Solutions to Homework 1
February 10, 2007
Note:
If you don’t understand how a question was graded then (1) ﬁrst
read the solutions, (2) check with the grader who graded it, and then (and
only then) see Prof. Halpern. There is no statute of limitations on complaints
about grading.
General comment:
You should try to make their proofs clear and suc
cint, and make sure that each step follows logically. Remember, the graders
can’t read your minds! You have to lay out your reasoning clearly for them.
Also, writing neatly and laying things out neatly helps. Remember, they’re
all grading about 85 solutions!
0.1, 11
No justiﬁcation was required for this problem because of its simplic
ity. However, points were deducted for an incorrect or unclear justiﬁcation
(sometimes less is more!). For those who are interested, here is a formal
proof:
A
∩
B
=
A
:
If
x
∈
(
A
∩
B
), then
x
∈
A
and
x
∈
B
(thus
x
∈
A
). If
x
∈
A
,
then
x
∈
B
since
A
⊆
B
. So
x
∈
A
and
x
∈
B
or
x
∈
(
A
∩
B
).
Therefore
A
∩
B
=
A
if
A
⊂
B
.
A
∪
B
=
B
:
If
x
∈
(
A
∪
B
), then
x
∈
A
or
x
∈
B
. If
x
∈
B
, then we are
done. If
x
∈
A
, then again, we are done. Since
A
is a subset of
B
,
x
∈
A
→
x
∈
B
.
Conversely, if
x
∈
B
, then from the deﬁnition of union,
x
∈
(
A
∪
B
).
Therefore
A
∪
B
=
B
if
A
⊂
B
.
1
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View Full DocumentNote:
in general, to prove set equality, show that for every
x
∈
A
,
x
∈
B
, and vice versa.
0.1, 14
A
∪
B
⊂
S
.
0.1, 27
(b)
{
(
A,B
)
} 
A
⊃
B
}
(this is the superset relation)
(c)
{
(
x,y
)

x
≤
y
}
Comments:
Note that the reverse of
{
(
x,y
)

x
≥
y
}
is not
{
(
y,x
)

y
≥
x
}
. They are just the same sets. Also, many
students seem to have problems with specifying relations formally.
I tried to be lenient with this, but some mistakes were penalized.
For example, some students failed to write a relation as a SET of
ORDERED PAIRS. For example, in 27(c) students often wrote:
{
x,y

x
≤
y
}
or
x
≤
y
or just
≤
. I considered the last of these
minimally acceptable, but the ﬁrst two were not.
Section 0.1, 33
(a) Injective; not surjective since 1 is not an image of anything.
(b) Injective; not surjective since 0 is not an image of anything.
(c) Bijective.
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 Spring '07
 
 Inductive Reasoning, Natural number, Transitive closure

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