Questões Resolvidas de Mdc

8/12/2019 Questes Resolvidas de Mdc

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GREATEST COMMON DIVISOR

DEFINITION:

The greatest common divisor (gcd) ofaand b, denoted by (a, b),is the largest common divisorof integers aand b.

THEOREM: Ifaand bare nonzero integers, then their gcd is a linear combination ofaandb,that is there exist integer numbers sand t such that

sa+tb= (a, b).

Proof: Let dbe the least positive integer that is a linear combination ofaand b.We write

d= sa+tb, (1)

wheres and tare integers.We first show that d| a.By the Division Algorithm we have

a= dq+r, where 0

r < d.

From this and (1) it follows that

r= a dq= a q(sa+tb) =a qsa qtb = (1 qs)a+ (qt)b.This shows thatr is a linear combination ofa and b. Since 0 r < d,and d is the least positivelinear combination of a and b, we conclude that r = 0, and hence d| a. In a similar manner,we can show that d| b.

We have shown that d is a common divisor ofaand b. We now show that d is the greatestcommon divisorof a and b. Assume to the contrary that (a, b) = d and d > d. Since d | a,d | b,and d= sa+tb, it follows that d | d,therefore d d.We obtain a contradiction. So, dis the greatest common divisor ofaand band this concludes the proof.

EUCLIDS LEMMA

THEOREM (Euclids Lemma): Ifp is a prime and p | ab,thenp | aor p | b.More generally,if a prime pdivides a product a1a2. . . an, then it must divide at least one of the factors ai.

Proof: Assume that p | a.We must show that p | b.By the theorem above, there are integerssand t with

sp+ta= (p, a).

Since p is prime and p | a,we have (p, a) = 1, and sosp+ta= 1.

Multiplying both sides by b, we getspb+tab= b. (2)

Since p| aband p| spb, it follows that p| (spb+tab). This and (2) give p| b.This completesthe proof of the first part of the theorem. The second part (generalization) easily follows byinduction on n 2.

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COROLLARY: Ifp is a prime and p| a2,then p| a.Proof: Put a= b in Euclids Lemma.

THEOREM: Let pbe a prime. Then

p is irrational.

Proof: Assume to the contrary that

pis rational, that is

p= a

b,where a and b are integers

and b

= 0.Moreover, let aand b have no common divisor >1. Then

p= a2

b2 pb2 =a2. (3)

Since pb2 is divisible by p, it follows that a2 is divisible by p. Then ais also divisible by p bythe Corollary above. This means that there exists q Z such that a = pq. Substituting thisinto (3), we getpb2 = (pq)2 b2 =pq2.Sincepq2 is divisible byp, it follows thatb2 is divisiblebyp.Then bis also divisible by pby the Corollary above. This is a contradiction.

FUNDAMENTAL THEOREM OF ARITHMETIC

THEOREM (Fundamental Theorem of Arithmetic): Assume that an integer a 2 hasfactorizations

a= p1. . . pm and a= q1. . . q n,

where theps andqs are primes. Thenn = m and the qsmay be reindexed so that qi= pi forall i.

Proof: We prove by induction on ,the larger ofmand n, i. e. = max(m, n).Step 1. If= 1,then the given equation is a= p1 = q1, and the result is obvious.Step 2. Suppose the theorem holds for some = k 1.Step 3. We prove it for = k + 1. Let

a= p1. . . pm= q1. . . q n, (4)

wheremax(m, n) =k + 1. (5)

From (4) it follows that pm| q1. . . q n, therefore by Euclids Lemma there is some qi suchthat pm| qi. But qi, being a prime, has no positive divisors other than 1, therefore pm =qi. Reindexing, we may assume that qn = pm. Canceling, we have p1. . . pm1 = q1. . . q n1.Moreover, max(m 1, n 1) = k by (5). Therefore by step 2 qs may be reindexed so thatqi = pi for all i; plus, m 1 =n 1, hence m= n.

COROLLARY: If a 2 is an integer, then there are unique distinct primes pi and uniqueintegers ei >0 such thata= pe11 . . . p

enn .

Proof: Just collect like terms in a prime factorization.

EXAMPLE: 120 = 23 3 5.

PROBLEM: Prove that log35 is irrational.

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EUCLIDEAN ALGORITHM

THEOREM (Euclidean Algorithm): Let a and b be positive integers. Then there is analgorithm that finds (a, b).

LEMMA: Ifa, b, q, r are integers and a= bq+r,then (a, b) = (b, r).

Proof: We have (a, b) = (bq+r, b) = (b, r). Proof of the Theorem: The idea is to keep repeating the division algorithm. We have:

a= bq1+r1, (a, b) = (b, r1)

b= r1q2+r2, (b, r1) = (r1, r2)

r1= r2q3+r3, (r1, r2) = (r2, r3)

r2= r3q4+r4, (r2, r3) = (r3, r4)

. . .

rn2= rn1qn+rn, (rn2, rn1) = (rn1, rn)

rn1= rnqn+1, (rn1, rn) =rn,

therefore

(a, b) = (b, r1) = (r1, r2) = (r2, r3) = (r3, r4) =. . .= (rn2, rn1) = (rn1, rn) =rn.

EXAMPLE: Find (252, 198).By the Euclidean Algorithm we have

252 = 198

1 + 54

198 = 54 3 + 3654 = 36 1 + 1836 = 18 2

therefore(252, 198) = 18.

PROBLEM: Find (35, 55) and (326, 78).

