Download MA1025 Solutions for Exam # 2, part 1 Mon. Aug 18th, 2008 Name

MA1025
Solutions for Exam # 2, part 1
Mon. Aug 18th, 2008
Name
Instructor: Dr. Ralucca Gera
Show all the necessary work in each problem to receive full credit, for a total up
to 50 points.
1. (20 points) Prove that 1 · 2 + 2 · 3 + 3 · 4 + . . . + n(n + 1) =
integers n.
n(n+1)(n+2)
3
for every positive
Proof: We prove by induction that Sn : 1 · 2 + 2 · 3 + 3 · 4 + . . . + n(n + 1) =
true for all natural numbers n.
The statement S1 : 1 · 2 =
1·2·3
3
n(n+1)(n+2)
3
is
is true.
Assume that Sk : 1 · 2 + 2 · 3 + 3 · 4 + . . . + k(k + 1) =
Sk+1 : 1 · 2 + 2 · 3 + 3 · 4 + . . . + (k + 1)(k + 2) =
k(k+1)(k+2)
3
(k+1)(k+2)(k+3)
3
is true and prove that
is true. Observe that
1 · 2 + 2 · 3 + 3 · 4 + . . . + (k + 1)(k + 2) = 1 · 2 + 2 · 3 + . . . + k(k + 1) + (k + 1)(k + 2)
=
k(k + 1)(k + 2)
+ (k + 1)(k + 2)
3
=
k(k + 1)(k + 2) + 3(k + 1)(k + 2)
3
=
(k + 1)(k + 2)(k + 3)
3
Thus by the Principle of Math Induction Sn is true for all natural numbers n.
Choose two problems (15 points each), or all three problems below (10 points each). Note
that you have a choice again in problem 3, and you must only choose one of (a) and (b).
2. (10/15 points) Prove that if a nonzero rational is divided by an irrational number, the
quotient is irrational.
Proof. Suppose, to the contrary, that a ∈ Q, a 6= 0, b ∈
/ Q and c = ab ∈ Q. As a ∈ Q, we
x
have that a = y , x, y ∈ Z, y 6= 0 (also note that x 6= 0 since a is nonzero rational). Also,
x
as c ∈ Q, we have that c =
which simplifies to
r
,
s
r
a
y
r, s ∈ Z, s 6= 0. Now, c = b 6= 0 is equivalent to = ,
b
s
b
rb
x
sx
= with s, y 6= 0. This implies that b = , ry 6= 0, which makes b
s
y
ry
a rational number. This is a contradiction since we assumed that b is irrational.
3. (10/15 points) Chose either (a) or (b) below:
(a) Prove that for every set A ⊆ R, there is a set B such that |P(A ∩ B)| = 1.
Proof. Let A be a set in R, and let B = A. Then A ∩ B = ∅ and so |P(A ∩ B)| =
20 = 1
Solution 2: Proof. Let A be a set in R, and let B = ∅. Then A ∩ B = ∅ and so
|P(A ∩ B)| = 20 = 1
(b) Every nonzero rational number can be written as the product of two irrational numbers.
Proof.
Lat a ∈ Q. Then we can write a as a product of two irrationals as a =
√
( 2)( √a2 ), where √a2 is irrational as product of a rational and an irrational.
4. (10/15 points) Disprove the following: For every two sets A and B, |P(A) ∩ P(B)| = 0.
Solution: The statement is false.
Counterexample: Let A = {1, 2} and B = {2, 3}. Then P(A ∩ B) = {∅, {2}} and so
|P(A) ∩ P(B)| = 2.
(Note: another counterexample Counterexample: Let A = {1} and B = {2}. Then P(A) =
{∅, {1}} and P(A) = {, {2}}. Thus P(A) ∩ P(B) = {∅} and so |P(A) ∩ P(B)| = 1.)
Extra credit. (10 points) Let m be a positive integer of the form m = 2s, where s is an
odd integer. Prove that there does not exist positive integers x and y such that x2 − y 2 = m.
Proof. Let m be a positive integer of the form m = 2s, where s is an odd integer, and assume to the contrary that there exists positive integers x and y such that x2 − y 2 = m. Then
s = 2k +1 and so m = 2(2k +1). Thus x2 −y 2 = m is equivalent to (x−y)(x+y) = 2(2k +1).
We consider the following cases.
(a) x is even, and y is odd. Then x − y and x + y are both odd, and so (x − y)(x + y) =
2(2k+1) is a contradiction since the left hand side is odd and the right hand side is even.
(b) x is odd, and y is even. Then x − y and x + y are both odd, and so (x − y)(x + y) =
2(2k + 1) is a contradiction since the left hand side is odd and the right hand side is
even. (Note that case (a) and (b) could have been combined if using without loss of
generality).
(c) x and y are odd. Then x − y and x + y are both even, say x − y = 2s and x + y = 2r,
r, s, ∈ Z. And so (x − y)(x + y) = 2(2k + 1) is equivalent to 4rs = 2(2k + 1) which
simplifies to 2rs = 2k + 1. This is a contradiction since the left hand side is even and
the right hand side is odd.
(d) x and y are even. Then x − y and x + y are both even, say x − y = 2s and x + y = 2r,
r, s, ∈ Z. And so (x − y)(x + y) = 2(2k + 1) is equivalent to 4rs = 2(2k + 1) which
simplifies to 2rs = 2k + 1. This is a contradiction since the left hand side is even and
the right hand side is odd.
Note that a much shorter proof could be obtained if we would consider these cases instead:
Proof. Let m be a positive integer of the form m = 2s, where s is an odd integer, and assume to the contrary that there exists positive integers x and y such that x2 − y 2 = m. Then
s = 2k +1 and so m = 2(2k +1). Thus x2 −y 2 = m is equivalent to (x−y)(x+y) = 2(2k +1).
We consider the following cases.
(a) x and y are of opposite parity. Then x−y and x+y are both odd, and so (x−y)(x+y) =
2(2k+1) is a contradiction since the left hand side is odd and the right hand side is even.
(d) x and y are of the same parity. Then x − y and x + y are both even, say x − y = 2s and
x + y = 2r, r, s, ∈ Z. And so (x − y)(x + y) = 2(2k + 1) is equivalent to 4rs = 2(2k + 1)
which simplifies to 2rs = 2k + 1. This is a contradiction since the left hand side is even
and the right hand side is odd.