Math 201 – Exam 1 - solutions
Problem 1 a) In the first Venn diagram below, shade in the area representing P ∩ (Ē ∪ R).
b) Express the area shaded in the second Venn diagram in (simple!) words.
c) Express the area shaded in the second Venn diagram in terms of P, R, E
using ∩, ∪, and the complement.
Solution. For a), see below. For b), you could say it’s the area contained
in at least two of the ellipses. For c), you could write (several possibilities,
the parentheses are important)
(P ∩ E) ∪ (P ∩ R) ∪ (E ∩ R).
R
P
E
Problem 2
Everything is awesome,
If you’re part of a team!
a) Express this statement in symbolic logic using quantifiers, the set S of
things, the set T of teams, the sentence A(x) saying x is awesome, and the
sentence P (t) saying you are part of team t.
b) Negate the above statement in symbolic logic.
c) Express the negated statement in plain English.
Solution. a) Several possible answers. Implicit all-quantification is not OK,
since the problem asked for quantifiers. Personally, I think the first answer
below is closest to the original wording, but all three are logically equivalent.
Watch the parentheses and the position of the quantifiers!
∀t ∈ T : (P (t) ⇒ (∀x ∈ S : A(x)))
∀t ∈ T, ∀x ∈ S : (P (t) ⇒ A(x))
(∃t ∈ T : P (t)) ⇒ (∀x ∈ S : A(x)).
b) ∃t ∈ T, ∃x ∈ S : (P (t) ∧ (∼A(x))).
c) You can be part of a team, but not everything is awesome.
Or more polished: it is possible that you are part of a team, but still not
everything is awesome.
Problem 3 Consider this statement about a function f (x).
S: If x > 1 and f (x) is not continuous, then f (x) ≥ 3.
a) Write the converse of S.
b) Express the contrapositive of S without any negative words.
Solution. a) If f (x) ≥ 3 then x > 1 and f (x) is not continuous.
b) If f (x) < 3 then x ≥ 1 or f (x) is continuous.
Problem 4 Let Tα be the triangle with vertices (0, 0), (1, 0), and (α, 1).
Sketch three of the triangles Tα for 0 ≤ α ≤ 1, and find the set
\
Tα .
α∈[0,1]
Solution. The points (α, 1) are all on the horizontal line y = 1. a) Answers
will vary. The three desired triangles all share the baseline from (0, 0) to
(1, 0) (unit interval on the x-axis).
b) The intersection of all Tα is the triangle with vertices (0, 0), (1, 0), and
(0.5, 0.5).
Problem 5 You are sitting in a back corner in the dignified, oak-panelled
ISU Math Department Faculty lounge. You overhear the following argument.
1. Dr Bergstein: If Non-Eulerian semirings are commutative, then PseudoRiemannian hypersquares do exist!
2. Dr Steinberg: I would not be so sure about that, but I can prove that
Quaternion pre-algebras are unitary, if Pseudo-Riemannian hypersquares do
not exist.
3. Dr B: Well, if Pseudo-Riemannian hypersquares exist, then Quaternion
pre-algebras are not unitary.
4. Dr S: Obviously! But if Non-Eulerian semirings are commutative, then
Quaternion pre-algebras are, indeed, unitary.
5. Dr B: Yeah, I like to set that as an exercise for my freshman algebra class.
But did you know that Non-Eulerian semirings are commutative, provided
that Quaternion pre-algebras are not unitary?
6. Dr S: Of course! But I wonder if we will ever discover which of these is
actually true.
You step out of the shadows and tell the professors: if all your conditional
statements are true, then I can tell you exactly which of these is true. But
first, what about my grade in Math 201?
a) Using the letters N, P, Q for the three building blocks, express the clues
1-5 in symbolic logic.
b) Diagram each of the clues as shown in class (shade the impossible areas),
and deduce which of N, P, Q is true and which is false. Or use a truth table,
if you prefer, crossing out impossible combinations.
Solution. For a): we let N stand for ”Non-Eulerian semirings are commutative”, P for ”Pseudo-Riemannian hypersquares exist”, Q for ”Quaternion
pre-algebras are unitary”.
1. N ⇒ P .
2. (∼P ) ⇒ Q.
3. P ⇒ (∼Q).
4. N ⇒ Q.
5. (∼Q) ⇒ N .
b) A truth table for a statement with 3 variables N, P, Q needs to have 8
lines. Equivalently, we can use a rectangle divided into 8 squares. After we
shade the impossible squares in diagrams for clues 1 - 5, we overlap these
diagrams and obtain a diagram with only 1 unshaded square, corresponding
to (∼N ) ∧ (∼P ) ∧ Q. This means the only possibility is that N, P are false
and Q is true.
P
P
~Q
N
~Q
N
Q
Q
~Q
~Q
P
P
~Q
N
~Q
N
Q
Q
~Q
~Q
P
P
~Q
N
~Q
N
Q
Q
~Q
~Q