51 NUMBER THEORY: DIVISIBILITY; PRIME & COMPOSITE NUMBERS 210 f8


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1 51 NUMBER THEORY: DIVISIBILITY; PRIME & COMPOSITE NUMBERS 210 f8 Note: Integers are the w hole numbers and their negatives (additive inverses). While our text discusses only whole numbers, all these ideas extend to the negative integers as well as positive, so the statements can be inclusive. For all statements below, variables a, b, c, d, and k are whole numbers (or integers).* Def' n: For k,b integers k*b (k divides b) if, & only if, b = kc for some integer* c. If k b w e say " k is a divisor or factor of b" " b is a multiple of k" equivalent statements b is divisible by k If k is not a factor of b, w e say k does not divide b, and w rite k b. for example 4 20 because 20 = 5 4 eg 4 17 EG 3 *12 (or 12 is a multiple of 3) 6 is a factor of 24 Is 42 divisible by 3? because 12 = = = 7 6 & 6 = 3 2 so 42 = & 4 is a w hole number Fact: If a W then a*0 because 0 = a 0 (or Z) and 1*a because a = 1 a and a*a because a = a 1 Fact: If d*a, then d*ab EG 150 must be divisible by 3, because 15 is... Pf : if d*a then a = dc Think: If 3 15 then 15 = 3 5 ab = (dc)b = d(cb), Then 15 w must be 3 5 w so ab is a multiple of d PQED So 15w must be a multiple of 3. Fact: If a*b and b*c then a*c (" divides" is transitive) Pf : a*b and b*c c = bm and b = an, for some m,n Z [ by def'n of " *" ] bm = (an)m & (an)m = a(nm) [ assoc prop x ] so c = a(nm) and c = a(some integer) [ closure prop of mult.]...thus proving that a*c PQED COMMENT: This is almost the same fact as above. E.G. Since 3 15, and , therefore Fact: If d*a and d*b, then d*(a+ b) and d*(a b). Pf: If d*a and d*b, then a = dc and b = de, so a+ b = dc + de = d(c+ e) PQED The second follow s immediately from the first, since a b = a + ( b). Put another way: if a and b are both multiples of d, then a+ b is also a multiple of d. For example: 7000 and 21 are both multiples of 7, therefore 7021 must be a multiple of 7. Fact: If d*a and d b, then d (a+ b). Pf: If d*a and d*(a+ b) then d must divide b since d*a d* a so d*((a+ b) + ( a)) and (a+ b) + ( a) = b. PQED Put another way: If a is a multiple of d, and b is not, then a+ b cannot be a multiple of d. For example: 90 is a multiple of 3, 2 is not. Therefore, 92 cannot be a multiple of 3. (We all know the multiples of 3 are spaced 3 units apart!!! ) Notice the tw o above facts together say: If d*a then d*(a+ b) iff d*b....that is: If d is already know n to divide a, then to see if d*(a+ b), just test w hether d*b.
2 TESTS for DIVISIBILITY One way to determine whether a divides b is to attempt to divide b by a. How ever, there are a number of " tests" for divisibility by certain values. We are expected to know these. DIV ISIBILITY TEST FOR 2 : Is the units digit even? (i.e. is the units digit divisible by 2)? This fact is often observed by children who learn to count by twos. Eg. It is obvious to all of us that 3574 is divisible by 2 because 2*4. This shortcut w orks because 3574 = is a multiple of 2, and 2*10 (so 2*357x10... ie 2*3570). In general, if u is the units digit, and v represents the rest of the value of a number, then the number is actually v+ u and v is a multiple of 10. Since 2*10 and 10*v, w e know automatically, w ithout inspecting v, that 2*v. If 2 also divides u, the units digit, then 2*(v+ u)....because of On the other hand, if 2 u, since w e know 2*v, 2 (v+ u)....because of DIVISIBILITY TEST FOR 5: Is the units digit 0 or 5? (i.e. is the units digit divisible by five)? This fact is genereally realized by children who learn to count by fives. Eg. That is divisible by 5 is obvious to anyone who has counted by fives. This shortcut works for the same reason that the test for divisibility by 2 works... because = and = 3117x10 and 5*10, so 5* In general, if u represents the units digit, and v represents the rest of the value of a number, then v is a multiple of 10 and the number is v+ u. Since 5*10 and 10*v, w e know automatically, w ithout inspecting v, that 5*v. Thus 5*(v+ u) if and only if 5*u....because of multiple of 10 DIVISIBILITY TEST FOR 10: Is the units digit 0? (divisible by ten)? E.g is obviously NOT divisible by 10. This w orks because = and = 3457 x 10 so 10*34570 but In general, if the units digit of a number is u, and v represents the remaining value of a number, then the entire value of the number is v+ u (just as = ). v is a multiple of 10. So w e need check only u (because of ). Since 10*v, 10*(v+ u) iff 10*u. DIVISIBILITY TEST FOR 4: Is the number represented by the two rightmost digits divisible by four? E.g is divisible by 4 since 4* is not divisible by 4 since Why it w orks: = = , and 4*100, so 4*56700 automatically. Since 4 also divides 12, 4 must divide Similarly, = Since 4*56700 and 4 18, it follows that In general, if " tu" represents the number represented by the tens & units digits, and v represents the remaining value of a number, then the number is v + "tu" and v is a multiple of 100. Since 100 is divisible by 4, 4*v. If 4 also divides " tu" then 4 must divide v + "tu". (because of ) If 4 "tu" then 4 cannot divide v + "tu". (because of ) DIVISIBILITY TEST FOR 8: check the last digits... 3 E.g is divisible by 8 since 8*. 120 This w orks because...
