Writing intention

The textbook introduces a simple "drawer problem", in which four pencils are put into three pencil boxes. Students can find a phenomenon in the process of operating objects: no matter how they are placed, there are always at least two pencils in a pencil box, which leads to questions and arouses the desire to seek answers. Here, "four pencils" means "four objects to be separated" and "three pencil cases" means "three drawers". This problem is described in the language of "drawer problem": if four objects are put into three drawers, there will always be at least two objects in one drawer.

In order to explain this phenomenon, the textbook presents two ways of thinking. The first method is to enumerate through operations. By placing pencils intuitively, it is found that four pencils are divided into three pencil boxes, and there are only four situations (only the existence problem is considered here, that is, no matter which pencil box four pencils are placed in, they are regarded as the same situation). In each case, there must be at least two pencils in the pencil box. List all the experimental results, and the above questions can be explained. In fact, from the decomposition of numbers, this method is equivalent to decomposing 4 into three numbers. * * There are four situations, namely (4,0,0), (3, 1, 0), (2,2,0), (2, 1, 65438+. The second method adopts the idea of "disproof" or "hypothesis", that is, it is assumed that 1 pencil is put in each pencil box first, and three pencils are put in three pencil boxes. There are 1 pencils left. If you put them in any pencil box, there will be two pencils in this pencil box. This method is more abstract and general than the first method. For example, if you want to answer the question "Why do you put (n+ 1) pencils in n pencil boxes? There are always at least two pencils in a pencil box", it is difficult to explain by enumerating, and it is easy to explain by "assuming".

In order to have a deeper understanding of this kind of "drawer problem", the textbook arranges a "pigeon nest problem" in "doing the problem", and students can use the method in the example to make analogy transfer and explain.

Teaching suggestion

Because the data in the example is very small, it provides a lot of space for students to explore independently. Therefore, in teaching, students can think for themselves, first "prove" in their own way, and then communicate. In addition to the two methods provided in textbooks, there will be other methods (such as decomposition), which should be encouraged as long as they are reasonable. In this process, teachers should also give appropriate guidance. For example, let students make it clear that we only need to solve the existing problems here. If some students mark the three pencil boxes with serial numbers when enumerating, and understand (4,0,0), (0,4,0) and (0,0,4) as three different situations, the teacher should point out that it is unnecessary to make such a distinction when studying such problems. Such guidance is helpful to cultivate students' mathematical thinking of concrete analysis of specific situations.

Students should consciously understand the "generalized model" of "drawer problem" in teaching. In teaching, on the basis of students' independent exploration, we can guide them to compare the two methods provided in textbooks, think about the advantages and limitations of enumeration method and the advantages of hypothesis method, so that students can gradually learn to think in general mathematical methods. After solving the problem of "three pencil boxes and four pencils", students can continue to think: if you put five pencils in four pencil boxes, there will always be at least two pencils in one pencil box. Why? If you put six pencils in five pencil boxes, will the result be the same? How about putting seven pencils into six pencil boxes? How about putting 10 pencils into 9 pencil boxes? How about putting 100 pencils in 99 pencil boxes? Lead the students to a general conclusion: As long as the number of pencils is more than the number of pencil boxes 1, there will always be at least two pencils in a pencil box. Then, you can continue to ask questions: What if the number of pencils to be put is 2, 3 or 4 more than the number of pencil cases? Guide the students to find that as long as there are more pencils than pencil boxes, this conclusion is valid. Through this teaching process, it is helpful to develop students' analogy ability and form more abstract mathematical thinking.

2. Example 2.

Writing intention

This example introduces another kind of "drawer problem", that is, "if more than kn objects are randomly put into n empty drawers (k is a positive integer), then there must be at least (k+ 1) objects in one drawer." In fact, if k= 1 is set, this "drawer problem" becomes the form of example 1. Therefore, these two "drawer problems" are essentially the same, and example 1 is only a special case of example 2.

