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Review: Mathematical Induction
Use induction to prove that the sum
of the first n odd integers is n2.
Prove a base case (n=1)
Base case (n=1): the sum of the first 1 odd integer
is 12. Yes, 1 = 12.
Prove P(k)P(k+1)
Assume P(k): the sum of the first k odd ints is k2. 1
+ 3 + … + (2k - 1) = k2
Prove that 1 + 3 + … + (2k - 1) + (2k + 1) = (k+1)2
1 + 3 + … + (2k-1) + (2k+1) = k2 + (2k + 1)
= (k+1)2
Networking
Platform
1 Extensible
- CSE 240 – Logic
and Discrete
Mathematics
Inductive
hypothesis
By inductive
hypothesis
By arithmetic
1
Mathematical Induction -
a cool example
Deficient Tiling
A 2n x 2n sized grid is deficient
if all but one cell is tiled.
2n
2n
Networking
Platform
2 Extensible
- CSE 240 – Logic
and Discrete
Mathematics
2
Mathematical Induction -
a cool example
• We want to show that all 2n x 2n sized
deficient grids can be tiled with tiles, called
triominoes, shaped like:
Networking
Platform
3 Extensible
- CSE 240 – Logic
and Discrete
Mathematics
3
Mathematical Induction -
a cool example
Yes!
• Is it true for all 21 x 21 grids?
Networking
Platform
4 Extensible
- CSE 240 – Logic
and Discrete
Mathematics
4
Mathematical Induction -
a cool example
Inductive Hypothesis:
We can tile any 2k x 2k deficient
board using our fancy designer
tiles.
Use this to prove:
We can tile any 2k+1 x 2k+1
deficient board using our fancy
designer tiles.
Networking
Platform
5 Extensible
- CSE 240 – Logic
and Discrete
Mathematics
5
Mathematical Induction -
2k
2k
2k
2k
?
?
OK!!
(by
IH)
Networking
Platform
6 Extensible
- CSE 240 – Logic
and Discrete
Mathematics
?
a cool example
2k+1
6
Mathematical Induction -
2k
2k
2k
OK!!
(by
IH)
OK!!
(by
IH)
a cool example
2k+1
2k
OK!!
(by
IH)
Networking
Platform
7 Extensible
- CSE 240 – Logic
and Discrete
Mathematics
OK!!
(by
IH)
7
Mathematical Induction -
Networking
Platform
8 Extensible
- CSE 240 – Logic
and Discrete
Mathematics
a cool example
8
Mathematical Induction -
why does it work?
Definition:
A set S is “well-ordered” if every
non-empty subset of S has a least
element.
Given (we take as an axiom): the set
of natural numbers (N) is wellordered.
Is the set of integers (Z) well
ordered?
Networking
Platform
9 Extensible
- CSE 240 – Logic
and Discrete
Mathematics
No.
{ x  Z : x < 0 } has no
least element.
9
Mathematical Induction -
why does it work?
Is the set of non-negative reals (R)
well ordered?
Extensible
Networking
Platform
10
- CSE 240 – Logic
and Discrete
Mathematics
No.
{ x  R : x > 1 } has no
least element.
10
Mathematical Induction -
why does it work?
Proof of Mathematical Induction:
We prove that (P(0)  (k P(k)  P(k+1)))
 (n P(n))
Assume
1. P(0)
2. k P(k)  P(k+1)
3. n P(n)
Networking
Platform
11Extensible
- CSE 240 – Logic
and Discrete
Mathematics
Proof by
contradiction.
n P(n)
11
Mathematical Induction Assume
1. P(0)
2. n P(n)  P(n+1)
3. n P(n)
Let S = { n : P(n) }
why does it work?
n P(n)
Since N is well ordered, S has a least
element. Call it k.
But by (2), P(k-1)  P(k).
Contradicts P(k-1) true, P(k)
false.
What do we know?
-P(k) is false because it’s in S.
-k  0 because P(0) is true.
-P(k-1) is true because P(k) is the least element
in S.
Done.
Extensible
Networking
Platform
12
- CSE 240 – Logic
and Discrete
Mathematics
12
Strong Mathematical Induction
If
P(0) and
n0 (P(0)  P(1)  …  P(n))  P(n+1)
