Proofs
by
Induction
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.1
An Example of Induction
Suppose we have a property (say color) of
the natural numbers:
0, 1, 2, 3, 4, 5, …
Showing that zero is red, and that
the successor of any red number is red,
proves that all numbers are red!
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.2
The Induction Rule
0 and (from n to n+1)
proves 0, 1, 2, 3,….
R(0), ∀n∈N [R(n)→R(n+1)]
∀m∈N R(m)
_____
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.3
Like Dominos…
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.4
Example Induction Proof
Let’s prove:
n +1
−1
1+ r + r +L + r =
r −1
2
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
n
r
September 21, 2005
L2-1.5
Proof by Induction
Statements in green form a template for
inductive proofs:
Proof: (by induction on n)
The induction hypothesis:
n +1
−1
P(n) ::= 1 + r + r + L + r =
r −1
2
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
n
September 21, 2005
r
L2-1.6
Example Induction Proof
Base Case (n = 0):
? r 0+1 − 1
2
0
r2444
r =
1144
+ r +4
+ L +3
r −1
1
r −1
=
Wait: divide by zero bug!
This is only true for r ≠ 1
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
r −1
September 21, 2005
=1
L2-1.7
An Example Proof
Revised Theorem:
n +1
−1
∀r ≠ 1 1 + r + r + L + r =
r −1
2
r
n
Revised Induction Hypothesis:
n +1
−1
P (n) ::= ∀r ≠ 1 1 + r + r + L + r =
r −1
2
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
n
r
L2-1.8
An Example Proof
Induction Step: Assume P(n) for
n ≥ 0 to prove P(n + 1):
∀r ≠ 1 1 + r + r + L + r
2
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
n +1
September 21, 2005
=
r
( n +1) +1
−1
r −1
L2-1.9
An Example Proof
Have P(n) by assumption:
n +1
−1
1+ r + r +L + r =
r −1
2
n
r
Adding r n+1 to both sides:
1+L + r + r
n
n +1
=
=
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
n +1
− 1 n +1
+r
r −1
n +1
n +1
r − 1 + r (r − 1)
r −1
r
September 21, 2005
L2-1.10
An Example Proof
Continued…
1+L + r + r
n
n +1
=
=
n +1
n +1
− 1 + r ⋅ r + −r
r −1
( n +1) +1
r
−1
r −1
r
n +1
Which is just P(n+1)
Therefore theorem is true
by induction. QED.
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.11
An Aside: Ellipses
Ellipses (…) mean that the reader is
supposed to infer a pattern.
• Can lead to confusion
• Summation notation gives more precision,
for example:
n
1+ r + r +L + r = ∑ r
2
n
i
i =0
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.12
Problems
Class Problem 1
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.13
The MIT Stata Center
Copyright © 2003, 2004, 2005 Norman Walsh. This work is licensed under a Creative Commons License.
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.14
The Stata Center Plaza
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.15
The Gehry/Gates Plaza
Goal: tile the squares, except one in the
middle for Bill.
Photo courtesy of Ricardo Stuckert/ABr.
2
n
2
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
n
September 21, 2005
L2-1.16
The Gehry/Gates Plaza
Gehry specifies L-shaped tiles covering
three squares:
For example, for 8 x 8 plaza might tile for Bill
this way:
Photo courtesy of Ricardo Stuckert/ABr.
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.17
The Gehry/Gates Plaza
Theorem: For any 2n × 2n plaza, we can
make Bill and Frank happy.
Proof: (by induction on n)
P(n) ::= can tile 2n × 2n with Bill in middle.
Base case: (n=0)
(no tiles needed)
Photo courtesy of Ricardo Stuckert/ABr.
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.18
The Gehry/Gates Plaza
Induction step: assume can tile 2n × 2n,
prove can handle 2n+1 × 2n+1.
2
2
n
n +1
Photo courtesy of Ricardo Stuckert/ABr.
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.19
The Gehry/Gates Plaza
Now what?
2
2
n
n +1
Photo courtesy of Ricardo Stuckert/ABr.
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.20
The Gehry/Gates Plaza
The fix:
Prove that we can always find
a tiling with Bill in the corner.
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.21
The Gehry/Gates Plaza
Note: Once have Bill in corner,
can get Bill in middle:
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
Photo courtesy of Ricardo Stuckert/ABr.
September 21, 2005
L2-1.22
The Gehry/Gates Plaza
Method:
Rotate the squares as indicated.
Photo courtesy of Ricardo Stuckert/ABr.
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.23
The Gehry/Gates Plaza
Method: after rotation have:
Photo courtesy of Ricardo Stuckert/ABr.
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.24
The Gehry/Gates Plaza
Method: Now group the 4 squares together,
and insert a tile.
Photo courtesy of Ricardo Stuckert/ABr.
Done!
Bill in
middle
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.25
The Gehry/Gates Plaza
Theorem: For any 2n × 2n plaza, we can
put Bill in the corner.
Proof: (by induction on n)
P(n) ::= can tile 2n × 2n with Bill in corner
Base case: (n=0)
(no tiles needed)
Photo courtesy of Ricardo Stuckert/ABr.
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.26
The Gehry/Gates Plaza
Induction step:
Assume we can get Bill in corner of 2n × 2n.
Prove we can get Bill in corner of 2n+1 × 2n+1.
2
n
2
n
Photo courtesy of Ricardo Stuckert/ABr.
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.27
The Gehry/Gates Plaza
Method: Rotate the squares as indicated.
Photo courtesy of Ricardo Stuckert/ABr.
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.28
The Gehry/Gates Plaza
Method: Rotate the squares as indicated.
after rotation have:
Photo courtesy of Ricardo Stuckert/ABr.
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.29
The Gehry/Gates Plaza
Method: Now group the squares together,
and fill the center with a tile.
Done!
Photo courtesy of Ricardo Stuckert/ABr.
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.30
Ingenious Induction
Hypotheses
Note 1: To prove
“Bill in middle,”
we proved something
else: “Bill in corner.”
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.31
Ingenious Induction
Hypotheses
Note 2: Other
times it helps to
choose a stronger hypothesis
than the desired result.
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.32
Inductive (Recursive)
Procedures
Note 3: The induction proof of
“Bill in corner” implicitly defines
a recursive procedure for
constructing a 2n+1 × 2n+1 corner
tiling from a 2n × 2n corner tiling.
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.33
Problems
Class Problem 2
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.34
A False Proof
Theorem: All horses are the same color.
Proof: (by induction on n)
Induction hypothesis:
P(n) ::= any set of n horses have the same color
Base case (n=0):
No horses so vacuously true!
…
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.35
A False Proof
(Inductive case)
Assume any n horses have the same color.
Prove that any n+1 horses have the same color.
…
n+1
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.36
A False Proof
(Inductive case)
Assume any n horses have the same color.
Prove that any n+1 horses have the same color.
Second set of n horses have the same color
…
First set of n horses have the same color
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.37
A False Proof
(Inductive case)
Assume any n horses have the same color.
Prove that any n+1 horses have the same color.
…
Therefore the set of n+1 have the same color!
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.38
A False Proof
What is wrong?
n =1
Proof that P(n) →P(n+1)
is false if n = 1, because the two
horse groups do not overlap.
Second set of n=1 horses
First set of n=1 horses
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.39
A False Proof
Proof that P(n) →P(n+1)
is false if n = 1, because the two
horse groups do not overlap.
(But proof works for all n ≠ 1)
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
September 21, 2005
L2-1.40