On the English templates of * * * and China, the name of an ordinary worker is engraved. He made contributions to the cleanliness and beauty of the capital in the spirit of "one person is dirty and ten thousand people are clean". This person is (pause) Shi Chuanxiang (students collectively answer).

Today, let's really get to know Shi Chuanxiang, an extraordinary figure.

Ask the students to show pictures and materials about Shi Chuanxiang, and then talk about their views on Shi Chuanxiang.

Shi Chuanxiang, 19 15, was born in a poor peasant family in Qihe County, Shandong Province. Due to the famine in his hometown, he fled to the suburbs of Beijing at the age of 14, and was forced by life to become a dung digger. At that time, the toilet cleaning in the city was mainly done by manpower, so the industry of "dung digger" appeared. Shi Chuanxiang's job is to dig with a dung spoon, carry it with a dung can, carry it with a dung bucket and transport it with a dung cart every day to clean up the dung in the city. The road to old Beijing is very difficult. Shi Chuanxiang pushes a broken wheeled cart to deliver dung every day, from Liukou to Guang 'anmen, and then to Yaogezhuang and Xiaojing. He travels twenty or thirty miles back and forth, often "step by step, step by step." Whether it is windy or rainy, cold or hot, he runs back and forth four times a day. The salary is pitiful, earning less than 3 silver dollars a month. The place where they live is even more humble. 13' s partner sleeps with a donkey, although this kind of residence is often impossible to live in. They often eat and sleep on the road, with half a brick on their heads and a pair of worn cotton trousers for eight years. In old China, city dwellers couldn't live without a dung digger, but they looked down on this profession very much. Especially the rich, often dismiss these dung diggers as "dung beetles". The dung digger is not only despised by the society, but also squeezed and exploited by some evil forces in the industry. Shi Chuanxiang worked under these dung tyrants for 20 years and suffered oppression and bullying. Once, he dug dung for a lawyer's house in Beijing, and wanted to get some water after digging. Who knows that the rich woman of that family actually hid the water ladle, covered the water tank tightly, and asked the maid to bring him some water from the basin for feeding the cat. During the Japanese puppet regime, Dunba forced him to dig dung in the Japanese barracks. When he entered the door, he couldn't take off his hat to salute the Japanese soldiers standing guard because he pushed the wheeled vehicle with his hand. He was beaten black and blue by Japanese soldiers with gun butts and leather boots. After Japan surrendered, American soldiers lived in this city again. They drove the jeep on the rampage in the street. Once, they deliberately knocked over Shi Chuanxiang's dung truck and hurt his leg.

After the founding of New China, the * * * production party and the people's government eliminated the evil forces such as dung tyrants, and Shi Chuanxiang really felt liberated. 65438-0952, joined the cleaning team in Chongwen District, Beijing, and continued to engage in urban cleaning work. At this time, in order to show respect for the work of cleaning workers, the Beijing Municipal People's Government not only stipulated that their wages were higher than those of other industries, but also tried to reduce the labor intensity of dung diggers and replaced all wheelbarrows that used to deliver dung with cars. The cleaning team in Chongwen District where Shi Chuanxiang is located has 1 1 car, and the cleaning workers only need to dig out the manure and load it on the car, and then transport it to the suburbs by car. After the improvement of transportation, Shi Chuanxiang calculated the working hours reasonably and tapped the potential, and changed a large class of seven people into a small class of five people. He led the class from 50 barrels to 80 barrels per class in the past, and he carried 90 barrels per class, with a maximum of 5 tons of manure per class. The residents in the community enjoy a clean and beautiful environment, but his right shoulder with dung on his back has been worn with thick calluses, which has won people's universal respect and many honors. 1954 was rated as an advanced producer, and 1956 was elected as a representative of Chongwen district people. In June of the same year, he joined the China Production Party. From 65438 to 0959, Shi Chuanxiang, as a national advanced producer, participated in the National Heroes' Meeting held in Beijing, and was elected as a member of the presidium of the Heroes' Meeting and a member of the Beijing Municipal Political Consultative Conference in the same year. 1964 was elected as a deputy to the Third National People's Congress. President Liu Shaoqi once held his hand and said, "As a cleaner, you are a public servant of the people, and as chairman, I am also a public servant of the people."

During the "Cultural Revolution", Shi Chuanxiang was hit by his close relationship with Liu Shaoqi. He was beaten as a "scab" and was sent back to his hometown in Shandong on 197 1. In August, 1973, Premier Zhou Enlai was very angry after learning the news, and instructed to send someone to take him back for treatment immediately. He was taken back to Beijing and died of illness on May 1975 at the age of 60. Before his death, he repeatedly told his son to continue his father's ambition and become a famous sanitation worker.

