Strong induction
Margaret M. Fleck
3 March 2010
This lecture presents proofs by “strong” induction, a slight variant on
normal mathematical induction.
1
Announcements
We aren’t quite ±nished grading the midterm. We’re hoping to have them
done late Thursday.
2
A geometrical example
As a warmup, let’s see another example of the basic induction outline, this
time on a geometrical application.
Tiling
some area of space with a certain
type of puzzle piece means that you ±t the puzzle pieces onto that area of
space exactly, with no overlaps or missing areas. A right triomino is a 2by2
square minus one of the four squares. (See pictures in Rosen pp. 277278.) I
then claim that
Claim 1
For any positive integer
n
, a
2
n
×
2
n
checkerboard with any one
square removed can be tiled using right triominoes.
Proof: by induction on
n
.
1
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View Full DocumentBase: Suppose
n
= 1. Then our 2
n
×
2
n
checkerboard with one
square remove is exactly one right triomino.
Induction: Suppose that the claim is true for some integer
k
.
That is a 2
k
×
2
k
checkerboard with any one square removed can
be tiled using right triominoes.
Suppose we have a 2
k
+1
×
2
k
+1
checkerboard
C
with any one
square removed. We can divide
C
into four 2
k
×
2
k
subcheckerboards
P
,
Q
,
R
, and
S
. One of these subcheckerboards is already miss
ing a square. Suppose without loss of generality that this one is
S
. Place a single right triomino in the middle of
C
so it covers
one square on each of
P
,
Q
, and
R
.
Now look at the areas remaining to be covered. In each of the
subcheckerboards, exactly one square is missing (
S
) or already
covered (
P
,
Q
, and
R
). So, by our inductive hypothesis, each
of these subcheckerboards minus one square can be tiled with
right triominoes. Combining these four tilings with the triomino
we put in the middle, we get a tiling for the whole of the larger
checkerboard
C
. This is what we needed to construct.
3
Strong induction
The inductive proofs you’ve seen so far have had the following outline:
Proof: We will show
P
(
n
) is true for all
n
, using induction on
n
.
Base
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 Spring '08
 Erickson
 Mathematical Induction, Natural number, Strong Induction, Mathematical proof

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