The textbook provides students with the scene of putting five books in two drawers. In the process of operation, students find that no matter how they put them, there are always at least three books in a drawer, which leads to the desire to explore the reasons. Students can still use enumeration method to decompose 5 into two numbers, including (5,0), (4, 1) and (3,2). In any result, there is always a number not less than 3. More generally, the hypothetical method is to divide five books into two parts on average. Using the division with the remainder of 5 ÷ 2 = 2 ... 1, we can find that if there are two books in each drawer, there is still 1 book left. Put the remaining 1 books in any drawer, and there will be three books in that drawer.

After studying the problem of "putting five books in two drawers", the textbook further puts forward "What if there are seven books and nine books in a * * *?" Ask the students to make an analogy with the previous method, and draw the conclusion that "seven books are put in two drawers, there is always one drawer for at least four books, two drawers for nine books, and there is always one drawer for at least five books".

On this basis, let students observe the characteristics of these "drawer problems" and look for the rules, so that students can have a general understanding of this kind of "pigeon coop principle". For example, students can conclude that "a (a is an odd number) should be put in two drawers. If A ÷ 2 = B... 1, then there must be at least (B+ 1) books in one drawer".

On page 7 1, the textbook "Do one thing" continues the situation of "Do one thing" on page 70, and expands it on the basis of Example 2, changing the number of drawers to three, requiring students to transfer and make analogy on the basis of the thinking method of Example 2.

Teaching suggestion

When teaching Example 2, students should still be encouraged to solve problems in various ways and sum up the "pigeon cage principle" by themselves. For example, when solving the problem of "two drawers and five books", because of the small amount of data, the method of students' hands-on operation or decomposition of numbers still has its intuitive and simple characteristics, which is also the easiest method for students to think of. However, due to the limitation of data, with the increase of books, teachers should give appropriate guidance. For example, you can ask students, "How about two drawers 125 books?" Because of the large amount of data, it is quite complicated to solve by enumeration method, which can urge students to consciously adopt a more general method, that is, hypothesis method. The core idea of the hypothesis method is to "divide" books into each drawer as much as possible and see how many books can be divided into each drawer. No matter which drawer the remaining books are put in, there is always one drawer with more than average 1 book. This core idea is expressed in the mathematical form of "division with remainder", which requires students to gradually understand and master by intuition.

When students use division with remainder to solve three specific problems in this example, teachers should guide students to sum up the general rules of this kind of "drawer problem". Put a certain number (odd number) of books in two drawers. As long as this number is divided by 2, there will always be at least 1 more books in one drawer than the quotient. For example, put 125 books in two drawers, 125 ÷ 2 = 62... 1, so there is always a drawer that can hold at least 63 books. If further summarized, it is: object A should be put into N drawers. If A ÷ n = B...C (C ≠ 0), then at least one drawer can hold (B+ 1) objects. This conclusion completely accords with the meaning of "if more than kn objects are randomly put into n empty drawers (k is a positive integer), then there must be at least (k+ 1) objects in one drawer".

When students finish "doing one thing", they can imitate Example 2, and use 8 ÷ 3 = 2...2 to know that there are always at least three pigeons in a pigeon house.

It should be noted that "how many books are there in a drawer at least" in Example 2 is the quotient plus "1" in the division formula, and the remainder in Example 2 is just 1, which is easy for students to misunderstand as quotient plus "remainder" and move to "doing one thing" and think that there is at least "2 (quotient) +2. In fact, as long as students understand the reasoning process of pigeon hole principle in essence, they can overcome this misunderstanding.

3. Example 3.

Writing intention

This example is a concrete application of the pigeon hole principle and a typical example of reverse thinking using the pigeon hole principle. How many balls do you need to find out two balls with the same color from four red balls and four blue balls? To solve this problem, we can think of the drawer problem in the first two examples. Because a red and blue ball can be regarded as two drawers, and the same color means the same drawer. In this way, the "ball-touching problem" can be transformed into the "drawer problem". Suppose you need to touch at least one ball, a ÷ 2 = 1...b, when b = 1, A is the smallest, and a=3. That is to say, at least three balls must be pulled out to ensure that the colors of the two balls are the same.

Through the dialogue of three students, the textbook points out that students can solve problems by guessing first and then verifying, and also reflects some difficulties that students may encounter when solving problems. For example, the "four red balls and four blue balls" in this example can easily interfere with students.

