Functions
Intro to Induction
Margaret M. Fleck
26 February 2010
This lecture finishes up functions and introduces mathematical induction
(section 4.1 of Rosen).
1
Announcements
Midterms should be graded mid next week. The last conflict exams will be
Tuesday afternoon. Solutions should be released soon after those are done.
Soon after midterm grades appear, I will also release mock averages, which
are weighted combinations of the first midterm, first quiz, and the first four
homeworks. We’ll also supply information on how to interpret these numbers
in terms of letter grades, so you can see where you stand at the moment.
2
Warmup using 2D points
Monday’s lecture probably went by really fast since you were all worrying
about the midterm. So, let’s remember what it means to prove that a specific
function is, say, onto.
Let
f
:
Z
2
→
Z
be defined by
f
(
x,y
) =
x
+
y
. I claim that
f
is onto.
First, let’s make sure we know how to read this definition.
f
:
Z
2
is
1
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shorthand for
Z
×
Z
, which is the set of pairs of integers. So
f
maps a pair
of integers to a single integer, which is the sum of the two coordinates.
To prove that
f
is onto, we need to pick some arbitrary element
y
in the
codomain. That is to say,
y
is an integer. Then we need to find a sample
value in the domain that maps onto
y
, i.e. a “preimage” of
y
. At this point,
it helps to fiddle around on our scratch paper, to come up with a suitable
preimage. In this case, (0
,y
) will work nicely. So our proof looks like:
Proof: Let
y
be an element of
Z
.
Then (0
,y
) is an element of
f
:
Z
2
and
f
(0
,y
) = 0 +
y
=
y
. Since this construction will work
for any choice of
y
, we’ve shown that
f
is onto.
3
Another proof involving composition
Last class, we saw a proof of the following fact:
Claim 1
For any sets
A
,
B
, and
C
and for any functions
f
:
A
→
B
and
g
:
B
→
C
, if
f
and
g
are injective, then
g
◦
f
is also injective.
Let’s prove a similar claim involving surjective:
Claim 2
For any sets
A
,
B
, and
C
and for any functions
f
:
A
→
B
and
g
:
B
→
C
, if
f
and
g
are surjective, then
g
◦
f
is also surjective.
Proof: Let
A
,
B
, and
C
be sets. Let
f
:
A
→
B
and
g
:
B
→
C
be functions. Suppose that
f
and
g
are surjective.
We need to show that
g
◦
f
is surjective. That is, we need to show
that for any element
x
in
C
, there is an element
y
in
A
such that
(
g
◦
f
)(
y
) =
x
.
So, pick some element
x
in
C
. Since
g
is surjective, there is an
element
z
in
B
such that
g
(
z
) =
x
. Since
f
is surjective, there is
an element
y
in
A
such that
f
(
y
) =
z
.
Substituting the value
f
(
y
) =
z
into the equation
g
(
z
) =
x
, we
get
g
(
f
(
y
)) =
x
. That is, (
g
◦
f
)(
y
) =
x
. So
y
is the element of
A
we needed to find.
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 Spring '08
 Erickson
 Mathematical Induction, Inductive Reasoning, Natural number, Mathematical proof

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