Junten Science Library 4

Revised English version based on the lecture on 2nd Dec. 2010 for

Junten Junior High School and Chua Chu Kan Higih School in Singapore 星の研究 A Study of Stars

Written and translated by Dr. Nakahara

－1－ Junten Science Library 4 A Study of Stars

§１ An n-pointed Star --- Our Definition of the Star Shape

Draw star shapes. What kind of star-shape did you draw?

These are the examples of the star shapes I found on the web. Left one is called a pentagram. It is the most common shape almost everyone reminds of when he hears the word “star-shape”. The right one is a hexagram. It is also called the David’s star and is considered as sacred in Judaism. You can find the one on the national flag of Israel (the picture on the right.) I did not expect to find the centre one. It is sometimes adopted by the Manga authors to represent a “girl’s bright eye” or “the star twinkling in the winter sky”. The stars in the real sky are either the planets or fixed stars. The planets wax and wane, changing their shapes. They are by no means of the star shapes. The fixed stars are similar to the sun and are round. It is not the star-shape either. We can not recognize the fixed stars as a disk with definite radius because they are too far from us. They are seen as the bright points. As our photoreceptor cells are of the finite size, the points smear to some shapes which may be star shapes.

To investigate the star shapes mathematically, we will have to define them first. Pentagram is a star shape with 5 vertices. Hexagram is the one with 6. When we go further to the decagram, the dodecagram and the icosapentagram, the Latin words for the large numbers are practically undecipherable. Therefore we will use the plainer words with Arabic figures. We call pentagram a 5-pointed star and hexagram a 6-pointed star. These names are less interesting but easier to generalize.

Now let us define an n-pointed star. n-pointed star ①n-pointed star is a closed graph(*) with n sides which connects all n vertices of a regular n-gon, the regular polygon(**) with n vertices, and is an Euler’s circuit(***). －2－ ②All the angles are of the same magnitude and all the sides are of the same length. ③No sides are shared with the regular polygon.

(*) closed graph A graph with no open segments of line. Every vertex is connected by more than one sides (**)regular polygon A polygon with all the sides are of the same length and all the angles are of the same magnitude. (***)Euler’s circuit A path you can draw by the pencil without lifting it.

Exercise 1 Find the one in the above pictures which is a n-pointed star.

§2 Drawing n-pointed stars Let us draw various stars with various polygons.

Exercise 2 The following are the polygons with 3,4,5,6,7,8,9 and 10 vertices. Draw as many different stars using these polygons. Can you draw stars with any kind of polygons? Can you draw two or more different stars with the same polygon?

3-pointed star 4-pointed star 5 pointed star 6-pointed star

7-pointed star

－3－

8-pointed star

9-pointed star

10-pointed star

Let us summarize the exercise. There are no such things as（ ）-pointed stars, （ ）pointed-stars or（ ）-pointed stars. There is only one each of（ ）-pointed star, （ ）-pointed star and （ ）-pointed star. There are two different shapes in （ ）-pointed stars and（ ）-pointed stars.

You may have a lot of questions unanswered at the moment. How many polygons with which we cannot draw any stars? We have two polygons with two different kinds of stars. Are they exceptions?

Let us proceed to the next polygon and find our idea is not quite right. In fact, there are as many as four different stars we can draw with 11-gon.

－4－ Exercise 3 Draw the four different stars with 11-gon.

It is now clear that the polygon with many different stars is not the exception at all. For example you can draw 8 different stars with 19-gon. How do we see the number of the different stars with a 360-gon? Our final aim of these lectures is to find the number of the different stars with a given polygon. If the only way we can see is just drawing the stars, our way to the goal will be very long and winding. Is there any systematic method of calculating the number of different stars from the number of vertices?

We need to look at the problem from the different angle to make a breakthrough.

§3 Stars and Modular Arithmetic

２

１ This is a 7-pointed star. I put the numbers to the vertices of the 7-gon. The vertex on the

３ right is 0. Then the number is getting larger as we move anti-clockwise and the final vertex is 6. Start from vertex 0 and travel along the ０ star sides to the direction shown by an arrow. As the star is an Eulerian circuit, we travel ４ through all the vertices of 7-gon and finally come back to the vertex 0. Let us write down the number in the order of our travel. ６ ５ ０→３→６→２→５→１→４→０

Look at the sequence of the numbers and you will see that the sequence is the same as the one appeared in the multiplication table of modulo 7 arithmetic.