3. Teachers ask questions and students discuss in groups.

Question 1: Why is the author looking for Shi Chuanxiang? Why is it called "spiritual plateau"?

The teacher concluded: Looking for Shi Chuanxiang is actually looking for the spirit of the times. That is, Shi Chuanxiang's spirit of "integrity and dedication" mentioned at the end of the article. Because a period of history is fading away, Shi Chuanxiang's spirit has also been forgotten. Modern cities no longer need dung diggers, as long as there is social division of labor. There will always be a hard, tired and dirty job. Therefore, Shi Chuanxiang's spirit of "one person is dirty for all" is still indispensable today. Moreover, Shi Chuanxiang's spirit is not only not afraid of hardship, fatigue and dirt, but also lies in his seriousness and dedication. Whether as a worker, an official, a businessman or a scholar, Shi Chuanxiang's spirit will never be out of date.

Question 2: There is such a passage in the article. A kindergarten teacher points to a sanitation worker's child to educate other children and says, "If you don't obey, you will sweep the streets and dig toilets like his parents in the future!" Nowadays, it is very common to have such an idea. How can we revive Shi Chuanxiang's spirit?

For this question, let the students write an article and talk about their views in the form of written homework.

4. Linguistic features of the article.

Not paying much attention to rhetoric, because of its authenticity, it still has moving power.

blackboard-writing design

Ishikawa Xiang Ishikawa Xiang Spirit

Manure workers and model workers are honest, dedicated and serve the public.

NPC deputies and scabs do clean work.

Honesty and dedication are really bad and inconvenient for everyone.

Positive description

Call for the return of human nature

Goldbach's Conjecture

order

According to relevant historical records, Goldbach conjecture has been put forward for more than 200 years. According to relevant mathematical authorities, many mathematical experts in the world are studying it day and night. However, no one has solved it so far. After I retired from 1980, I studied it for about twenty years. During this period, I explored a brand-new idea-to determine the latest arguments and create unique arguments, thus solving the ancient mathematical mystery of Goldbach conjecture. The full text is divided into three chapters.

The first chapter identifies two "latest arguments"

These two "latest arguments" are: ① s+h+f = n/2+[1-1) n]/4 (that is, when n is even,

S+H+F=n/2, and when n is odd, s+h+f = n/2+1/2); ② S ≠ 0 in the "latest argument" ① [N is an arbitrary positive integer, S, H and F are numbers (r+r), (h+h) and (r+h) equal to (2n+4) respectively, R is a prime number and H is an odd number].

In order to concretize the above-mentioned "latest argument" and make it initially recognized, it was randomly selected.

As follows:

When n = 13, only (7+23) and (119) of (r+r) are equal to 30.

And (13+ 17), that is, its number S = 3；; (h+h) equals 30 only (9+2 1) and (15+ 15), that is, its number h = 2；; (r+h) equals 30, and only (3+27) and (5+25) exist, that is, its number f = 2.

Substitute the above n= 13, s = 3, H=F=2 into the "latest argument": ① Left = 3+2+2 = 7, right.

= 13/2+ 1/2 = 7, that is, the left and right sides of formula ① are equal; (2) Left = 3, right = 0, that is, (2) left and right are not equal.

When n= 16, (2n+4) = 36-(r+r) equals 36, only (5+3 1), (7+29),

(13+23) and (17+ 19), that is, the number s = 4；; (h+h) equals 36 only (9+27) and (15+2 1), that is, its number h = 2；; (r+h) equals 36, and only (3+33) and (1 1+25) exist, that is, its number f = 2.

Substitute the above n = 16, s = 4, h = f = 2 into the "latest argument": ① left = 4+2+2 = 8, right.

Side = 16/2 = 8, that is, the left and right sides of formula ① are equal; (2) Left = 4, right = 0, that is, (2) left and right are not equal.

The second chapter creates a "unique argument" to demonstrate two "latest arguments"

The first section is based on "s+h+f = n/2+[1-(-1) n]/4"

First, prove a related "latest proposition"-in the first n items of the sequence 3, 5, 7, 9,1,... (2n+ 1), ..., all two items equal to the average of the first two items and the last two items or the absolute value of the difference between the two items are added separately (the items themselves are added) ② nothing more than (r+R+H)；, (h+h) and (r+h); ③ There is {n/2+[1-1) n]/4} (n is an arbitrary positive integer, r is a prime number and h is an odd number). When n→∞, this proposition still holds.

Known and verified.

It is proved that ① ∶ the known sequence is an infinite arithmetic progression with the first term of 3 and the tolerance of 2.

