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MATH 135
Solutions to
Assignment 4
W5
This assignment is due at 8:30 a.m. on Wednesday, February 3, in the drop box outside the
Tutorial Centre, MC 4067.
Remember to show all of your computational work.
1. [
5 marks
]
Find a test for divisibility by 8.
Answer
: We notice that 10
≡
2(mod 8), 10
2
≡
4(mod 8), and 10
3
≡
0(mod 8). This means
that 10
n
≡
0(mod 8) for any
n
≥
3, so given an integer
a
r
a
r

1
. . . a
1
a
0
, we have
10
r
a
r
+
. . .
+ 10
3
a
3
+ 10
2
a
2
+ 10
a
1
+
a
0
≡
0 +
. . .
+ 0 + 10
2
a
2
+ 10
a
1
+
a
0
(mod 8)
≡
10
2
a
2
+ 10
a
1
+
a
0
(mod 8)
Therefore, our integer will be divisible by 8 if and only if the number one gets by considering
only its last three digits (10
2
a
2
+ 10
a
1
+
a
0
) is divisible by 8.
2. [
9 marks total: 3 marks for each of (a), (b), and (c)
]
For each of the following
relations, determine whether it is reﬂexive, symmetric, and transitive and explain why in
each case.
(a) Let
S
be the integers
Z
and let
aRb
if and only if

a

b
 ≥
5.
Answer
:
R
is symmetric because

a

b

=

b

a

for all integers
a
and
b
.
R
is not
reﬂexive since

0

0
 6≥
5, and it is not transitive since 0
R
5 and 5
R
(

1), but 0 is not
related to

1 (or any other counterexamples).
(b) Let
S
be the real numbers
R
and let
xRy
if and only if

x

+

y

= 1.
Answer
:
R
is symmetric because

x

+

y

=

y

+

x

for any reals
x
and
y
. However, it
is not reﬂexive because 1 is not related to 1 (or any other counterexample), and it is not
transitive because 0
R
1 and 1
R
0, but 0 is not related to 0 (or any other counterexample).
(c) Let
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This note was uploaded on 02/21/2010 for the course MATH 135 taught by Professor Andrewchilds during the Spring '08 term at Waterloo.
 Spring '08
 ANDREWCHILDS
 Math

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