Next, the textbook guides students to further popularize this conclusion, pointing out that "as long as there are more balls than their colors 1, two balls can be guaranteed to be the same color." For example, balls have three colors, and at least four balls must be touched to ensure that two of them are the same color. This situation is described in the second question of "doing" on page 72 of the textbook.

The 1 question of "Do it" is also a typical example of "Pigeon Hole Principle". Among them, "2 out of 370 students must have the same birthday" is the same as the "Pigeon Cave Principle" in Example 1, and "5 out of 49 students must have the same birth month" is the same as Example 2.

Teaching suggestion

When teaching Example 3, first of all, students should be guided to think about whether the problems in this example are related to the pigeon hole principle mentioned above, what is the connection, what should be regarded as a drawer and what should be opened separately. However, when students think about these problems, they may lack the direction of thinking from the beginning, and it is difficult to find a breakthrough point. At this point, students can guess freely first and then verify. For example, some students will guess that "only touching two balls can ensure that the two balls are the same color", and this guess can be overturned by giving a counterexample. For example, if two balls happen to be one red and one blue, the conditions cannot be met. For another example, due to the interference of the condition of "four red balls and four blue balls" in the title, many students guess that the number of balls to be touched is only more than the number of one color 1, that is, "at least five balls must be touched to ensure that two of them are the same color". In order to test this conjecture, students will consciously associate the "touch ball problem" with the "drawer problem" and regard the two colors as two drawers. According to 5 ÷ 2 = 2... 1, we can know that at least three of the five balls are of the same color. So there is no need to touch five balls.

On the basis of students' guessing and verification, gradually guide students to turn specific problems into "drawer problems", find out what "drawers" are here and how many drawers there are, and then apply the "pigeon hole principle" learned earlier to carry out reverse reasoning. For example, in this case, according to the conclusion in example 1 that "as long as the number of divided objects is greater than the number of drawers, there must be at least two balls in a drawer", it can be inferred that "there must be at least two balls in a drawer, and the number of divided objects is at least 1". Now that the number of drawers is the number of colors, the conclusion becomes: "To ensure that two balls of the same color are found, the number of balls found must be at least more than the number of colors 1." Therefore, it is necessary to ensure that two balls of the same color are drawn from the balls of two colors, and at least three balls must be drawn. Applying this conclusion, we can directly solve the problem of "doing one thing" in the second question.

In the process of teaching, it is not easy to bridge the gap between practical problems and "drawer problems". If students have great difficulty in understanding, they can also be guided to think that the ball has two colors. If you only take two balls, there will be three situations: two red balls, one red ball, one blue ball and two blue balls. If you take another ball, whether it is a red ball or a blue ball, you can guarantee that two of the three balls must be of the same color.

When you finish the question 1 on page 72, guide the students to turn the "birthday question" into a "drawer question". Because there are at most 366 days in a year, if these 366 days are regarded as 366 drawers, and 370 students are put into 366 drawers, the number of people is greater than the number of drawers, then there are always at least two people in a drawer, that is, their birthdays are the same day. There are 12 months in a year. If this 12 month is regarded as 12 drawers, and 49 students are put into 12 drawers, 49 ÷ 12 = 4... 1 Therefore, there are always at least five in one drawer.

4. Explanations of some exercises in Exercise 12 and teaching suggestions.

In the 1 topic, students can use playing cards to see if the experimental results are consistent with the topic description, and then think about the reasons. We can use the pigeon hole principle to explain this phenomenon: there are 54 playing cards in a deck, and two trump cards are removed, leaving only four colors: diamonds, hearts, clubs and spades. We regard four colors as four drawers, and five playing cards are put in the four drawers. There must be at least two playing cards in a drawer, that is, at least two of the same suit.

Question 2: It is equivalent to dividing the 4 1 ring into five drawers (representing five darts). According to 4 1 ÷ 5 = 8... 1, a drawer must have at least 9 rings (that is, 8+ 1).

The first of the third questions is the same as that in Example 3. As long as a * * * has three colors, at least four sticks can be taken out to ensure that there must be two sticks of the same color.

Question 4: Take two colors as two drawers, take six faces of a cube as objects, and assign six faces to two drawers. 6÷2=3. At least three sides should be painted with the same color.

I hope it works for you!