∃y ∃z ∀w ((z ≠ y) ∧ T(Sam, y) ∧ T(Sam, z) ∧ ((w ≠ y ∧ w ≠ z) → ¬T(Sam, w)))

Question #12

## Question #12

Solve the following questions from the Discrete Math zyBook:

Exercise 1.8.2

b: Every patient was given the medication or the placebo or both.

∀x (D(x) ∨ P(x))

Negation: ¬ ∀x (D(x) ∨ P(x))

Applying De Morgan’s law: ∃x (¬D(x) ∧ ¬P(x))

English: There is a patient who was not given the medication and not given the placebo.

c: There is a patient who took the medication and had migraines.

∃x (D(x) ∧ M(x))

Negation: ¬ ∃x (D(x) ∧ M(x))

Applying De Morgan’s law: ∀x (¬D(x) ∨ ¬M(x))

English: Every patient did not get the medication or did not have migraines or both.

d: Every patient who took the placebo had migraines. (Hint: you will need to apply the conditional identity, p → q ≡ ¬p ∨ q.)

∀x (P(x) → M(x))

Negation: ¬ ∀ x (P(x) → M(x))

Applying De Morgan’s law: ∃x (P(x) ∧ ¬M(x))

English: Some patient took the placebo and did not have migraines

e: There is a patient who had migraines and was given the placebo.

∃x (M(x) ∧ P(x))

Negation: ¬ ∃x (M(x) ∧ P(x))

Applying De Morgan’s law: ∀x (¬ M(x) ∨ ¬P(x))

English: Every patient either did not have migraine or was not given the placebo or both.

Exercise 1.9.4

c: ∃x ∀y (P(x, y) → Q(x, y))

Negation: ∀x ∃y (P(x, y) ∧ ¬Q(x, y))

d: ∃x ∀y (P(x, y) ↔ P(y, x))

¬ (∃x ∀y ((P(x,y) → P(y,x)) ∧ (P(y,x) → P(x,y))))

¬ (∃x ∀y ((¬P(x,y) ∨ P(y,x)) ∧ (¬P(y,x) ∨ P(x,y))))

∀x ∃y (¬ (¬P(x,y) ∨ P(y,x)) ∨ ¬ (¬P(y,x) ∨ P(x,y)) )

Negation: ∀x ∃y ((P(x,y) ∧ ¬P(y,x)) ∨ (P(y,x) ∧ ¬P(x,y)) )

e: ∃x ∃y P(x, y) ∀x ∀y Q(x, y)

¬ (∃x ∃y P(x, y) ∧ ∀x ∀y Q(x, y))

Negation: ∀x ∀y ¬P(x, y) ∨ ∃x ∃y ¬Q(x, y)

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