Section 5.5. Greatest common factors

Transcription

1 Section 5.5 Greatest common factors

2 Definition GCF The greatest common factor (GCF) is the largest factor common to both of the specified numbers. Ex. GCF(36,84) List factors 36:1,2,3,4,6,9,12,18,36 84:1,2,3,4,6,7,9,12,14,21,28,42,84 12 is the largest number to appear in each of these lists. So we say the greatest common denominator of 36 and 84 is 12 or GCF(36,84)=12.

3 Compute GCF An easier way to compute the GCF is to prime factor both numbers. Then take the most common prime factors and multiply. 36=2x2x3x3 84=2x2x3x7 There are two 2s and one 3 common to both, so GCF(36,84)=2x2x3=12. Why does this work??

4 GCF of 3+ numbers Find GCF(24,60,90). The factors of these are 24:1,2,3,4,6,8,12,24 60:1,2,3,4,5,6,10,12,15,30,60 90:1,2,3,5,6,9,10,15,18,30,45,90 6 is the greatest number common to all three of these lists.

5 GCF of 3+ numbers Find GCF(24,60,90) using prime factors. 24=2x2x2x3 60=2x2 x3 x5 90=2 x3x3x5 The only prime factors common to all three of these are 2 and 3, so the greatest common factor is 6.

6 Relatively prime When a pair of numbers have greatest common factor equal to one, they are called relatively prime.

7 Properties If a b, then GCF(a,b)=a. Why? If a=bq+r, then GCF(a,b)=GCF(b,r). To prove this, observe that a factor of a and b is also a factor of r. Similarly, a factor of b and r is also a factor of a.

8 Simplification Find GCF(348,72). Simplify by dividing: 348=4(72)+60. So GCF(348,72)=GCF(72,60). Simplify again: 72=1(60)+12. So GCF(72,60)=GCF(60,12). Now since 12 60, GCF(60,12)=12.

9 Euclidean Algor. By repeatedly dividing and taking remainders as in the previous example, we will always eventually get one number is a multiple of another, then the smaller number is the greatest common divisor. This process is called the Euclidean Algorithm.

10 Practice Use the Euclidean Algorithm to solve: GCF(91,52) GCF(812,336) GCF( ,59675) You can use a calculator.

11 Lemma 5.13 For any whole numbers, a and b, there is a multiple of a and a multiple of b which differ by GCF(a,b). Said another way, GCF(a,b)=ma-nb, for some positive integers m and n. This follows from the Euclidean Algorithm. Just keep track of where all of the numbers come from.

12 Lemma 5.13 For example, GCF(1785,546)= =3(546) =3(147) =1(105) =2(42)+21 Stop since Subtract each line 21=105-2(42) 42=147-1(105) 105=546-3(147) 147=1785-3(546)

13 Lemma =105-2(42) 42=147-1(105) 105=546-3(147) 147=1785-3(546)

14 Lemma =105-2(147-1(105)) 105=546-3(147) 147=1785-3(546)

15 Lemma =105-2(147)+2(105) 105=546-3(147) 147=1785-3(546)

16 Lemma =105-2(147)+2(105) 105=546-3(147) 147=1785-3(546)

17 Lemma =3(105)-2(147) 105=546-3(147) 147=1785-3(546)

18 Lemma =3(546-3(147))-2(147) 147=1785-3(546)

19 Lemma =3(546)-9(147)-2(147) 147=1785-3(546)

20 Lemma =3(546)-11(147) 147=1785-3(546)

21 Lemma =3(546)-11(1785-3(546))

22 Lemma =3(546)-11(1785)+33(546)

23 Lemma =-11(1785)+36(546)

24 Lemma 5.13 So we get that 21=36(546)-11(1785) Even if you don't remember this process, the idea is that all number come from some (linear) combination of the two numbers we start with.

25 Lemma 5.14 If p is a prime factor of ab, then p is a factor of either a or b. Proof: Suppose to the contrary that p does not divide a or b. Then GCF(p,a) can't be p. So it must be 1. So by Lemma 5.13, 1=mp-na. Multiplying both sides by b, we get b=mpb-nab. Now, p nab and p mpb, so p b, by the Divisibility Lemma. This contradicts our assumption. So it must instead be true that p divides either a or b.

26 5.14 & FTA Using Lemma 5.14, we can finally prove the uniqueness statement of the Fundamental Theorem of Arithmetic. Suppose you have two prime factorizations of a number a=(p1)(p2)...(pn)=(q1)(q2)...(qm) Then p1 a, so by Lemma 5.14, p1 qi, for some i. So cancel these terms and repeat. Eventually, we've matched all the primes.

