Express each of these statements using quantifiers. Then form the negation of the statement so that no negation is to the left of a quantifier. Next, express the negation in simple English. In each case, identify the domain and specify the predicates. a. No one has lost more than one thousand dollars playing the lottery.
b. There is a student in this class who has chatted with exactly one other student.
c. No student in this class has sent e-mail to exactly two other students in this class.
d. One student has solved every exercise in this book.
e. No student has solved at least one exercise in every section of this book.

Answer and Step-by-step explanation:

Not p = ¬p

P or q = p ∨ q  

P and q = p ∧ q

If p then q = p → q

P if and only if q = p ↔ q

Existential quantification:  There exist an element x in the domain such that p(x).

Universal quantification: p(x) for all values of x in the domain.

(a)  No one has lost more than one thousand dollars playing the lottery.

Let A(x) means ‘x has lost more than one dollars playing the lottery’

It can also write as “there does not exists a person that lost more than one thousand dollars playing”

                     ¬Ǝ x A (x)

Negation of this statement:  

By using double negation law:

                               ¬ [¬Ǝ x A (x)]  ≡ Ǝ x A(x)

(b)There is a student in this class who has chatted with exactly one other student.

Let B(x,y) means “ x has chatted with y” and domain is all students of this class.

We can write the given sentence as:

“There is a student in the class who has chatted with one student and this student is not himself and for all people the student chatted with, this student has to be himself or the one student he chatted with”

Ǝ x Ǝ y[B ( x, y) ∧ x ≠ y ∧ ∀ z (B(x,y) → ( z = x v z = y))]

The negation:

               ¬ Ǝ x Ǝ y[B ( x, y) ∧ x ≠ y ∧ ∀ z (B(x ,y) → ( z = x v z = y))]

By using De Morgan’s law for quantifiers:

≡∀x ¬ Ǝ y [B ( x, y) ∧ x ≠ y ∧ ∀ z (B(x ,y) → ( z = x v z = y))]

≡∀x ∀ y [B ( x, y) ∧ x ≠ y ∧ ∀ z (B(x ,y) → ( z = x v z = y))]

De Morgan’s law:

≡∀x ∀ y [¬  B ( x, y) v  ¬ ( x ≠ y) v ∀ z (B(x ,y) → ( z = x v z = y))]

By using De Morgan’s law for quantifiers:

≡∀x ∀ y [¬  B ( x, y) v  x=  y  v Ǝ z¬ (B(x ,z) → ( z = x v z = y))]

(c)  No student in this class has sent e-mail to exactly two other students in this class

Let c(x, y) means “ x has sent email to y” and the domain is all student of class.

Using double negation law:

Ǝ x Ǝ y Ǝ z [c(x, y) ∧c(x ,z) ∧ x≠ y ∧ x ≠z ∧ y ≠ z ∀ w (c(x,w) → ( w = x v w = y v w = z)]

There is a student in class that has sent email to exaxtly two other students in class.

(d)  One student has solved every exercise in this book

Let D(x , y) mean student x has solved exercise y in this book.

The negation:  

Ǝx∀yD(x,y)

Use De Morgan’s law for qualifiers:

    ≡∀ x Ǝ y ¬D(x, y)  

(e). No student has solved at least one exercise in every section of this book.

Let E(x, y,z) be student x has solved exercise y in section z of this book.

We can write “there does not exist a student that solved at least one exercise in all sections of this book”

¬Ǝ x Ǝ y ∀ Z E(x, y, z)  

Negation:

                      ≡¬ [¬ Ǝ x Ǝ y ∀ Z E(x, y, z)  ]

Use double negation law:

                                     ≡ Ǝ x Ǝ y ∀ Z E(x, y, z)  

false

step-by-step explanation:

answer: 75°

step-by-step explanation:

∠esk is the supplement of ∠edk, so is 100°. ∠esd = 1/2·arc ed, so is 25°.

since ∠esk = ∠esd + ∠dsk

you have 100° = 25° + ∠dsk . .

fill in known values 75° = ∠dsk . . . . .

subtract 25°

step-by-step explanation: