# Math 2200 Quiz 1 Solutions

**Problem 1.** (15 points) Show carefully that the
compound proposition

is a tautology.

Proof. We use the laws of propositional logic (also a table of truth values would work).

We used the resolution of the implication, one of deMorgan
laws, the double

negation, associativity, and

**Problem 2.** (15 points) Is the following argument
valid? (Carefully

express it in propositional logic, and show the rules of inference used at

each step .)

“You can score well in the GRE only if you have good
analytical skills.

Every student who takes Discrete Math has good analytical skills or good

memory. Maggie doesn’t have good memory. Therefore, if Maggie takes

Discrete Math, then Maggie will score well in the GRE.”

Proof. Define the following propositional functions, where
the domain of the

variable x is “all students”:

• P(x) = x scores well in the GRE;

• Q(x) = x has good analytical skills;

• R(x) = x takes Discrete Math;

• S(x) = x has good memory.

Then the premises of the argument are:

and the conclusion is

(d) R(Maggie) → P(Maggie).

We can use instantiation for (a) and (b) to get

(a’) P(Maggie) → Q(Maggie);

(b’) R(Maggie) → Q(Maggie) ∨ S(Maggie).

From (b’) and (c), we get

(b”) R(Maggie) → Q(Maggie).

But (a) and (b”) do not imply the conclusion (d). The argument is invalid.

**Problem 3.** (17 points)

(a) Express the following sentence using quantifiers (and logical
propositions ):

“There is a student in this class who has taken some course in every

department in the school of science.”

You should use the variables : s student, c course, d department, and these

should be the only variables. Give the domain of each variable.

(b) Find the logical negation of the quantifier expression that you obtained

in (a).

n (a).

(c) Translate the negation you obtained in (b) into an English sentence.

Proof. (a) | Variable | Domain |

s student | this class | |

c course | all courses | |

d department | departments in the school of science |

We define the propositional function:

P(s, c, d) = s took course c in department d.

Then the sentence can be written as

(b) The negation is

(c) In English: “Every student in this class has not taken any course in

some department in the school of science.”

**Problem 4.** (18 points)

(a) List the elements of the set

(b) For every integer i ≥ 1, setFind

Proof. (a) If (x+1)(2−x) ≥ 0, then −1 ≤ x ≤ 2. Since x is also an integer,

we have S = {−1, 0, 1, 2}.

since every nonnegative integer is in
every A_{i}, and if n is a

negative integer , then

since A_{1} is
a subset of every A_{i}.

**Problem 5.** (18 points)

(a) Is the function ,
given by f(x) = 3x^2 + 2 one-to-one?

What is its image(range)

(b) Consider g :What is
the preimage

Proof. (a) f is one-to-one (note that the domain is positive numbers ). To see

this, prove the contrapositive:

and since x_{1} > 0, x_{2} >
0, this is equivalent to x_{1} = x_{2}

For the image, set y = 3x^2 + 2, thenThe
image

is the interval (2,∞). (In particular, f is not onto.)

(b) Recall thatfor
every real number y. Then

−1, if and only ifor, equivalently

Similarlyif and only if

Therefore, the preimage

**Problem 6**. (17 points) Prove the following
statements:

(a) For every x > 0, if x is irrational, then
is irrational.

(b) Between any two distinct rational numbers , there exist infinitely many

rational numbers .

Proof. (a) (proof by contrapositive) If
is rational, then x is rational.

Write = m/n , for some integers m, n, n ≠ 0.
Then by squaring both sides,

. Since n ≠ 0, then n^2 ≠ 0, moreover, both
m^2, n^2 are integers, since

m, n are. This proves that x is rational.

(b) (proof by contradiction) Assume that there exist two
rational numbers

a, b, a < b, such that between a and b there are only finitely many rational
numbers.

Then there must be a rational number closest to a among

them, call it r. We have a < r and no rational numbers between a and r.

ConsiderThis is a rational number, andThis
gives the

contradiction.

Prev | Next |