27 5.14 & FTA Lemma 5.14 can be proven easily if you assume the Fundamental Theorem of Arithmetic. p ab, implies that p is a prime factor of ab. Each prime factor of ab must belong to a or b, by the uniqueness of factorizations. So we have (5.14) FTA and FTA (5.14). We say that (5.14) and FTA are logically equivalent. The validity of one implies the validity of the other. It is necessary to prove one without assuming the other in order to prove both.

28 LCM Related to the idea of the GCF is the idea of the least common multiple (LCM). For example, what is LCM(6,15)? List the multiples 6:6,12,18,24,30,36,42,48,60,... 15:15,30,45,60,75,... The two have common multiples, but 30 is the least common multiple. Note, we only consider positive multiples.

29 Existence of LCM Given two positive integers a and b, ab is a common multiple. So any two positive integers must have a least common multiple and it must be smallest than their product. (This helps us search.)

30 Common multiples A pair of positive integers have infinitely many common multiples. Why?

31 An identity The following identity is useful for computing the LCM: LCM(a,b)=ab/GCD(a,b) This is true because ab is the product of all prime factors of a times the product of all prime factors in b. The denominator cancels one of each prime that is repeated (between the two). This new number is still a multiple of each a and b. If we cancel any more though, we will no longer have a multiple of one of a or b, making it the least common multiple.

32 An identity For example consider LCM(12,60) 12=2*2*3 and 60=2*3*5 Any multiple of 12 must have for its prime factors two 2s and a 3, but it may have more. Any multiple of 60 must have for its prime factors a 2, a 3, and a 5, but it may have more. We might just list all of these. 2*2*3*2*3*5

33 An identity For example consider LCM(12,60) 12=2*2*3 and 60=2*3*5 Although 2*2*3*2*3*5 is a common multiple, it is not the least common multiple because we do not need both 3s. We also don't need a yellow 2 and a cyan 2. So we get a smaller (smallest) common multiple. 2*2*3*5

34 An identity For example consider LCM(12,60) 12=2*2*3 and 60=2*3*5 Now (2*2*3)*5=12*5 and 2*(2*3*5)=2*60 We can't get rid of any more factors and still have a multiple of both 12 and 60. So this is our least common multiple. We got it by dividing by all common prime factors. We saw earlier that the product of all common prime factors is the GCF.

35 Example Use the identity along with the Euclidean Algorithm to determine LCM(8361,75)

36 Exercise 1 Prove that if GCF(a,b)=1 and a c and b c, then (ab) c. Intuition: If GCF(a,b)=1, then LCM(a,b)=ab. Since c is a common multiple, we want to say that it is a multiple of the least common multiple. It is. It always will be. Lemma 5.13 helps us prove this.

37 Exercise 1 Prove that if GCF(a,b)=1 and a c and b c, then (ab) c. We can write c=ja, c=kb, and am-bn=1. Try multiplying both sides by jkab jkaabm-jkabbn=jkab So, ab(jkam-jkbn)=(ja)(kb) So, ab((ja)km-(kb)jn)=cc So, ab(ckm-cjn)=cc So, abc(km-jn)=cc Cancelling the c's gives us that ab*(integer)=c, as desired.

38 Exercise 2 Prove that if GCF(c,a)=1 and c (ab), then c b. Intuition: c (ab) means that all of the prime factors of c are in (ab). But GCF(c,a)=1 means that none of those primes are prime factors of a. So they must all be prime factors of b.

39 Exercise 2 Prove that if GCF(c,a)=1 and c (ab), then c b. We again use Lemma 5.13 cm-an=1 and ck=ab. If we start with cm-an=1 (we usually start here), we want a b to show up. Try multiplying both sides by b. cmb-abn=b Try making the other equation appear. cmb-(ab)n=b cmb-(ck)n=b. So c(mb-kn)=b.

40 Exercise recap We saw as a general rule, we can prove facts about the GCF by starting with Lemma 3.14 and manipulating.

41 Exercise 3 Prove that GCF(n,n+1)=1 for all n. What are some possible values for GCF(n,n+2)? GCF(n,n+6)?

42 What to know You should what the GCF and LCM are. You should be able to compute these using the Euclidean Algorithm and the identity. You should know the statements of Lemma 3.14 and 3.15, but do not need to know the proofs. You should be able to prove things about the GCF. Exercises 2 and 3 above are not unreasonable.

43 To try p. 130 #2,3,6,8,10,12 Modify Exercise 2 above to prove the following: Prove that if GCF(c,a)=d and c (ab), then c bd.