3 M ORE TESTS for DIVISIBILITY DIVISIBILITY TEST FOR 3: Is the SUM OF THE DIGITS divisible by three? Eg is divisible by 3 since = 21 and 21 is divisible by 3. This test w orks because = = 8 ( ) + 6(99 + 1) + 5 (9+ 1) = = ( ) Multiple of 9, thus of 3 21, a multiple of 3 In general: The value of each digit in a numeral (in our numeration system) is its face value times a pow er of ten. But each pow er of ten is exactly 1 + (a multiple of nine). 3*9 so 3*(any multiple of 9). Thus 3 divides the given number if, and only if, 3 divides the sum of the digits, as seen in the example above. The same argument demonstrates the reason behind the divisibility test for nine, w hich follow s... DIVISIBILITY TEST FOR 9: Is the SUM OF THE DIGITS divisible by nine? EG: because the sum of the digits is 21, not a multiple of because = because = 27, a multiple of 9. This w orks because... (see above). DIVISIBILITY TEST FOR 11: Is the "ALTERNATING SUM" OF THE DIGITS divisible by eleven? Eg is divisible by 11 since = 0 and 0 is divisible by 11***. This works because = = 5 (11 1) (11 1) (11 1) (11 1) + 8 (11 1) (11 1) = 5 (11j 1) + 6 (11k+ 1) + 2 (11m 1) + 2 (11n+ 1) + 8 (11 1) = 5 (11j) + 6 (11k) + 2 (11m) + 2 (11n) + 8 (11) + ( ) As we can see, 11 divides the first terms, so divisibility of by 11 depends on the divisibility of the last part the alternating sum of the digits, (Q: Do w e start w ith + or? A: Try both & see: ) For the follow ing, think of factor trees. Eg 36 What numbers divide (are factors of) 36? 4 9 1, 2, 3, 4, 6, 9, 12, 18, DIVISIBILITY "TEST" FOR 6: DIVISIBILITY TEST FOR 12: True or false: If a number is divisible by 2 and by 9, then it is divisible by 18. (A) True or false: If a number is divisible by 3 and by 6, then it is divisible by 1 8. (B) (Statement A above is true. To see that Statement B is false, consider the number 6. Or 1 2. Or 2 4. Or 3 0. ) * * * As noted previously, if a is any w hole number (or any integer), then a 0. ( because a 0 = 0 ) That is, 0 is a multiple of EVERY whole number. Put another w ay: 0 belongs on the list of multiples of 2. (, 6, 4, 2, 0, 2, 4, 6, ) 0 belongs on the list of multiples of 3. (, 9, 6, 3, 0, 3, 6, 9, ) 0 belongs on the list of multiples of any w hole number a. (, 3a, 2a, a, 0, m, 2m, 3m, )
4 53 PRIME NUMBERS 210 f8 Def'n: A PRIM E NUM BER is a natural number that has EXACTLY TWO natural number divisors 1 & itself. A COMPOSITE number is a natural number that has natural number divisors other than 1 and itself. Eg. 5 is a prime number, divisible by 1 and by 5, and no other elements of N. 6 is composite, since it is divisible by 2 and 3, in addition to 1 and 6. 1 is neither prime nor composite! ( 1 is called a UNIT. ) Prime or composite? 123 is (You can probably tell right aw ay, because 123 is clearly divisible by... ) 223 is In testing whether 223 is prime, must we test for divisibility...by 2? 3? 4? 5? 6? 7? 8?...by 19? because 3 is not even because = 7 Do w e need to test for divisibility by 4? No, if 223 were divisible by 4, it would be divisible by because... Do we need to test for divisibility by 6? No, if 223 were divisible by 6, it would be... So we need to test for divisibility by only primes only, right? because 223 = because 223 = because 223 = hmmm... do w e have to test every prime up to 223? Well, it turns out NOT. We have to test only up to the square root of 223, which is less than 17, in fact less than 15. WHY?... Consider the nontrivial* factor pairs of = 2 36 = 3 24 * Every w hole number w is the product of 1 & itself w = 1 w. This is called the trivial factorization...and w and 1 are called trivial factors. = 4 18 = 6 12 = 8 9 In this example n = 72. In the following, think n= 72: If a is a factor of n, then n/a = c is a factor of n. That is: n = a c. If a is greater than n, then c is smaller than n. (Eg 18 > 7 2, but then c = 3, much less than 7 2. Not both of a and c can be greater than n (if both > n, then a c > n n a contradiction of the fact that n = a c). So if n has any factors other than n and 1, there must be at least one n. If none exist, then n has no factors other than the trivial factors, 1 and n and must be a prime. To test whether a given number "n" is prime, test for divisibility by up to THERE ARE INFINITELY M ANY PRIM ES. [see text! It does a great job of showing you that if you think you have them all, then there must be one more that is greater than any in your list... therefore, you can never have them all! ] A composite number may be expressed as a product of two or more nontrivial factors For any n, there exists a run of n consecutive composite numbers: Let m = (n+ 1). Then m+ 2 is composite because 2*m and 2*2; m+ 3 is composite because... Similarly, m+ k is composite for k = 2,3,4,...,n+ 1. PQED
5 53 More on PRIME NUMBERS 210 f8 Reminder: A composite number may be expressed as a product of tw o or more nontrivial factors (the trivial factors are 1 and the number itself). Such an expression is called a FACTORIZATION of the number. Eg. The composite number 12 may be expressed as or as or as = 3 12 = 4 9 =... = 6 6 = 2 18 = = = 2 30 =... = Intuitively w e can see that a composite number can be factored & if the factors are not all prime, they can be factored again... until the factors are all primes, no further factorization is possible: we call this the PRIME FACTORIZATION of the number. There is only one prime factorization of the number, as stated in the: FUNDAMENTAL THEOREM OF ARITHM ETIC: Every natural number > 1 has a unique prime factorization. Example: Find the prime factorization of 132. Tree: = = = Find the prime factorizations of Up for a challenge? Find n and m such that = n 3 m Think prime! Find the prime factorizations of these. (You might want to use a calculator to save some time.) 70,224 70,
6 55 GCF & LCM: Greatest Common Factor and Least Common Multiple 210 f8 If a natural number d divides every one of a set of integers, then d is a common factor or common divisor of those integers. Eg. 72 and 60 have common factors: 1, 2, 3, 4, 6, and 12. Eg. 22 and 27 have common factors: 1 Eg. 12, 30, and 75 have common factors: 1 and 3 The largest or maximum common factor of tw o w hole numbers is called the GREATEST COMMON FACTOR, or GCF, of the numbers, called GCF(a,b)...also called the greatest common divisor, or GCD. Eg. GCF( 72, 60) = 12 GCF( 22, 27) = 1 GCF( 12, 30, 75) = 3 If the GCF of tw o numbers is 1, the numbers are said to be relatively prime. Eg. 22 and 27 are relatively prime; they have no factors, other than 1, in common. Three Methods for obtaining GCF(a,b) M1. List the factors of the numbers: Eg. GCF(72,60) factors of 72: {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72} factors of 60: {1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60} Find the common factors: {1, 2, 3, 4, 6, 12} Find the greatest: M2. Write the prime factorization of the numbers: 72 = = Build the largest possible factor of both by taking the low est pow er of each common prime factor: GCF( 72, 60) = GCF( 2 3, 2 3 5) = 2 3 = 12 Fill in the Exponents! GCF( 336, 240, 990) = GCF( 2 3 7, 2 3 5, ) = = GCF( 5040, 960) = GCF(, ) = = M3. Euclidean Algorithm Fact: For any a,b in N with a > mb for some m in N, GCF(a,b) = GCF(b,a b) = GCF(b,a mb). Pf: If d = GCF(a,b) then d*a and d*b, so d*(a b) and d is in the set of common factors of a b and b. Conversely, if d*(a b) and d*b then d*((a b)+ b), that is d*a, so d is in the set of common factors of a and b. Thus the set of common factors of (a,b) and the set of common factors of (a b,b) are identical. Induction on m N extends this to GCF(a mb,b). Notice that a mb is the remainder obtained by subtracting m " b's" from a. So if mb is the largest multiple of b that is a, then a mb must be the remainder obtained w hen a is divided by b. This implies that GCF(a,b) = GCF(r,b) w here r is the remainder obtained when a is divided by b! Eg. GCF(30,12) = GCF(6,12) = 6. Eg. GCF(72,30) = GCF(12,30)... but GCF(30,12) = GCF(6,12) = 6... So GCF(72,30) = 6. We have just discovered the secret of the Euclidean algorithm for finding the GCF of two numbers: Divide and use the remainder... Eg. GCF(676,182): 676 = GCF(676,182) = GCF(182,130) What 182 = GCF(182,130) = GCF(130,52)...Why you do 130 = GCF(130,52) = GCF(52,26) it w orks. 52 = GCF(52,26) = 26 Note: To use the Euclidean algorithm for more than two numbers: find the GCF of any two; then find the GCF of that result w ith the next value, or w ith the GCF of another pair; etc. Each of the above methods (M1, M2, M3) has advantages. What is an advantage of each method?
