Download Exam II Solutions

Math 311 Exam II
Name:
Fall 2011
Exam is due in class on Friday, October 28, 2011
Problem
Number
1
2
3
4
5
6
Total
Possible Score
Points
10
10
10
10
10
10
50
Directions—Please Read Carefully! This is a take-home exam. You may use calculators, any notes that you have taken in class, and your textbook. I will also be available to
clarify any questions. Make sure to use correct mathematical notation. To receive full credit
on a problem, you will need to justify your answers carefully unless indicated otherwise—
unsubstantiated answers will receive little or no credit. Please be sure to write neatly—
illegible answers will receive little or no credit. If more space is needed, use the back of the
previous page to continue your work. Be sure to make a note of this on the problem page so
that the grader knows where to find your answers.
You may not use any resources other than those listed above, including speaking with other faculty or students and consulting any sources not listed. You
are not allowed to use the Internet.
Good Luck!!!
Do any five of the following six problems.
1. (10 points) Prove that 10n+1 + 10n + 1 is divisible by 3 for n ∈ N.
Solution: If n = 1, then 10n+1 + 10n + 1 = 111 is divisible by 3. Assume that
10k+1 + 10k + 1 is divisible by 3 for 1 ≤ k ≤ n. Then 10(n+1)+1 + 10n+1 + 1 =
10(10n+1 + 10n + 1) − 9 is divisible by 3. Therefore, the statement is true for all
natural numbers by the Principle of Mathematical Induction.
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2. (10 points) Do union and intersection distribute over set difference?
(a) Show that
A ∪ (B \ C) = (A ∪ B) \ (A ∪ C).
is true by providing a proof or provide a counterexample showing it to be false.
Solution: This statement is not true. Let A = {a, b, c}, B = {b}, and C = {c}.
Then
A ∪ (B \ C) = A ∪ (B ∩ C C ) = A = {a, b, c}.
However,
(A ∪ B) \ (A ∪ C) = A \ A = A ∩ AC = ∅.
Other counterexamples are possible.
(b) Show that
A ∩ (B \ C) = (A ∩ B) \ (A ∩ C).
is true by providing a proof or provide a counterexample showing it to be false.
Solution: The statement is true. Consider
(A ∩ B) \ (A ∩ C) = (A ∩ B) ∩ (A ∩ C)C
= (A ∩ B) ∩ (AC ∪ C C )
= (A ∩ B ∩ AC ) ∪ (A ∩ B ∩ C C )
= ∅ ∪ [A ∩ (B ∩ C C )]
= A ∩ (B \ C).
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3. (10 points) Let m and n be natural numbers. We say that n divides m if there exists
a ∈ N such that m = an. We denote this relation by n | m.
(a) Prove that N is partially ordered under the relation |.
Solution: The relation is certainly reflexive since a | a for all a ∈ N. If m | n
and n | m, then m = n; hence, the relation is also antisymmetric. The relation
is transitive, because if m | n and n | p, then m | p.
(b) Is | a total order on N? Explain.
Solution: No. Indeed, if | was a total order either m | n or n | m for all
m, n ∈ N. However, for m = 2 and n = 3, we have 2 - 3 and 3 - 2.
(c) Draw a lattice diagram that depicts the order | on the set {1, 2, 3, 4, 6, 8, 12, 24}.
Solution:
24
8
12
4
6
2
3
1
(d) Does the set X = {2, 3, 4, 5, . . .} have any minimal or maximal elements with respect
to the order | ? Explain.
Solution: There are no maximal elements. If m is maximal, then there is no
number larger than m that is divisible by m. This is not the case since m | 2m.
The minimal elements in X are the prime numbers p since p is only divisible by
itself and one and 1 ∈
/ X.
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4. (10 points) Define a relation on R2 by (x1 , y1 ) ∼ (x2 , y2 ) if x21 + y12 = x22 + y22 .
(a) Prove that ∼ defines an equivalence relation on R2 .
Solution: The relation is reflexive since x21 + y12 = x21 + y12 . If (x1 , y1 ) ∼ (x2 , y2 ),
then x21 + y12 = x22 + y22 . Consequently, x22 + y22 = x21 + y12 and (x2 , y2 ) ∼ (x1 , y1 )
and the relation is symmetric. Finally, suppose that (x1 , y1 ) ∼ (x2 , y2 ) and
(x2 , y2 ) ∼ (x3 , y3 ). Then x21 + y12 = x22 + y22 and x22 + y22 = x23 + y32 and x21 + y12 =
x23 + y32 . Thus, (x1 , y1 ) ∼ (x3 , y3 ) and the relation is transitive.
(b) What are the corresponding equivalence classes? Give a geometric description.
Solution: The equivalence classes are circles centered at the origin.
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5. (10 points) Define a relation on the integers by m ∼ n if mn > 0, where m, n ∈ Z.
(a) Show that this relation is reflexive by providing a proof or provide a counterexample
that shows the relation is not reflexive.
Solution: The relation is not reflexive since 0 · 0 is not greater than zero and
0 0.
(b) Show that this relation is symmetric by providing a proof or provide a counterexample that shows the relation is not symmetric.
Solution: If m ∼ n, then mn > 0 and nm > 0. Consequently, n ∼ m and the
relation is symmetric.
(c) Show that this relation is antisymmetric by providing a proof or provide a counterexample that shows the relation is not antisymmetric.
Solution: The relation is not antisymmetric since 2 ∼ 3 and 3 ∼ 2 but 2 6= 3.
(d) Show that this relation is transitive by providing a proof or provide a counterexample that shows the relation is not reflexive.
Solution: Suppose that m ∼ n and n ∼ p. Then mn > 0 and np > 0.
Therefore, (mn)(np) = mn2 p > 0. Since n2 > 0, we know that mp > 0 and
m ∼ p. Thus, the relation is transitive.
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6. (10 points) If x is a nonnegative real number, then show that (1 + x)n − 1 ≥ nx for
n ∈ N.
Solution: If n = 1, then (1 + x)1 − 1 = x ≥ 1 · x. Assume that (1 + x)k − 1 ≥ kx
for k ≥ 1. We must show that (1 + x)k+1 − 1 ≥ (k + 1)x. Multiplying the left side
of the expression (1 + x)k − 1 ≥ kx by x + 1 and noting that x + 1 > 1, we obtain
(1 + x)k+1 − (x + 1) > kx.
Adding x to both sides of the inequality, we have
(1 + x)k+1 − 1 ≥ (k + 1)x.
By the Principle of Mathematical Induction,
(1 + x)n − 1 ≥ nx
for n ∈ N.
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