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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 onetoone?
What is its image(range)
(b) Consider g :What is
the preimage
Proof. (a) f is onetoone (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.
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