－5－

Shown here is the multiplication table of Mod ０ １ ２ ３ ４ ５ ６ the modulo 7 arithmetic. Each entry ７× represents a class of the numbers whose ０ ０ ０ ０ ０ ０ ０ ０ remainders are the same when they are divided by 7. The class 2 contains the １ ０ １ ２ ３ ４ ５ ６ numbers 2, 9, 16, 23 and so on as well as -5, -12, -19. The table is the multiplication ２ ０ ２ ４ ６ １ ３ ５ table. You can find the result of the multiplication among the classes. The ３ ０ ３ ６ ２ ５ １ ４ table shows that the class 4 multiplied by the class 3 makes the class 2, which ４ ０ ４ １ ５ ２ ６ ３ means, a number belonging to the class 4 multiplied by a number belonging to the ５ ０ ５ ３ １ ６ ４ ２ class 3 makes a number belonging to the class 2. The row in pink colour ６ ０ ６ ５ ４ ３ ２ １ represents the multiples of 3, that is:０× ３ １×３ ２×３ ３×３ ４×３ ５×３ ６×３, and is exactly the same as the sequence we have seen in the previous page.

The result suggests us that the other rows may correspond to one of the 7-pointed stars.

Exercise ４ Look at the other rows of the multiplication table and draw the graphs which correspond to the rows.

The result tells us that the row 1 corresponds to 7-gon and the row 2 corresponds to another 7-pointed star which is different from the one corresponded by the row 3.

Let us try with the row 4. What kind of shape do we get? The classes attached to the vertices are ０→４→１→５→２→６→３→０ in the order of the path. When we compare this sequence to the one with the row 3, which is ０→３→６→２→５→１→４→０ you will see that the order of the classes is in the opposite direction. Therefore the shapes drawn with these two sequences are exactly the same although the directions we draw them are opposite. －6－ Similarly the star drawn using the row 5 is as same as the one with row 2, and we have the 7-gon again with the row 6 just like with the row 1.

Exercise 5 Complete the multiplication table of modulo 10 arithmetic and find the shape drawn using each row. We know that there is only one 10-pointed star. Find the rows which we can draw the 10-pointed star with. Investigate why we can not draw the stars with the other rows.

Mod10 ０ １ ２ ３ ４ ５ ６ ７ ８ ９ ×

０

１

２

３

４

５

６

７

８

９

You saw that in the rows we can not draw the stars the same class appears more than once and certain classes never appear. On the other hand, in the row corresponding to the star every class from 0 up to 9 appears and it does only once. This property means the n-pointed stars are Eulerian circuit.

§４ Counting the stars

Let us summarize the conclusion of the previous section. (1)Every n-pointed star can be drawn with a row of the multiplication table of modulo n arithmetic connecting the vertices of the n-gon in the order of the number in the row. (2)If we can draw an n-pointed star with a certain row (let us call it the row r) then the same star can be drawn with the row n-r. In this case the pen travels in the opposite direction to the one

－7－ with the row r. (3)If a certain row contains all of the classes from 0 up to n-1, then the corresponding shape will be an Eulerian circuit connecting all of the vertices of n-gon. Otherwise, the path reaches the original vertex labelled 0 again before it connects all of the vertices. (4)With the row 1 and the row n-1, we can not draw the n-pointed stars despite that these two rows satisfy all the conditions above, because the shape we can draw with them is the n-gon itself.

Of these four conditions, the condition (3) is the most complicated one to state. It needs further investigation. We would like to know whether a certain row contains all the classes without writing down the whole row. In the multiplication table of modulo 10 arithmetic the rows 1, 3, 7 and 9 contain all the classes.

Exercise 6 Find the greatest common measure (g.c.m.) of 10 and r when r is a positive integer less than 10.

r １ ２ ３ ４ ５ ６ ７ ８ ９ g.c.m. of 10 and r 2

Exercise 6 implies that the row r contains all of the classes if and only if the g.c.m. of 10 and r is 1. If the g.c.m. of two integers m and n is 1 then we say that the integers m and n are co-prime.

Exercise 7 Show that if two numbers n and r are not co-prime then the row r does not include class 1.

Let us now prove that the row r contains all the classes if n and r are co-prime.

First of all let us show that if no class appears in the row r more than once, then all the classes appear in it. It is because there are n places in the row r and there are n classes in modulo n arithmetic. Now we will prove the main statement using the method called reductio ad absurdum. Assume that a certain class appears in the row r twice. Then ｒ×ｐ＝ｒ×ｑ where ｐ and ｑ are two different classes. The equality implies that the remainder of both rp and rq are the same when divided by n. Therefore rq  p  nk, where k is an integer. As r and n are co-prime, q  p must be a multiple of n. Hence ｐ＝ｑ. The conclusion ｐ＝ｑ contradicts with the assumption; ｐ and ｑ are two different classes. －8－ The contradiction shows that the assumption was wrong. Therefore the statement that no classes appear in the same row twice is proved. Q.E.D

Summary If the number r labelling the row ｒ in the multiplication table and the modulus n are co-prime, then all the classes appear in the row ｒ, and then the shape drawn using the row ｒ is the Eulerian circuit connecting all the vertices of the n-gon. The shape is the n-pointed star except the case with the row １ or n-1 which are the n-gon itself. The n-pointed star with the row ｒ and the one with the row n-r are the same star.