THEOREM: Let a= pe11 . . . penn and b= p

f11 . . . p

fnn be positive integers. Then

(a, b) =pmin(e1,f1)1 . . . p

min(en,fn)n .

EXAMPLE: Since 720 = 24 32 5 and 2100 = 22 3 52 7, we have:(720, 2100) = 22 3 5 = 60.

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PROBLEMS

1. (a) Find (34, 55) using the Euclidean Algorithm.

(b) Find (22335577, 27355372).

(c) Find (2273195237, 27355372).

2. Find the prime factorization of the following numbers:(a) 100

(b) 222

(c) 5040

(d) 8000

(e) 111111

(f) 20!

3. Let a Z

. Prove that:(a) (a, 2a) =a.

(b) (a, a+ 1) = 1.

(c) (a, a+ 2) = 1 or 2.

(d) (8a+ 3, 5a+ 2) = 1.

(e) (3a+ 2, 5a+ 3) = 1.

(f) (am 1, an 1) =a(m,n) 1, where a >1 and m, n Z+.

4. Let a, b

Zand (a, b) = 1.Prove that:

(a) (a+b, a b) = 1 or 2.(b) (a+ 2b, 2a+b) = 1 or 3.

(c) (a2 +b2, a+b) = 1 or 2.

5. Prove that all of the powers in the prime-power factorization of an integer n are even ifand only ifnis a perfect square.

6. Which positive integers have exactly three positive divisors? Which have exactly fourpositive divisors?

7. Prove that ifaand bare positive integers and a2

|b2,then a

|b.

8. Prove that ifaand bare positive integers and a3 | b2,then a| b.9. Prove that ifa and b are positive integers with (a, b) = 1,then (an, bn) = 1 for all n Z+.10. Prove that if a and b are positive integers with (a, b) = 1 and ab= cn, then there are

positive integers dand esuch that a= dn and b= en.11. Prove that the set of all numbers of the form a+ b

5, where a and b are integers,does not enjoy the property of unique factorization.

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THEOREMS, EXAMPLES, AND HINTS

THEOREM 1: Ifa, b, k are integers, then (a+kb, b) = (a, b).

THEOREM 2 (Euclids Lemma): If p is a prime and p| ab, then p| a or p| b. Moregenerally, if a prime p divides a product a1a2. . . an, then it must divide at least one of the

factorsai.

THEOREM 3: If a 2 is an integer, then there are unique distinct primes pi and uniqueintegers ei >0 such that

a= pe11 . . . penn .

EXAMPLE 1: Let a Z.Prove that (2a+ 3, a+ 2) = 1.Proof: By Theorem 1 we have

(2a+ 3, a+ 2) = (a+ 1 +a+ 2, a+ 2) = (a+ 1, a+ 2) = (a+ 1, a+ 1 + 1) = (a+ 1, 1) = 1.

EXAMPLE 2: Let a, b Zand (a, b) = 1.Prove that (a+ 3b, a+b) = 1 or 2.Proof: By Theorem 1 we have

(a+ 3b, a+b) = (a+b+ 2b, a+b) = (2b, a+b).

It is easy to see that (2b, a+ b) equals (b, a+ b) or 2(b, a+ b). From this and Theorem 1 itfollows that

(2b, a+b) = (b, a+b) = (b, a) = 1

or (2b, a+b) = 2(b, a+b) = 2(b, a) = 2.

EXAMPLE 3: Prove that ifa and b are positive integers with (a, b) = 1,then (a2, b2) = 1 forall n Z+.Proof 1: Assume to the contrary that (a2, b2) =n >1.Then there is a primep such thatp | a2and p| b2. From this by Euclids Lemma it follows that p| a and p| b, therefore (a, b) p.This is a contradiction.

Proof 2 (Hint): Just use Theorem 3.

HINTS: In problems 5, 7, 8, 10 use Theorem 3. In problem 11 it is enough to provide anexample.

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Step 1. If = 1, then the given equa-

tion is a = p1 = q1, and the result isobvious.

Step 2. Suppose the theorem holds forsome = k 1.

Step 3. We prove it for = k+ 1. Let

a= p1 . . . pm = q1 . . . qn, ()where

max(m,n) = k+ 1. ()

From () it follows that pm | q1 . . . qn,therefore by Euclids Lemma there issome qi such that pm | qi. But qi, beinga prime, has no positive divisors otherthan 1, therefore pm = qi. Reindexing,we may assume that qn = pm. Cancel-ing, we have p1 . . . pm1 = q1 . . . qn1.Moreover, max(m 1, n 1) = k by(). Therefore by step 2 qs may bereindexed so that qi = pi for all i; plus,m1 = n1, hence m= n.


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LEMMA: If a,b,q,r are integers and

a= bq+r, then (a, b) = (b, r).

Proof: We have

(a, b) = (bq+r, b) = (b, r).

Proof (Euclidean Algorithm): We have:

a = bq1+r1, (a, b) = (b, r1)

b= r1q2+r2, (b, r1) = (r1, r2)

r1 = r2q3+r3, (r1, r2) = (r2, r3)

r2 = r3q4+r4, (r2, r3) = (r3, r4)

. . .rn2 = rn1qn+rn, (rn2, rn1) = (rn1, rn)

rn1 = rnqn+1, (rn1, rn) = rn,

therefore (a, b) = (b, r1) = (r1, r2) =(r

2, r

3) = (r

3, r

4) = . . .= (r

n2, r

n1) =

(rn1, rn) = rn.

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