∴ The sum of the first n terms in this series is equal to [3+(2n+ 1)] = (2n+4), {(3+2)+[(2n+ 1)-2]} = (. That is, "all are equal to (2n+4)" (w is a non-negative integer and less than n/2);

②: In a known sequence, all terms can be divided into prime numbers and odd numbers. Therefore, when the first n items of this series are added according to the above-mentioned "latest proposition", they can only be added separately and mutually.

∴ The sum of the two numbers "is nothing more than (r+r), (h+h), (r+h)";

(3) The first n terms of the known series are added according to the above-mentioned "latest proposition", and the number of the sum of the two numbers obtained is obviously equal to half of the number of terms and is a positive integer. So the sum of two numbers * * * has n/2 (when n is even) or (n/2+1/2) (when n is odd). That is, "* * has {n/2+[1-(-1) n/4]}".

It is known that n is an arbitrary positive integer, so when n→∞, the above "latest proposition" still holds. And call it "unique argument ①".

Based on the unique argument ①, the first "latest argument" is deduced ―― from the above "latest propositions ② and ①, it can be concluded that (r+r), (h+h) and (r+h) are all equal to (2n+4) (n is an arbitrary positive integer); From the above "latest propositions" ② and ③, it can be concluded that (r+r), (h+h) and (r+h) * * have {n/2+[1-(-1) n/4]} (n is any positive integer).

It is known that when n is an arbitrary positive integer, s, h and f are numbers (r+r), (h+h) and (r+h) equal to (2n+4) respectively (notes in two "latest arguments" in Chapter 1).

Therefore, s+h+f = n/2+[1-1) n]/4 (omitted), that is, the first "latest argument" holds.

The second section is about the basis of "S≠0" in "s+h+f = n/2+[1-1) n]/4".

The first step is to reveal the distribution law of R-H.

Definition of 1.R type H: an odd number H that can only be divisible by a prime number R or a prime number not less than R is called R type H. For example, 9 is H and can only be divisible by a prime number 3. Therefore, 9 is 3 H type; Another example: 35 is h, which can be divisible by prime numbers 5 and 7 respectively, and 7 and 5 are not less than 5. So 35 is 5 H type, not 7 H type.

Second, the distribution law of R-type H: As can be seen from the above definition, in the infinite sequence {(2n+ 1)}, any H that can be divisible by prime number 3 is of type 3 H. From the fourth term, it has 1 in every three terms. That is, from (3× 3- 1)/2, every three items have (2- 1). Such as: 9, 15, 2 1, ... (9+6m) (m is a non-negative integer, the same below); Where h is divisible by prime number 5, and every 5 terms have 1 from item 7. Such as: 15, 25, 35, ... (15+ 10m). So there are three items per 15. However, the 3 H type accounts for 1/3. Therefore, from item 7, there are only two kinds of H in each item 15. That is, from (3×5- 1)/2, there is only (2- 1) × (3- 1) every 3×5. Such as: 15, 25, 35; 45,55,65; 75,85,95; ..., (15+30m), (25+30m), (35+30m) (indicating that the above figures or formulas should be removed, the same below); Where h is divisible by prime number 7, and from 10, there is 1 in every 7 items. Such as: 2 1, 35, 49, ... (2 1+ 14m). So every 105 has 15. However, 3 H type accounts for 1/3, and 5 H type accounts for 2/ 15. Therefore, starting from the item 10, there are only 8 kinds of H in each item 105. That is, from (3× 7- 1)/2, every 3×5×7, there is only (2-1) × (3-1) × (5-1). Such as: 2 1, 35, 49, 63, 77, 9 1, 105,19, 133, 147. 23 1,245,259,273,287,30 1,3 15,329,343,357,37 1,385,399,4 13,427; ……(2 1+2 10m)，(35+2 10m)，(49+2 10m)，(63+2 10m)，(77+2 10m)，(9 1+2 10m)，( 105+2 10m)，( 1 10m)

Similarly, every H divisible by the prime number 1 1 has 1 from 16, so every 1 1. However, 3 H type accounts for 1/3, 5 H type accounts for 2/ 15, and 7 H type accounts for 8/ 105. So starting from 1 1, there are only 48 h's in each item 1 155. That is, from (3× 1 1- 1)/2, every 3×5×7× 1 1, there is only (2-1)× (3-/kloc-) Where H is divisible by the prime number 13, there is 1 in every 13 from the item 19, so there is155 in every155. However, 3 H type accounts for 1/3, 5 H type accounts for 2/ 15, 7 H type accounts for 8/ 105,1h type accounts for 48/ 1 155. Therefore, starting from the item 13, each item 150 15 has only 480 H. That is, from (3× 13- 1)/2, every 3× 5× 7×1×13, only (2-1)× (3- ……

Generally, in the infinite sequence {(2n+ 1)}, the R-type H starts from (3r- 1)/2, and only (2-1) × (3-1)× (5-1)×.

