Estrategias Básicas Para La Toma De Decisiones

First, here is an application of the Fundamental Theorem of Arithmetic that looks rather more obvious than it really is.

PRIME FACTORIZATION 93 PROPOSITION 11.1

Let n = pa₁1pa₂2. . .p_mam_{, where the pis are prime, p1<p2< . . . <pm and}

the a_is are positive integers. If m divides n, then

m = pb1

1 pb22. . .pmbm with 0 ≤ bi≤ ai for alli .

For example, the only divisors of 2100₃2_{are the numbers 2}a₃b_{, where 0 ≤}

a ≤ 100, 0 ≤ b ≤ 2.

PROOF Ifm|n, then n = mc for some integer c. Let m = qc₁1. . .qc_kk,c =

rd1

1 . . .rldl be the prime factorizations of m,c. Then n = mc gives the equation

pa1

1 pa22. . .pamm=qc11. . .qckkr1d1. . .rdll .

By the Fundamental Theorem 11.1, the primes, and the powers to which they occur, must be identical on both sides. Hence, each q_i is equal to some p_j, and its power c_i is at most a_j. In other words, the conclusion of the proposition holds.

We can use this to prove some further obvious-looking facts about integers. Define the least common multiple of two positive integers a and b, denoted by

lcm(a,b), to be the smallest positive integer that is divisible by both a and b.

For example, lcm(15,21) = 105. PROPOSITION 11.2

Let a,b ≥ 2 be integers with prime factorizations

a = pr1

1pr22. . .prmm, b = p1s1ps22. . .psmm

where the p_i are distinct primes and all r_i,s_{i≥ 0 (we allow some of the}

ri ands_i to be0). Then

(i)hc f (a,b) = pmin(r₁ 1,s1). . .pmin(r_m m,sm) (ii) lcm(a,b) = pmax(r₁ 1,s1). . .pmax(r_m m,sm) (iii)lcm(a,b) = ab/hc f (a,b).

PROOF In part (i), the product on the right-hand side divides both

a and b and is the largest such integer, by Proposition 11.1. And in part

(ii), the product on the right-hand side is a multiple of botha and b and is the smallest such positive integer, again by Proposition 11.1. Finally, if we take the product of the right-hand sides in (i) and (ii), then we

94 A CONCISE INTRODUCTION TO PURE MATHEMATICS

obtain

pmin(r1,s₁)+max(r₁,s₁)

1 . . .pmin(rm m,sm)+max(rm,sm),

which is equal toab since min(r_i,s_i) +max(r_i,s_i) =r_i+s_i.

Here is our next application of the Fundamental Theorem of Arithmetic. PROPOSITION 11.3

Let n be a positive integer. Then √n is rational if and only if n is a perfect square (i.e., n = m2 for some integerm).

PROOF The right-to-left implication is obvious: if n = m2 with

m ∈ Z, then √n = |m| ∈ Z is certainly rational.

The left-to-right implication is much less clear. Suppose√n is rational, say

√n = r_s

wherer,s ∈ Z. Squaring, we get ns2=r2. Now consider prime factoriza- tions. Each prime in the factorization of r2 appears to an even power (since if r = pa₁1. . .pa_kk then r2=p2a₁1. . .p2a_k k). The same holds for the primes in the factorization of s2. Hence, by the Fundamental Theo- rem, each prime factor of n must also occur to an even power — say

n = q2b1

1 . . .q2bl l. Thenn = m2, wherem = qb11. . .qbll ∈ Z.

A similar argument applies to the rationality of cube roots, and more gener- ally, nth_{roots (see Exercise 5 at the end of the chapter).}

Now for our final consequence of the Fundamental Theorem 11.1. Again it looks rather innocent, but in the example following the proposition we shall give a striking application of it.

In the statement, when we say a positive integer is a square (or an nth_power), we mean that it is the square of an integer (or the nth_{power of an integer).}

PROPOSITION 11.4

Let a and b be positive integers that are coprime to each other. (a) If ab is a square, then both a and b are also squares.

(b) More generally, if ab is an nth _{power (for some positive integern),} then botha and b are also nth powers.

PROOF (a) Let the prime factorizations of a,b be

a = pd1

PRIME FACTORIZATION 95 (wherep1< . . . <p_kandq1< . . . <q_l). Ifab is a square, then ab = c2for some integerc; let c have prime factorization c = r₁f1. . .r_mfm. Thenab = c2 gives the equation

pd1

1 . . .pdkkqe11. . .qlel =r2 f11. . .rm2 fm .

Sincea and b are coprime to each other, none of the p_is are equal to any of the q_is. Hence, the Fundamental Theorem 11.1 implies that each p_i is equal to somerj, and the corresponding powersdi and2 fj are equal; and likewise for theqis and their powers.

We conclude that all the powers d_i,e_i are even numbers — sayd_i= 2d0

i,ei=2e0_i. This means that

a =pd01 1 . . .pd 0 k k 2 , b =qe₁01. . .qe_l0l2 , soa and b are squares.