§５ Euler’s totient function

In the number theory, the totient n of a positive integer n is defined as the number of positive integers less than or equal to n that are co-prime to n. In particular 1 1 since 1 is co-prime to itself (1 being the only natural number with this property). For example, 9  6 since the six numbers 1, 2, 4, 5, 7 and 8 are co-prime to 9. The function  so defined is the Euler’s totient function, or Euler’s function.

The investigation we have done in the previous section shows that the number of the different n-pointed stars can be described using Euler’s function. If n=7, the positive integers which are co-prime to 7 are 1,2,3,4,5 and 6. As there are 6 such numbers the value of the Euler’s function is 6. That is 7  6 .

There are 6 rows in which all the classes appear, in the multiplication table of modulo 7 arithmetic. Of 6 rows, the rows １ and ６ correspond to the 7-gon. The remaining 7  2  4 rows correspond to the stars. Two each rows are designated to the same star. Hence the number of the different stars are the half of 7  2 . The number of the different n-pointed stars S7 is calculated by the expression 1 1 S7  7  2  6  2  2 2 2 For 10-pointed stars we can find that 10  4 as there are 4 positive integers less than 10 which are co-prime to 10, namely 1,3,7 and 9. Therefore the number of the different 10-pointed stars is 1 1 S10  10  2  4  2 1 2 2 In general, the number of the different n-pointed stars is 1 Sn  n  2…(＊) 2 －9－ If we can find the value of the Euler’s function when n  360, then we can find the number of the different the 360-pointed stars S360 1 S360  360  2 2

Now our problem has reduced to the finding the values of Euler’s function n when n is given. How can we find the values of Euler’s function then? If the only way to do so is to count the all the integers co-prime to n, our task remains as enormous as before. Is there any more efficient way?

§６ Properties of the Euler’s function

●As ７ is a prime number, the only measures of 7 are 1 and 7 itself. All the positive integers less than 7 are co-prime to 7. Therefore the value of the Euler’s function is one less than 7. 7  7 1 6 The same argument applies to any prime numbers. Therefore we get the following property. p  p 1, where p is a prime number…①

●When n  74 , the positive integers not co-prime to n must be multiples of 7. When we check the integers from 1 in ascending order, the multiple of 7 occurs once in 7 times. 7 1, 7  2, 7  3, , 74 The number of the positive multiples of 7 less than 7 4 is one seventh of . The other integers are co-prime to . 74  74  74  7  74  73  7 1 73 As the same argument applies to the other powers of a prime number, we get the following property. p n  p 1p n1 , where p is a prime number and n is a positive integer…②

Now we can find the values of the Euler’s function n when n is a prime number or a power of a prime number without counting all the integers co-prime to n. Most difficult part of the theory is to expand the property to the n which consists of more than one prime factor. ●For the example of the number with more than one prime factor, let us investigate n  36. As n  36 4 9  22  32 , n has two different prime factors: 2 and 3. We shall look here at the remainders of the numbers when they are divided by 4 and 9. Suppose that an integer r is co-prime to 36, say r=23. Then we get the remainders, 23 3 mod4, 23 5 mod9. Note that the remainders 3 and 5 are co-prime to the moduli 4 and 9 respectively. Since r is co-prime to 36 two remainders must be co-prime to the moduli. There is only one integer less than 36 which is congruent to 3 modulus 4 and is congruent to 5 －10－ modulus 9. The reason is as follows; Let us list all the integers less than 36 which are congruent to 3 modulus 4. They are: ３、 ３＋４＝７、 ３＋４×２＝１１、 ３＋４×３＝１５ ３＋４×４＝１９ ３＋４×５＝２３ ３＋４×６＝２７ ３＋４×７＝３１ ３＋４×８＝３５. As you can see, there are nine of them. The remainders of these integers when they are divided by 9 are ３，７，２，６，１，５，０，４，８ respectively. Note that the remainders are all different. This is the consequence of the fact that the row ４ in the multiplication table of modulo 9 arithmetic contains one (and only one) each of all the classes. Since the remainders are all different, 5 appears only once.