The second step is to discuss the general trend of (H+F)/n when n→∞

A, according to the "distribution law of R-type H", in the first n terms of infinite series {(2n+ 1)}, types 3 H, 5 H, 7 H,1h, 13 H, ... 48/ 1 155, 480/ 150 15, ... 1×. In addition, these share ratios appear intermittently. Therefore, the occupation rate of the total number of H types in the first n terms of this series (set as G) is G/N ≈1/3+2/15+8/105+48/155+480/. Obviously, when n→∞, G/n increases infinitely as a whole. And call it an important argument.

It can be inferred from the "important argument" that in the sequence {(2n+ 1)}, when n→∞, G/2/n generally increases infinitely. And call it an "important inference."

B, it is known that in the first n items of the infinite sequence {(2n+ 1)}, according to the condition of the "latest proposition" (the first section of Chapter 2), the numbers in the sum of two numbers (h+h) and (r+h) equal to (2n+4) are H and F (two in Chapter 1) respectively. Moreover, there is 2hor (2h-1) in h (h+h) [when only one (h+h) represents the sum of two identical odd numbers. For example: (15+ 15) heteroh, and there are f heteroh in F (r+h).

In addition, for the first n terms of the infinite sequence {(2n+ 1)} in the above A, the total number of each type H has been set to g, so 2h+f = g or (2h-1)+f = g.

Add f and (f+ 1) to the left of the above two formulas, and it can be concluded that in the first n terms of infinite sequence {(2n+ 1)}, h+f ≥ g/2 (n is any positive integer, and h, f and g are all non-negative integers).

It is known that n is an arbitrary positive integer, so when n→∞, the above formula still holds.

Divide both sides of the above formula by n to get: (H+F)/n≥G/2/n(n is any positive integer).

From the "important inference" (step 2 A in the second section of Chapter 2), it can be concluded that in the sequence {(2n+ 1)}, when n→∞, (H+F)/n increases infinitely as a whole. And call it "unique argument ②".

The third step is to prove the second "latest argument" (the original text is omitted)

Known: s+h+f = n/2+[1-(-1) n]/4 (omitted).

Verification: S≠0 under known conditions.

Proof: Assuming that S = 0, then H+F = n/2+[1-(-1) n]/4.

That is, when n is an even number, h+f = n/2; When n is odd, h+f = n/2+ 1/2 or (n+ 1)/2.

Divide the above two formulas by n and (n+ 1) respectively, and you can get:

(H+F)/n = 1/2； ①

(H+F)/(n+ 1)= 1/2 .②

① In the formula, when n→∞, the limit of [(h+f)/n] is 1/2.

It is known that "(H+F)/n increases infinitely as a whole when n→∞" (Chapter 2, Section 2, Section 2

"Unique parameter ②" in step b).

Obviously, (H+F)/n is not a constant.

Therefore, (H+F)/n≠ 1/2.

Therefore, (h+f)/n < 1/2. ③

② In the formula: (H+F)/(n+ 1) cannot be greater than (H+F)/n,

∴(H+F)/(n+ 1)< 1/2。 ④

Obviously, ① ② types are in contradiction with ③ ④ types respectively. That is, ① and ② cannot be established.

Therefore, it is impossible to "assume that S=0 under known conditions".

Therefore, S≠0 is given (omitted). That is, the second "latest argument" holds.

The third chapter proves Goldbach conjecture.

-Every even number not less than 6 is the sum of two prime numbers.

It is known that when n is an arbitrary positive integer, (2n+4) represents an even number not less than 6, and (r+r) represents the sum of two prime numbers.

It is proved that (2n+4) is (r+r) when n is an arbitrary positive integer.

It is proved that when n is an arbitrary positive integer, the number of s (r+r) is equal to (2n+4) (notes in two "latest arguments" in the first chapter).

It can be concluded that s here can only be a non-negative integer [∵ If (r+r) equals (2n+4) does not exist, then s is 0, otherwise s is a positive integer].

S≠0 is known here (Chapter 2, Section 2, Step 3).

Therefore, s here is a positive integer.

That is, when n is an arbitrary positive integer, the number of (r+r) equal to (2n+4) is at least 1.

Therefore, when n is any positive integer, (2n+4) is (r+r).

That is, every even number not less than 6 is the sum of two prime numbers.

That is, the proof of Goldbach's conjecture.