Then
n0 P(n)
In our proofs, to show P(k+1), our
inductive hypothesis assumes
that ALL of P(0), P(1), … P(k) are
true, so we can use ANY of them
to make the inference.
Extensible
Networking
Platform
13
- CSE 240 – Logic
and Discrete
Mathematics
13
Game with Matches
• Two players take turns removing any
number of matches from one of two piles
of matches. The player who removes the
last match wins
• Show that if two piles contain the same
number of matches initially, then the
second player is guaranteed a win
Extensible
Networking
Platform
14
- CSE 240 – Logic
and Discrete
Mathematics
14
Strategy for Second Player
• Let P(n) denote the statement “the second
player wins when they are initially n matches in
each pile”
• Basis step: P(1) is true, because only 1 match in
each pile, first player must remove one match
from one pile. Second player removes other
match and wins
• Inductive step: suppose P(j) is True for all j
1<=j <= k.
• Prove that P(k+1) is true, that is the second
player wins when each piles contains k+1
matches
Extensible
Networking
Platform
15
- CSE 240 – Logic
and Discrete
Mathematics
15
Strategy for Second Player
• Suppose that the first player removes r
matches from one pile, leaving k+1 –r
matches there
• By removing the same number of matches
from the other pile the second player
creates the situation of two piles with
k+1-r matches in each. Apply the
inductive hypothesis and the second
How is this
player wins each time.
Extensible
Networking
Platform
16
- CSE 240 – Logic
and Discrete
Mathematics
different than
regular
induction?
16
Postage Stamp Example
• Prove that every amount of postage of 12
cents or more can be formed using just
4-cent and 5-cent stamps
• P(n) : Postage of n cents can be formed
using 4-cent and 5-cent stamps
• All n >= 12, P(n) is true
Extensible
Networking
Platform
17
- CSE 240 – Logic
and Discrete
Mathematics
17
Postage Stamp Proof
• Base Case: n = 12, n = 13, n = 14, n = 15
– We can form postage of 12 cents using 3, 4-cent stamps
– We can form postage of 13 cents using 2, 4- cent stamps
and 1 5-cent stamp
– We can form postage of 14 cents using 1, 4-cent stamp and
2 5-cent stamps
– We can form postage of 15 cents using 3, 5-cent stamps
• Induction Step
– Let n >= 15
– Assume P(k) is true for 12 <= k <= n, that is postage of k
cents can be formed with 4-cent and 5-cent stamps
(Inductive Hypothesis)
– Prove P(n+1)
– To form postage of n +1 cents, use the stamps that form
postage of n-3 cents (from I.H) with a 4-cent stamp
Extensible
Networking
Platform
18
- CSE 240 – Logic
and Discrete
Mathematics
Why does
this work?
18
Recursive Definitions
We completely understand the function f(n)
= n!, right?
As a reminder, here’s the definition:
n! = 1 · 2 · 3 · … · (n-1) · n, n  1
But equivalently, we could define it like this:
Recursive Case
Base Case
 n  (n  1)! if n  1
n! 
1 if n  0

Extensible
Networking
Platform
19
- CSE 240 – Logic
and Discrete
Mathematics
Inductive
(Recursive)
Definition
19
Recursive Definitions
Another VERY common example:
Fibonacci Numbers
if n  0
0

f (n)  1
if n  1
 f (n  1)  f (n  2) if n  1

Is there a non-recursive
definition for the Fibonacci
Numbers?
Extensible
Networking
Platform
20
- CSE 240 – Logic
and Discrete
Mathematics
Base Cases
Recursive Case
n
nù
éæ
ö
æ
ö
1 ê 1+ 5
1- 5 ú
f (n) =
ç
÷ ç
÷
2
2
5 êëè
ø è
ø úû
20