If an integer r which is less than 36 is co-prime to 36 then the remainders a and b when r is divided by 4 and 9 are co-prime to 4 and 6 respectively, and conversely if r has remainders a and b co-prime to 4 and 9 when r is divided by 4 and 9 respectively then r is co-prime to 36 and there is only one such an integer which is less than 36. Now we have one to one correspondence between the set of all the integers less than 36 which are co-prime to 36 and the set of all the pairs of integers (a,b) co-prime to 4 and 9 respectively.

Exercise 8 Complete the table showing the correspondence of the integers co-prime to 36 and the pairs of integers co-prime to 4 and 9 respectively. Each row is labelled by a, while each column is labelled by b. Each entry is the integer r whose remainders are a and b when divided by 4 and 9 respectively.

b １ ２ ４ ５ ７ ８ a

１ ３ ２３ In the table, all the integers co-prime to 36 appear once. Therefore the number of the integers co-prime to 36 is the number of the integers co-prime to 9 times that of the integers co-prime to 4. If we remind the definition of the Euler’s function we will get 36 49.

－11－ Note that the fact that two moduli 4 and 9 are co-prime is essential. If the two moduli are not co-prime as 3 and 12, the integers do not correspond to the pairs of remainders. There are twelve integers less than 36 which are congruent to 2 modulus 3. They are 2,5,8,11,14,17,20,23,26,29,32 and 35. The remainders when divided by 12 are 2,5,8,11,2,5,8,11,2,5,8,11 respectively. We can see that the same remainder appears three times and there are some remainders such as 1 and 7 which do not appear in the sequence. This is because only 4 classes appear in the row ３ in the multiplication table of the modulo 12 arithmetic. When divided by 3 and 12, the pair of the remainders (2,7) does not correspond to any of the integers co-prime to 36 although 2 and 7 are co-prime to 3 and 12 respectively. Hence the number of the pairs of the remainders is not necessarily equal to the number of the integers co-prime to 36. 36  312

Theorem If the integer n is the product of two integers a and b co-prime to each other then n ab ab…③

§７ Calculating the Euler’s function

We discovered three important properties of Euler’s function in the §6. If p is a prime number then p  p 1…① If p is a prime number and n is an integer then p n  p 1p n1 …② If a,b are the integers co-prime to each other then ab ab…③ We can now calculate any values of Euler’s function making use of these properties.

●40 can be found by the following procedure (1)Get the prime factorization of 40. 40 8 5  23  5 (2)Calculate value of each Euler’s function of the factor with powers of one prime number. As 8  23 , using the property ②, 8 23  2 1 231 1 22  4 The factor 5 being a prime number, property ① shows us that 5  5 1 4 (3)Finally, the product of all the values found in (2) will make the value of the Euler’s function. We can do so since two factors are co-prime to each other. 40 8 585 4 4 16 Let us check our result by writing down all the integers co-prime to 40. 1，3，7，9，11，13，17，19，21，13，27，29，31，33，37 and 39. There are in fact 16 of them.

－12－ ●Let us find the value 360: 360 23  32  523 32 5  2 1231  3 1321  5 1 4 6 4  96

Exercise 9 Find the following values of the Euler’s function (1)11

(2)64

(3)77

(4)100

(5)900

§8 The number of different n-pointed stars

We can finally count the number Sn of the different n-pointed stars, with all the result of the previous sections. The formula (＊) on page 9 shows that 1 Sn  n  2 2 ●The number of different 36-pointed star is: 1 1 S36  36  2  12 2  5 2 2 ●The number of different 360-pointed star is: 1 1 S360  360  2  96 2  47 2 2 We can now conclude that there are as many as 47 different 360-pointed stars.

Exercise 10 Find the number of the following stars. (1) 15-pointed stars

(2) 18-pointed stars

(3) 24-pointed stars

Exercise 11 List all the possible integer n so that there is n-gon but that there are no n-pointed stars.

－13－

Exercise 12 List all the possible integer n so that there is only one n-pointed star.

Exercise 13 List all the possible integer n so that there are just two different n-pointed stars.

§9 Conclusion Connecting every r-th vertex of the n-gon, we can draw an Euler’s circuit if n and r are co-prime, where n is an integer greater than 2 and r is a positive integer less than n. Of these cases, with r=1 and r=n-1 we will draw n-gon, and otherwise we will draw an n-pointed star. The n-pointed star we draw with r is the same one as the one with n-r. The number Sn of the different n-pointed stars can be found by the formula 1 Sn  n  2. 2 A 81-pointed star with the row ３１

In the part II, we will investigate the angles of the n-pointed stars.

－14－