# Mathematical Induction by babbian

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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
Inductive
hypothesis
Prove that 1 + 3 + … + (2k - 1) + (2k + 1) = (k+1)2
By inductive
1 + 3 + … + (2k-1) + (2k+1) = k2 + (2k + 1)     hypothesis
= (k+1)2
By arithmetic
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Mathematical Induction -                        a cool example

Deficient Tiling
A 2n x 2n sized grid is deficient
if all but one cell is tiled.
2n

2n

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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:

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Mathematical Induction -                    a cool example

Yes!

• Is it true for all 21 x 21 grids?

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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.

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Mathematical Induction -                             a cool example

2k   2k

2k
?         ?
2k+1

2k
OK!!
(by
IH)        ?
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Mathematical Induction -                               a cool example

2k    2k

OK!!       OK!!
2k                    (by        (by
IH)        IH)

2k+1
OK!!       OK!!
2k                    (by        (by
IH)        IH)

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Mathematical Induction -                   a cool example

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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 well-
ordered.
No.
{ x  Z : x < 0 } has no
Is the set of integers (Z) well              least element.
ordered?
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Mathematical Induction -                             why does it work?

Is the set of non-negative reals (R)
well ordered?

No.
{ x  R : x > 1 } has no
least element.

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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
Proof by
2. k P(k)  P(k+1)
3. n P(n)                                 n P(n)

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Mathematical Induction -                                         why does it work?

Assume
1. P(0)
2. n P(n)  P(n+1)
3. n P(n)                                 n P(n)

Let S = { n : P(n) }                          Since N is well ordered, S has a least
element. Call it k.

But by (2), P(k-1)  P(k).
What do we know?                                          Contradicts P(k-1) true, P(k)
-P(k) is false because it’s in S.                                    false.
-k  0 because P(0) is true.
-P(k-1) is true because P(k) is the least element
in S.                          Done.
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Strong Mathematical Induction

If
P(0) and
n0 (P(0)  P(1)  …  P(n))  P(n+1)

Then
In our proofs, to show P(k+1), our
n0 P(n)                                 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.

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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

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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

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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
player wins each time.            How is this
different than
regular
induction?
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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

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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

Why does
this work?

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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:
Inductive
(Recursive)
Recursive Case                                              Definition
 n  (n  1)! if n  1
n! 
Base Case                          1 if n  0

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Recursive Definitions

Another VERY common example:

Fibonacci Numbers
0                       if n  0
                                              Base Cases
f (n)  1                       if n  1
 f (n  1)  f (n  2) if n  1
                                            Recursive Case

éæ    ön æ1 - 5 ön ù
Is there a non-recursive                             1 ê 1+ 5
definition for the Fibonacci
f (n) =     ç    ÷ -ç      ÷ú
5 êè 2 ø è 2 ø ú
ë                  û
Numbers?

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