(b) The argument for (b) is the same as for (a): an equation ab = cn gives an equality

pd1

1 . . .pdkkqe11. . .qlel =rn f11. . .rmn fm .

The Fundamental Theorem implies that each power d_i,e_i is a multiple ofn, and hence a,b are both nth powers.

Example 11.1

Here is an innocent little question about the integers:

Can a non-zero even square exceed a cube by 1?

(The non-zero even squares are of course the integers4,16,64,100,144,... and the cubes are . . . ,_{−8,−1,0,1,8,27,....)}

In other words, we are asking whether the equation

4x2=y3+1 (11.1) has any solutions withx,y both non-zero integers. This is an example of aDiophantine equation. In general, a Diophantine equation is an equation for which the solutions are required to be integers. Most Diophantine equations are very hard, or impossible, to solve — for instance, even the equationx2=y3+k has not been completely solved for all values of

k. However, I have chosen a nice example, in that Equation (11.1) can

be solved fairly easily (as you will see), but the solution is not totally trivial and involves use of the consequence 11.4 of the Fundamental Theorem 11.1.

96 A CONCISE INTRODUCTION TO PURE MATHEMATICS

Let us then go about solving Equation (11.1) for x,y ∈ Z. First we rewrite it asy3=4x2− 1 and then cleverly factorize the right-hand side to get

y3_{= (2x+1)(2x−1) .}

The factors2x+1,2x−1 are both odd integers, and their highest common factor divides their difference, which is 2. Hence

hcf(2x+1,2x−1) = 1 .

Thus, 2x + 1 and 2x − 1 are coprime to each other, and their prod- uct is y3, a cube. By Proposition 11.4(b), it follows that 2x + 1 and 2x − 1 are themselves both cubes. However, from the list of cubes . . . ,_{−8,−1,0,1,8,27,... it is apparent that the only two cubes that dif-} fer by 2 are 1,−1. Therefore, x = 0 and we have shown that the only even square that exceeds a cube by 1 is 0. In other words, there are no non-zero such squares.

Exercises for Chapter 11

1. Find the prime factorization of 111111.

2. (a) Which positive integers have exactly three positive divisors? (b) Which positive integers have exactly four positive divisors?

(c) Suppose n ≥ 2 is an integer with the property that whenever a prime

p divides n, p2_{also divides n (i.e., all primes in the prime factorization}

of n appear at least to the power 2). Prove that n can be written as the product of a square and a cube.

3. Suppose that n is a positive integer such that p = 2n_{− 1 is prime. (The} first few such primes are 3,7,31,....) Define

N = 2n−1_{p .}

List all positive integers that divide N. Prove that the sum of all these divisors, including 1 but not N itself, is equal to N.

A positive integer that is equal to the sum of all its divisors (including 1 but not itself) is called a perfect number. Write down four perfect numbers.

4. Prove that lcm(a,b) = ab/hc f (a,b) for any positive integers a,b without using prime factorization.

PRIME FACTORIZATION 97 5. (a) Prove that 213 and 313 are irrational.

(b) Let m and n be positive integers. Prove that m1n is rational if and only

if m is an nth_{power (i.e., m = c}n_{for some integer c).}

6. Let E be the set of all positive even integers. We call a number e in E “prima” if e cannot be expressed as a product of two other members of

E.

(i) Show that 6 is prima but 4 is not.

(ii) What is the general form of a prima in E?

(iii) Prove that every element of E is equal to a product of primas. (iv) Give an example to show that E does not satisfy a “unique prima

factorization theorem” (i.e., find an element of E that has two dif- ferent factorizations as a product of primas).

7. (a) Which pairs of positive integers m,n have hc f (m,n) = 50 and lcm(m,n) =1500?

(b) Show that if m,n are positive integers, then hc f (m,n) divides lcm(m,n). When does hc f (m,n) = lcm(m,n)?

(c) Show that if m,n are positive integers, then there are coprime integers

x,y such that x divides m, y divides n, and xy = lcm(m,n).

8. Find all solutions x,y ∈ Z to the following Diophantine equations: (a) x2_=y3

(b) x2_{− x = y}3

(c) x2_=y4_{− 77}

(d) x3_=4y2_+4y−3.

9. Languishing in his prison cell, critic Ivor Smallbrain is dreaming. In his dream he is on the Pacific island of Nefertiti, eating coconuts on a beach by a calm blue lagoon. Suddenly the king of Nefertiti approaches him, saying, “Your head will be chopped off unless you answer this riddle: Is it possible for the sixth power of an integer to exceed the fifth power of another integer by 16?” Feverishly, Ivor writes some calculations in the sand and eventually answers, “Oh, Great King, no it is not possible.” The king rejoinders, “You are correct, but you will be beheaded anyway.” The executioner’s axe is just coming down when Ivor wakes up. He wonders whether his answer to the king was really correct.

Chapter 12