CONCORDIA UNIVERSITY
DEPARTMENT OF COMPUTER SCIENCE AND SOFTWARE ENGINEERING
COMP232Section DD
MATHEMATICS FOR COMPUTER SCIENCE
ASSIGNMENT 2
SOLUTIONS
FALL 2009
1. A
perfect number
is a positive integer that equals the sum of all of its divisors, except the
number itself. Give a proof by contradiction that there is no perfect prime number.
PROOF: Suppose
p
is a perfect number as well as a prime number. Since
p
is prime, its
only divisers are 1 and
p
itself. Since
p
is perfect it is the sum of its divisors, not including
p
itself. Thus
p
= 1. But 1 not a prime number. Contradiction.
2. Prove that if
x
3
is irrational then
x
is irrational, by proving the contrapositive.
PROOF: The contrapositive is “
If
x
is rational then
x
3
is rational
”. To prove this assume
that
x
=
p
q
,where
p
and
q
are integers. Then
x
3
=
p
3
q
3
, which is a rational number.
3. Give a proof by cases to show that there are no integer solutions to the equation
2
x
2
+5
y
2
=1
4
.
PROOF: We have 2
x
2
≤
14,
i.e.
,
x
2
≤
7,
i.e.
,

x
≤
2(since
x
is an integer), and 5
y
2
≤
14,
i.e.
,
y
2
≤
2
.
8,
i.e.
,

y
≤
1. There remain 6 cases to be considered, namely,
x
=0
,
1
,
2,
combined with
y
,
1. (Negative
x
and
y
need not be considered separately.) It is easily
checked that none of these 6 combinations of
x
and
y
satisfy the equation.
4. Give a proof by contradiction to show that the cube root of 2 is an irrational number.
PROOF: Suppose 2
1
/
3
is rational,
i.e.
,2
1
/
3
=
p/q
p
and
q
are relatively prime (
i.e.
,
p
and
q
only have 1 as common factor.) Then 2 =
p
3
/q
3
,or
,
p
3
=2
q
3
.T
h
u
s
p
3
is even.
Hence
p
is even, say,
p
k
for some integer
k
h
e
n2
q
3
=(2
k
)
3
=8
k
3
,o
r
,
q
3
=4
k
3
.
Thus
q
3
is even, hence
q
is even. So
p
and
q
have 2 as common factor. Contradiction.
NOTE: We used the fact that if
n
3
is even then
n
is even, whose contrapositive is easily
proved: If
n
is odd then
n
k
+ 1, hence
n
3
=2(4
k
3
+6
k
2
+3
k
) + 1 is odd.
5. Give a proof by contradiction to show that if the integers 1, 2,
···
, 99, 100 , are
placed randomly around a circle (without repetition), then there must exist three adjacent
numbers along the circle whose sum is greater than 152.