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Directory:

- Introduction to Mathematical Inductive method
- Incomplete inductive method
- Definition of incomplete inductive method
- An example of incomplete inductive method
- Strong Inductive method
- Strong inductive method definition
- Example of strong inductive method
- Recursive definition and structural inductive method
- Recursive definition
- Structural inductive method

Introduction to Mathematical Inductive method

Mathematical induction is a powerful proving tool. Mathematical induction is a tool for proving theorems, but it is not possible to find theorems. That is to say, mathematical induction is a means of proving.

Mathematics Induction High School We have also studied, that time study is incomplete induction method. In discrete mathematics, we will learn three mathematical induction methods:

- Incomplete induction (i.e. inductive method of high school description)
- Strong Inductive method
- Structural inductive method

There are also some interesting applications for recursive knowledge and mathematical induction.

The reference book in this chapter is "Discrete Mathematics and its application".

Incomplete inductive method

First of all we are familiar with incomplete induction, generally we call it a mathematical induction method.

About incomplete induction, although the name has "incomplete" three words, but this inductive method is precisely not incomplete. This naming is only frustration: because the strong inductive method is also called the complete induction method, so the induction method here is called the incomplete induction method.

For the incomplete inductive method, most books are illustrated by dominoes. Here we use the way to climb ladders to illustrate:

For ladders with n sections, we want to crawl from the first section to section N, using the following method:

- Can crawl to the first section
- If we can crawl to section K, then we can crawl to section k+1.

Now let's assume that the above two sentences are set up, so that you can use recursion to crawl to the nth staircase: first you can climb to the 1th staircase, and then make K=1, from the second sentence we can climb to the 2nd section, then make k=2, you can climb to the 3rd section. And so on, you can crawl to section N.

Incomplete inductive definition:

The above two steps to climb stairs are actually described using mathematical induction. The definition of incomplete inductive method is as follows:

In order to prove that all positive integers n,p (n) are true, where P (n) is a propositional function, it is necessary to complete the two-step proof:

Basic steps: proof P (1) is true

Induction step: Prove that for each integer k, the implication P (k) →p (K+1) is established. where P (k) is the inductive hypothesis.

This definition must be familiar to everyone, that is, the induction of high school in the definition of a discrete language to describe the chant.

For the stair problem above, P (n) is "able to climb to the nth staircase". The first step, then, is to get to the first stair, p (1), and then prove that the P (k+1) section ladder can be reached if P (k) can be reached. Complete the above two proofs to reach any staircase.

The above example is an abstract life example, below we give some of the mathematical use of incomplete inductive method of proof:

Examples of incomplete inductive methods:

For the proof of the summation formula:

A widely used area of incomplete generalization is the proof of the summation formula. It is important to note, however, that mathematical induction can only be used to prove, not to find, a theorem, so it is impossible to obtain a formula by means of a mathematical inductive method. Can only prove the correctness of the existing formula.

Example 1: proving the summation formula 1+2+3+...+n=n (n+1)/2 is correct

Prove:

First the Propositional function P (n) is "1+2+3+...+n=n (n+1)/2", then the first is the basic step: The basic step: proof P (1) is true because P (1) is 1 =1*(1+1)/2=1, the equation is equal on both sides, and the proposition is proven. Induction steps: When P (k) is established, p (k+1) is True P (k) for:1+2+3+...+k=k (k +1)/2P (k+1For1+2+3+.. +k+1= (k +1) (k +2)/2Here's a way to turn P (k) into P (k)+1), very simple, with P (k) on the left and right side plus K +1, then it is: P (k)+k+1=k (k +1)/2+k+1= (k +1) (k +2)/2=p (k +1) since P (k) becomes P (k+1) then the right-hand equals p (k +1) on the right-hand side. And since P (k) is true, p1To be true, a proposition is to be proven.prove

There are a lot of questions to prove the summation formula, here first one, the other topic and this is similar, but there may be more skills.

For the proof of inequalities:

Example 2: proving that N<2^n is true for all positive integers

P (N) is"N<2^n is set for all positive integers"Basic steps: P (1) for 1<2^1=2 Establishment Induction step: P (k) is K<2^k;p (k +1) for K +1<2^ (k +1) Convert P (k) to P (k)+1), plus 1 on both sides, P (k)+1 for k +1<2^k+1and since 2^k+1<2(k +1) so K+1<2^k+1<2^ (k +1) that is K+1<2^ (k +1) was established, P (k +1) set up, the proposition to be proved.prove

Proof of the Divide problem:

This division problem high School also baffled me for some time ah, here we look at how to solve the problem.

Example 3: when n is an integer, prove that n^3-n can be divisible by 3

proveBasic steps: P (1) for 1^3-1=0 divisible by 3 inductive steps: P (k) for K^3-k,p (k +1) for (k + 1 ^3-(+1) expand (k+1) ^3-(+ k1) k^3 +3k^2+3k+1-k-1= (k^3-K) +3* (k^2+k) This equation can be divisible by 3, the proof.

The peculiar application of inductive method:

Mathematical induction can sometimes be used in some unexpected situations to look at the following questions:

Example 4: use the right triple domino to cover a 2^nx2^n chessboard and leave only one space on the board

This is the right triple domino, to rotate the words can also get another three kinds:

So first to prove the case of P (1)

Basic steps: When the k=1, the chessboard is 2x2, then only need any of the above the domino can be covered, the proof of the proposition.

Induction step: Meet the conditions under the 2^kx2^k checkerboard, then 2^ (k+1) x2^ (k+1) can also meet the conditions

We divide the board into four parts:

Known by P (k), the lower right corner of the sub-chessboard must be able to use the Domino's only one space left:

Next we put a domino:

In this way, the extra spaces in the remaining three boards are filled up so that P (k) can be fully filled. So the final whole board will only have a space next to the next, the proposition to be proof.

Strong inductive method (fully inductive method)

Sometimes the use of incomplete induction is not easy