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1. DISCRETE MATHEMATIC
LEC-03:
Induction
Fall, 2023

2. Problem
)
(
, k
P
N
k 

n
r
P
A
N
n )
1
(
, 



A compounded amount (including interest)
P principle (the original amount)
r annual rate
n number of year
Example

3. More examples
 
2
1
,
0
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

 
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k
i
N
k
k
i
n
n
n
n
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n 
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,
1
,

4. Three steps

5. Example 1
2
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1
0
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0
0
0


i
2
)
1
(
0


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k
k
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k
2
)
2
)(
1
(
1
0
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k
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k
 
2
1
,
0



 

k
k
i
N
k
k
i
Base Case: P(0)
Inductive Hypothesis: P(k)
Inductive Step: P(k+1)
)
1
(
0
1
0
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2
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1
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2
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1
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k

6. Example 2
2
2
!
2 
k
k
k 
!
1
)
1
(
)!
1
( 


 k
k
k
Base Case: P(2)
Inductive Hypothesis: P(k)
Inductive Step: P(k+1)
!
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1
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1
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 k
k
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n
n
n
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,
1
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7. Four color theorem
Any map can be colored with
four colors such that any two
adjacent countries must have
different colors.
The problem was first posed
in 1852. It is very difficult to
prove.
It was not until 1976 that the
theorem was finally proved
(with the aid of a computer)
by Appel and Haken.

8. Two color theorem
We divide the plane
into regions by drawing
straight lines.
R
R R
B
B
B
B
Suppose a map is drawn using only lines that extend
to infinity in both directions; two colors are sufficient
to color the countries so that no pair of countries with
a common border have the same color.

9. Step 1: Base case
The plane is divided
into 2 parts by 1 line
R
B

10. Step 2: Inductive hypothesis
Assume a map with n
lines can be two-
colored
R
R
B
B

11. Assume a map with n
lines can be two-
colored
Step 3: Inductive step
Leave colors on one
side of the (n+1)th line
unchanged.
On the other side of the
line, swap red <-> blue
R
R B
R
B
B
R
B
R
B
The (n+1)th line

12. Compound interest
n
r
P
A
N
n )
1
(
, 



A compounded amount (including interest)
P principle (the original amount)
r annual rate
n number of year
Example

13. Base case: P(0)
P
r
P
A 

 0
)
1
(
In year 0 you receive no interest, the
compounded amount is equal to the
principle.
A compounded amount (including interest)
P principle (the original amount)
r annual rate

14. Inductive hypothesis: P(k)
k
r
P
A )
1
( 

Assume this is the compounded amount
after k year(s).
A compounded amount (including interest)
P principle (the original amount)
r annual rate

15. Inductive step: P(k+1)
1
)
1
( 

 k
r
P
A
We must show this is the compounded
amount after k+1 year(s).
A compounded amount (including interest)
P principle (the original amount)
r annual rate
1
)
1
(
)
1
(
)
1
( 





 k
k
k
r
P
r
r
P
r
P
A

16. L-shaped tiles
L-shaped tiles: 2 x 2 square tile with a missing
1 x 1 square.
We want to tile a 2n x 2n square with a missing
1 x 1 square in the middle.
Example

17. Base case: P(1)
A 2 x 2 square can be tiled with L-shaped tiles
with a missing 1 x 1 square in the middle.
Example

18. Inductive hypothesis: P(n)
Assume we can tile a 2n x 2n square with a
missing 1 x 1 square in the middle.
Example

19. Inductive step: P(n+1)
A 2n+1 x 2n+1 square can be broken up into 4
smaller squares of size 2n x 2n.
We can tile the square with the hole being
anywhere.
Example

20. Strong induction
Inductive hypothesis: Instead of just assuming
P(k) is true, we assume a stronger statement
that P(0), P(1), …, and P(k) are all true.

21. Example
Base Case: P(2)
Inductive Hypothesis: P(2),P(3),…,and P(k)
Inductive Step: P(k+1)
Every natural number n > 1 can be written as a product of primes.
2 is a prime number.
Every natural number 2 <= n <= k can be written as a product of
primes.
k + 1 can be written as a product of primes.
Case 1: k + 1 is a prime
Case 2: k + 1 = xy, 1 < x,y < k

22. Example
Base Case: P(8)
Inductive Hypothesis: P(8),…,and P(k)
Inductive Step: P(k+1)
Any integers from 8 upwards can be composed from 3 and 5.
8 = 3 + 5
Every natural number 8 <= n <= k can be composed from 3 and 5.
k+1 = (k-2) + 3
P(9), P(10)

23. Strong induction vs. simple induction
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0
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0
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Q 
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(
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1
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P
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(
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k
Q 
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Base Case:
Inductive Hypothesis:
Inductive Step:

24. Induction and Recursion

25. Exercises

26. Exercises

27. Homework
Problem Set 3