Download Exam 1 solutions

MA 347 Exam # 1
Name:
SOLUTION
• Please put all work that you want graded on these sheets.
• You may not use a calculator or any notes for this exam.
• All statements must be justified.
Number Possible Actual
1
20
2
20
3
20
4
20
5
20
Total
100
1. (20 pts.) Prove
n
X
i=
i=1
n(n + 1)
2
for all n ∈ N.
Solution: We will prove this by induction. Let
P (n) = “
n
X
i=
i=1
n(n + 1)
”.
2
We will show first that P (1) is true, and then that P (k) =⇒ P (k + 1).
First, we have
P (1) = “1 =
1∗2
”
2
which is true.
Now, assume P (k). We compute
k+1
X
i=
i=1
k
X
i + (k + 1)
i=1
k(k + 1)
+k+1
2
k 2 + k + 2k + 1
=
2
2
k + 3k + 1
=
2
(k + 1)(k + 2)
.
=
2
=
2. (20 pts.) Throughout this problem x is a real number. We define A, B, C, D to be
mathematical statements as follows:
A is the statement “x ∈ Z”, B is the statement “1 ≤ x < 3”, C is the statement
“x2 = 4”, D is the statement “x = 1”.
Which of the following statements are true for all x ∈ R?
(a) B =⇒ A
(b) A ∧ B ∧ ¬C =⇒ D
(c) A =⇒ B ∨ C ∨ D
(d) C =⇒ A
Solution:
(a) This is false, and to prove this we need only show one counterexample. Notice
that if we set x = 3/2, then B is true, but A is false.
(b) This is true. Let us assume that A ∧ B ∧ ¬C is true. If A is true, we know that
x is an integer. If A ∧ B is true, then x must be either 1 or 2, and if C is false
then x cannot be 2, thus we must have x = 1. But then this means D is true.
(c) This is false. To show this, we need only find one integer which does not satisfy
any of B, C, D. So choose x = 10.
(d) This is true. If C is true, we must have x2 = 4, or x = ±2. Either of these
numbers is an integer, so it follows that A is true.
3. (20 pts.)
(a)
(b)
(c)
(d)
Prove
Prove
Prove
Prove
that
that
that
that
¬(P ∧ Q) is equivalent to ¬P ∨ ¬Q.
¬(P =⇒ Q) is equivalent to P ∧ ¬Q.
(∀n ∈ N)(n even =⇒ n2 even).
P =⇒ Q is not equivalent to Q =⇒ P .
Hint: You might find truth tables useful for some of these.
Solution:
(a) Compute the truth table
P
T
T
F
F
Q P ∧ Q ¬(P ∧ Q) ¬P ¬Q (¬P ) ∨ (¬Q)
T
T
F
F
F
F
F
F
T
F
T
T
T
F
T
T
F
T
F
F
T
T
T
T
The fourth and seventh columns are the same.
(b) Compute the truth table
P
T
T
F
F
Q P =⇒ Q ¬(P =⇒ Q) ¬Q P ∧ (¬Q)
T
T
F
F
F
F
F
T
T
T
T
T
F
F
F
F
T
F
T
F
Comparing the fourth and sixth columns does it.
(c) We will show this directly, by assuming that n is even and showing that n2 must
be even. If n is even, then by definition n = 2k for some k ∈ Z. Then we have
n2 = (2k)2 = 4k 2 = 2(2k 2). But if k ∈ Z, then so is 2k 2, so we have that n2 is
twice as integer, and therefore n2 is even.
(d) Compute
P
T
T
F
F
Q P =⇒ Q Q =⇒ P
T
T
T
F
F
T
T
T
F
F
T
T
The last two columns are different, so these are not the same.
4. (20 pts.) For each of the following functions, determine whether or not the function
is injective, whether or not it is surjective, and whether or not it is bijective. Justify
your statements.
(a) f : Z → Z,
f (x) = 2x
(b) f : R → R,
f (x) = 2x
(c) f : N → N,
f (x) = x3
(d) f : {days of the week} → N, f (day) = number of letters in name of day
(e.g.: f (Monday) = 6, f (Tuesday) = 7, etc.)
Solution.
(a) This function is injective. The best way to see this is if we let x 6= y, then
2x 6= 2y, and therefore f (x) 6= f (y).
This function is not surjective. The only possible outputs are even numbers, and
thus the range will not be all of Z. For example, 2x = 3 has no solution for x ∈ Z.
Since the function is not surjective, it is not bijective.
(b) This function is injective, see reasoning in part (a).
This function is surjective. Choose any y ∈ R, and notice that to solve 2x = y
we choose x = y/2. If y is a real number so is x = y/2, so we have an element of
R which maps to y.
Since the function is both injective and surjective, then it is bijective.
(c) This function is injective. To see this, let x, y be two natural numbers with
0 < x < y.
Then multiplying this inquality by x and then by y gives the two inqualities
0 < x2 < xy,
0 < xy < y 2 ,
so 0 < x2 < xy < y 2 . Pushing this idea one step further, we can multiply this set
of inequalities by x and by y and obtain
0 < x3 < x2y < xy 2 ,
0 < x2y < xy 2 < y 3 .
From this we get x3 < y 3 and we are done.
This function is not surjective. To see this, note that x3 = 2 has no integer
solution.
Finally, since this function is not surjective it is not bijective.
(d) This function is not injective, since f (Monday) = f (Friday) = 6.
This function is not surjective, because there is no day with 22 letters.
Since it is neither injective or surjective, it is not bijective.
5. (20 pts.) We define the sets A = {1, 2, 4, 7}, B = {2, 3, 4, 5}, C = {1, 2, 3}.
(a) What is A ∩ B?
(b) Show A ∩ B ⊆ A. Explain in words why this should be so.
(c) Is C ⊆ A ∪ B?
(d) Is C = A ∪ B?
Solution.
(a) A ∩ B = {1, 2}.
(b) Since both 1 and 2 are in A, we have A ∩ B ⊆ A. In general, this will always be
a true statement, since if an element is in A ∩ B, it must be in A and in B, so it
is in A.
(c) Yes. We see that A ∪ B = {1, 2, 3, 4, 5, 7}, and we can see that every element in
C is also in A ∪ B, so C ⊆ A ∪ B.
(d) No. As we just saw, A ∪ B has six elements and is thus a bigger set than C.