U. of Illinois MATH 347H Test 1 – Spring 2017
Answer as many problems as you can. Each of the five questions is
worth 6 points (total of 30). Show your work. An answer with no
explanation will receive no credit. Write your name on the top
right corner of each page.
No external assistance permitted.
[Total time: 50 minutes]
Below are some details from the textbook you might find helpful. I
do not promise you will need all of them or even that it is a complete
list.
Axioms of arithmetic
(A1) If a, b ∈ Z then a + b ∈ Z
(M1) If a, b ∈ Z then a · b ∈ Z.
(A2) If a, b, c ∈ Z, then a + (b + c) = (a + b) + c
(M2) If a, b, c ∈ Z, then a · (b · c) = (a · b) · c
(A3) If a, b ∈ Z, then a + b = b + a
(M3) If a, b ∈ Z, then a · b = b · a
(A4) ∃0 ∈ Z such that for all a ∈ Z, a + 0 = 0 + a = a
(M4) ∃1 ∈ Z such that 1 6= 0 and for all a ∈ Z, a · 1 = 1 · a = a
(A5) ∀a ∈ Z, ∃ − a ∈ Z such that a + (−a) = (−a) + a = 0
(C) If a, b, c ∈ Z with a 6= 0 and ab = ac, then b = c
(O1) If a, b ∈ Z then one and only one of the following holds: a < b,
a = b or b > a
(O2) If a, b, c ∈ Z with a < b and b < c, then a < c
(O3) If a, b, c ∈ Z and a < b then a + c < b + c
(O4) If a, b, c ∈ Z, a < b, and 0 < c, then ac < bc.
(W) Well-ordering principle (you recall it)
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1. Short answer:
(a) Suppose A =⇒ B is a mathematical statement. Then you take
the contrapositive and then the negation. Is the result the same as first
taking the negation and then the contrapositive? If you think “yes”,
give a brief explanation. Otherwise, give a counterexample.
The answer is “yes”. The contrapositive gives a logically equivalent
statement not B =⇒ not A. Thus contrapositive is true iff the original
statement is true iff the negation (of either) is false.
(b) What is wrong with the following proof that “n + 1 < n for all
integers n”?
Assume the claim holds for n = k. We then show it for
n = k + 1. The induction hypothesis says k + 1 < k. By
(O3), k + 1 + 1 < k + 1, i.e., k + 2 < k + 1, as desired.
The proof has no base case.
(c) For a set X, recall P(X) is the power set (=set of all subsets of X).
What is #P(P({1, 2, 3, . . . , n}))?
22
n
(d) Write the contrapositive of the statement:
“If Lucus studies hard then Lucas will get an A.”
“If Lucas does not get an A, then Lucus did not study hard.”
(e) Suppose ∼ is an equivalence relation on a set X, and a ∼ b but b 6∼ c
but (b is not equivalent to c). Is a ∼ c possible?
No. By transitivity, if a ∼ c then b ∼ c, a contradiction.
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2. Let X = R and suppose we define x ∼ y if x − y ∈ Z. Is ∼ an
equivalence relation? If you think it is, give a proof. If you think it is
not, explain.
Proof: It is an equivalence relation. We must establish the three properties.
Reflexive: Since x − x = 0 ∈ Z, we have x ∼ x.
Symmetric: If x ∼ y then x − y ∈ Q. Thus y − x = −(x − y) ∈ Z
(A5). Hence y ∼ x.
Transitive: Suppose x ∼ y; this means x − y ∈ Z. Also suppose
y ∼ z; this means y − z ∈ Z. Hence x − z = (x − y) + (y − z) ∈ Z (by
(A1)). Hence x ∼ Z.
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3. Determine (with proof): for which positive integers is n2 ≤ n!.
Proof: On computes that for 1 ≤ n ≤ 3 in fact n2 > n!. We prove for
n ≥ 4 that n2 ≤ n!. Now,
(n + 1)2 < (n + 1)! ⇐⇒ (n + 1)2 < n!(n + 1) (definition of n!)
⇐⇒ (n + 1) < n! (n ≥ 4 so n + 1 6= 0).
However for n ≥ 4 this last statement is true since n+1
= 1 + n1 < 2
n
whereas (n − 1)! > 2. (The assertion (n − 1)! > 2 can be proved by an
easy induction.)
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4. Prove A ∪ (B ∪ C) = (A ∪ B) ∪ C.
Proof: This was a homework problem.
We must show that A ∪ (B ∪ C) ⊆ (A ∪ B) ∪ C:
x ∈ A ∪ (B ∪ C) ⇐⇒
x ∈ A ∨ x ∈ B ∪ C ⇐⇒ x ∈ A ∨ x ∈ B ∨ x ∈ C
⇐⇒ x ∈ A ∪ B ∨ x ∈ C ⇐⇒ x ∈ (A ∪ B) ∪ C.
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5. Compute the number of sets (A, B, C) such that
∅ ⊆ A ⊆ B ⊆ C ⊆ {1, 2, . . . , n}.
(Hint: this is close to a homework problem.)
Solution: The answer is 4n . Every element of {1, 2, 3, 4, . . . , n} is in one
of the following four cases:
• It is in C but not B
• It is in B but not A
• It is in A
• It is not in A, B or C.
Now use the fundamental counting principle.
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