7 55 GCF & LCM: Least Common Multiple 210 f8 If a natural number d is a multiple of every one of a particular set of integers, then d is a common multiple of those integers. The smallest or minimum common multiple of two whole numbers is called the LEAST COM M ON M ULTIPLE, or LCM, of the numbers, notated LCM(a,b,...). Eg. multiples of 12 = {12, 24, 36, 48, 60, 72, 84, 96, 108,... } multiples of 16 = {16, 32, 48, 64, 80, 96, 112, 128, 144,... } 12 and 16 have common multiples: {48, 96, 144,...} LCM( 12, 16) = 48 Eg. LCM( 12, 60 ) = 60 LCM( 24, 15 ) = 120 LCM( 6, 21, 10 ) = 210 Three Methods for obtaining LCM(a,b) M1. List the multiples of the numbers: Eg. LCM( 72, 60 ) multiples of 72: { 72, 144, 216, 288, 360, 432, 504, 576,... } multiples of 60: { 60, 120, 180, 240, 300, 360, 420, 480, 540,...} Find the common multiples: { 360, 720, 10800,... } Find the least (common mulitple): 360 M2. Write the prime factorization of the numbers: = = Build the least possible multiple of both by taking the highest pow er of each prime factor: LCM( 72, 60 ) = LCM( 2 3, ) = = 360 LCM(12,14,30) = LCM( 2 2 3, 2 7, ) = = LCM(18,60,54) = LCM( 2 3, 2 3 5, 2 3 ) = = LCM(336,240,990) = LCM( 2 3 7, 2 3 5, ) = = Fill in the Exponents! M3. Use the GCF Fact: For any a,b in N the product of their LCM & GCF is ab: GCF(a,b) LCM(a,b) = ab. r q Pf: Relies on the fact that if a prime appears in the factorization of a or b, then a = p m and b = p n r q and ab = p p m n. GCF contains p to the pow er min(r,q) and LCM contains p to the pow er max(r,q). min(r,q) + max (r,q) = r + q, the pow er on p in the factorization of ab. In short, all the pow ers of primes in the factorizations of a and b end up in either the GCF or the LCM... Multiplying the tw o puts them altogether, reconstituting ab. Thus we may use the Euclidean algorithm LCM(a,b) = a b to find the LCM of two numbers, since GCF(a,b) Eg. LCM( 72, 60 ) = / GCF(72,60) = 72 60/12 = 6 60 = 360 Eg. LCM( 676, 182 ) = Note: To find the LCM of more than two numbers via method (3), find the LCM of pairs, then find the LCM of the results (until every original number has been included). Something to think about! Given the prime factorization of a number, how can w e find the number of factors of the number? Here are some questions that may lead to an answ er: 3 8 = 2 W hat numbers divide 8? How many factors has 8? (Ans: four: 1, 2, 4, 8) What options do you have w hen "building" a factor of 8? (Ans: these four: 2, 2, 2, 2 ) = = 2 3 What numbers divide 36? (2? 3? 5? 8?) What goes into a factor of 36; how can you " build" a factor of 36? What choices do you have to make w hen building a factor of 36? How many factors of 36 exist in N? (Ans: 9) Write the prime factorization of 72. Write the prime factorizations of all the factors of 72 (these w ere listed on page 55). What makes up the factors of 72? How many different w ays can you build a number w hich divides 72? (If you are " building" a factor of 72, can you use 5? [no]...can you use 2? 4? 8? 16? no 2' s at all?...
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