Chapter 2, Quantificational Logic Video Solutions, How to Prove It: A Structured Approach

Problem 1

Analyze the logical forms of the following statements.
(a) Anyone who has forgiven at least one person is a saint.
(b) Nobody in the calculus class is smarter than everybody in the discrete math class.
(c) Anyone who has bought a Rolls Royce with cash must have a rich uncle.
(d) If anyone in the dorm has the measles, then everyone who has a friend in the dorm will have to be quarantined.
(e) Everyone likes Mary, except Mary herself.
(f) Jane saw a bear, and Roger saw one too.
(g) Jane saw a bear, and Roger saw it too.
(h) If anyone can do it, Jones can.
(i) If Jones can do it, anyone can.

Madi Sousa

Numerade Educator

01:39

Problem 2

Analyze the logical forms of the following statements. The universe of discourse is $\mathbb{R}$. What are the free variables in each statement?
(a) Every number that is larger than $x$ is larger than $y$.
(b) For every number $a$, the equation $a x^2+4 x-2=0$ has at least one solution iff $a \geq-2$.
(c) All solutions of the inequality $x^3-3 x<3$ are smaller than 10 .
(d) If there is a number $x$ such that $x^2+5 x=w$ and there is a number $y$ such that $4-y^2=w$, then $w$ is between -10 and 10 .

Vysakh M

Numerade Educator

01:46

Problem 3

Translate the following statements into idiomatic English.
(a) $\forall x[(H(x) \wedge \forall y \neg M(x, y)) \rightarrow U(x)]$, where $H(x)$ means $x$ is a man, $M(x, y)$ means $x$ is married to $y$, and $U(x)$ means $x$ is unhappy.
(b) $\exists z(P(z, x) \wedge S(z, y) \wedge W(y))$, where $P(z, x)$ means $z$ is a parent of $x$, $S(z, y)$ means $z$ and $y$ are siblings, and $W(y)$ means $y$ is a woman.

Christopher Stanley

Numerade Educator

01:29

Problem 4

Are these statements true or false? The universe of discourse is $\mathbb{N}$.
(a) $\forall x \exists y(2 x-y=0)$.
(b) $\exists y \forall x(2 x-y=0)$.
(c) $\forall x \exists y(x-2 y=0)$.
(d) $\forall x(x<10 \rightarrow \forall y(y<x \rightarrow y<9))$.
(e) $\exists y \exists z(y+z=100)$.
(f) $\forall x \exists y(y>x \wedge \exists z(y+z=100))$.

Vysakh M

Numerade Educator

00:41

Problem 5

Same as exercise 4 but with $\mathbb{R}$ as the universe of discourse.

Natalie Anderson

Numerade Educator

00:41

Problem 6

Same as exercise 4 but with $\mathbb{Z}$ as the universe of discourse.

Natalie Anderson

Numerade Educator