HOMEWORK 1  SOLUTIONS
1. By induction on
n
:
•
Base Case
(
n
= 1):
f
n
=
f
1
=
f
which is surjective by assumption.
•
Inductive Step
: Suppose
f
n
is surjective for all 1
≤
n
≤
k
. We want to show that
f
k
+1
is surjective.
Let
x
∈
X
be arbitrary.
We want to find
y
∈
X
such that
f
k
+1
(
y
) =
x
. Well
f
k
+1
(
y
) =
f
(
f
k
(
y
)) and we know
f
is surjective, so first pick
some
z
∈
X
such that
f
(
z
) =
x
.
Next, by our inductive hypothesis, we know
that
f
k
is surjective so pick some
y
such that
f
k
(
y
) =
z
. This choice of
y
does it,
since:
f
k
+1
(
y
) =
f
(
f
k
(
y
)) =
f
(
z
) =
x
2. (GRADED  EASY). By induction on

X

:
•
Base Case
(
n
= 0): If

X

= 0, then it’s the empty set and has only one subset,
namely itself, the empty set. And indeed:
P
(
X
)

=
{∅}
= 1 = 2
0
= 2

X

•
Inductive Step
: Assume that for all
n
≤
k
and every set
X
such that

X

=
n
,
P
(
X
)

= 2

X

.
Now suppose
X
is a set with

X

=
k
+ 1.
Let’s write
X
=
{
x
1
, . . . , x
k
+1
}
. Now every subset of
X
either contains
x
k
+1
or it doesn’t, and it’s
easy to see that there are an equal number of subsets of
X
which contain
x
k
+1
as there are which don’t. Thus the number of subsets of
X
is double the number
of subsets of
{
x
1
, . . . , x
k
}
. This latter set has cardinality
k
, so we can apply the
inductive hypothesis to obtain:
P
(
X
)

= 2
· P
(
{
x
1
, . . . , x
k
}
)

= 2
·
2
k
= 2
k
+1
3. The formula is
z

1
=
z
*

z

2
. To check that this works, we need to check that
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 Fall '08
 GUREVITCH
 Math, Complex number, inductive hypothesis, inductive step, quadratic polynomial x2

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