Mathematical Induction

The Principle of Mathematical Induction

Application in the Series

Application in divisibility

The Principle of Mathematical Induction

數學歸納法﹝ Mathematical Induction﹞是用來証明某些與

自然數 n 有關的數學命題的一種方法。它的步驟是：

1. 驗証 n = 1 時命題成立﹝這叫歸納的基礎﹞；

2. 假設 n = k 時命題成立﹝這叫歸納假設﹞，

在這假設下証明 n= k+1 時命題成立。

        根據 1、 2 可以斷定命題對一切自然數都 成立。

Application in Series

2n1n2531

Firstly, we need to know the names in the series clearly.

The first term

The n-th term

1n2531 How many terms in the above series?

Can you deduce the (n+1)-th term?

1n2

12n2

11n2

Answer is:

Example 21nn1n3n1037241

Let the proposition is S(n),can you write down S(k)? 21kk1k3k1037241

Also, what is S(k+1)?

k+1

1 2

k

WHO AM I？

211k1k

11k31k1k3k

1037241

In steps of calculation, the meaning of principle of mathematical induction is follows:

nSnnnExample

2

121:

2

1kkk21

Steps means: 2

1kkk21

2

11k1k1kk21

Explanation nSnn

nExample 2

121:

Firstly, prove S(1) is true:

.)1(

12

111..

1..

,1

trueisS

SHR

SHL

nWhen

Assume S(k) is true,

2

1kkk21

Use S(k) is true,prove S(k+1) is also true.

Following steps is the hardest part in the mathematical induction.

2

11k1k1kk21

SHR

kk

kk

kkk

kkk

..2

1112

212

121

12

1

Because S(k) is true

121.. kkSHL

.1 trueiskS By the principle of mathematical induction, S(n) is true for all positive integers n.

Example

3n2n1nn4

12n1nn543432321

Let the proposition is S(n).When n = 1,

.1

6

31211114

1...

6

21111...

trueisS

SHR

SHL

Assume S(k) is true, i.e.,

3k2k1kk4

12k1kk543432321

When n = k+1,

.1

...4

31211114

43214

3214321

3213214

1

32121543432321

trueiskS

SHR

kkkk

kkkk

kkkkkkk

kkkkkkk

kkkkkk

By the principle of mathematical induction, S(n) is true for all positive integers n.

Application in divisibility

The definition of divisibility

0bIf a and b be two integers with

Then, Integer a is divisible by b if

bMa where M is an integer.

Example

6 30 bydivisibleis

5636 where 5 is an integer.

6 32 bydivisiblenotis

3

16632 where is an integer.

3

16

ExplanationProve, by M.I., is divisible by 8 for all natural numbers n.

132 n

Let P(n) be the proposition

8132 bydivisibleisn

18

1313

,1212

nWhen

P(1) is true

Assume that P(k) is true,

183

8132

2

M

Mk

k

M is an integer.

When n = k+1,

198

898

1189

139

13132

2212

M

M

M

k

kk

9M + 1 is an

integer.

P(k+1) is true.

By the principle of mathematical induction, is divisible by 8 for all natural numbers n.

132 n

Further ExampleProve, by M.I., is divisible by 5 for all natural numbers n.

222 23 nn

Let P(n) be the proposition ‘is divisible by 5. ’

222 23 nn

Show P(1) is true.

,1nwhen

25

10

2323 0221212

Assume that P(k) is true.

Mkk 523 22

where M is an integer.

Consider P(k+1) is true or not.

So hard!!

22

22

2222222

222

22222

21212

295

2559

2429239

4293

23

23

k

k

kkkk

kk

kk

kk

M

M

k is positive

integer.

P(k+1) is true.

1k

By the principle of mathematical induction,

is divisible by 5

for all natural numbers n.

222 23 nn

2229 kM is an integer.