1. (10 points) Let A = {{0}, 0, 1} and B = {0, 1}.
(Note: A Yes or No answer will suffice.)
(a) Is A 
B? No
(b) Is B A? Yes
(c) Is A A? Yes
(d) Is {0} ∈ A? Yes
(e) Is {0} ∈B? No
2. (14 points)
(a) Fill in the blanks in the fol lowing statement of the Quotient Remainder
Theorem.
Given any integer n and any positive integer d, there exist unique integers
q and r such that
n = dq + r and
(b) Let n = −17 and d = 4. Find the unique q and r that show that -17 and 4
satisfy the
Quotient Remiander Theorem.
−17 = 4 · −5 + 3
Thus q = −5 and r = 3.
(c) Calculate the following .
47 mod 6 = 5
47 div 6 = 7
Since 47 = 6 · 7 + 5.
3. (10 points) As sume that p and q are integers with p ≠ 0
and q ≠ 0. Explain why 
is a
rational number .
(Note: Be sure to make explicit reference to the definition of rational number.)
The products of integers are integers, so 3q, 4p and q2 are integers.
The difference of integers is an integer, so 3q − 4p is an integer.
Since q ≠ 0 the zero product properties implies that q2 ≠ 0
So is the ratio of two integers with non- zero
denominator , hence a rational number.
4. (12 points)
(a) Write the ex pression below using summation notation so that the lower limit
of the
summation is 1, i.e., fill in the blanks.
(b) Transform your resulting from part (a) using the
change of variables j = i + 1.
j = i + 1 so i = j − 1.
Thus 
; and when i = 1, j = 2; and when i = n,
j = n + 1.
Plugging these into the summation above we have:
5. (14 points) Consider the following statement:
For all integers a, b and c if
then
.
(Note: Recall 
means “does not divide.”)
(Note: This problem requires you to provide only statements not proofs.)
(a) If you were to prove the above statement by contrapostion, what statement
would you
prove?
For all integers a, b and c if a | b then a | bc.
(b) If you were to prove the above statement by
contradiction, what statement would you
show leads to a contradiction?
There exist integers a, b and c such that
and a | b.
6. (20 points) Prove that for any integers a, b and c, if
a | b and a | c then a | (2b − 3c).
proof
Let a, b and c be integers such that a | b and a | c.
By the definition of divisibility there exist integers k and l such that b = ak
and c = al.
2b − 3c = 2ak − 3al = a(2k − 3l)
2k − 3l is an integer so a | (2b − 3c).
7. (20 points) Prove that the following statement by
mathematical induction .
For all integers n ≥1, 2 + 4 + 6 + · · · + 2n = n2 + n.
proof(by induction)
Let P(n) : 2 + 4 + 6 + · · · + 2n = n2 + n.
• Verify P(1)
The left hand side equals 2 and the right hand side equals 12 + 1 = 2, so P(1)
is true.
• Assume P(k) : 2 + 4 + 6 + · · · + 2k = k2 + k.
(2 + 4 + 6 + · · · + 2k) + 2(k + 1)
= (k2 + k) + 2(k + 1)
= k2 + k + 2k + 2
= k2 + 2k + 1 + k + 1
= (k + 1)2 + (k + 1)
Thus P(k + 1) : 2 + 4 + 6 + · · · + 2(k + 1) = (k + 1)2 + (k + 1) holds.
So by the principle of mathematical induction P(n) : 2+4+6+· · ·+2n = n2 +n is
true for
all n ≥1.
