Solucionario algebra de baldor

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The Real And Complex Number Systems

Integers

1.1 Prove that there is no largest prime.
Proof : Suppose p is the largest prime. Then p! + 1 is NOT a prime. So, there exists a prime q such that q |p! + 1 ⇒ q |1 which is impossible. So, there is no largest prime. Remark: There are many and many proofs about it. The proof that we give comes from Archimedes 287-212 B. C. In addition, EulerLeonhard (1707-1783) find another method to show it. The method is important since it develops to study the theory of numbers by analytic method. The reader can see the book, An Introduction To The Theory Of Numbers by Loo-Keng Hua, pp 91-93. (Chinese Version)

1.2 If n is a positive integer, prove the algebraic identity
n−1

a − b = (a − b)
k=0

n

n

ak bn−1−k

Proof : It suffices to showthat
n−1

x − 1 = (x − 1)
k=0

n

xk .

1

Consider the right hand side, we have
n−1 n−1 n−1

(x − 1)
k=0

x =
k=0 n

k

x

k+1



xk

=
k=1 n

xk −

k=0 n−1 k

x

k=0

= x − 1.

− 1 is a prime, prove that n is prime. A prime of the form 2 − 1, where p is prime, is called a Mersenne prime.
p

1.3 If 2

n

Proof : If n is not a prime, then say n= ab, where a > 1 and b > 1. So, we have
b−1

2ab − 1 = (2a − 1)
k=0

(2a )k

which is not a prime by Exercise 1.2. So, n must be a prime. Remark: The study of Mersenne prime is important; it is related with so called Perfect number. In addition, there are some OPEN problem about it. For example, is there infinitely many Mersenne nembers? The reader can see the book, An Introduction To TheTheory Of Numbers by Loo-Keng Hua, pp 13-15. (Chinese Version)

1.4 If 2
form 2 So,
2m

+ 1 is a prime, prove that n is a power of 2. A prime of the + 1 is called a Fermat prime. Hint. Use exercise 1.2.

n

Proof : If n is a not a power of 2, say n = ab, where b is an odd integer. 2a + 1 2ab + 1 and 2a + 1 < 2ab + 1. It implies that 2n + 1 is not a prime. So, n must be a power of 2.Remark: (1) In the proof, we use the identity
2n−2

x

2n−1

+ 1 = (x + 1)
k=0

(−1)k xk .

2

Proof : Consider
2n−2 2n−2 2n−2

(x + 1)
k=0

(−1)k xk =
k=0 2n−1

(−1)k xk+1 +
k=0 2n−2

(−1)k xk (−1)k xk
k=0

= =x

(−1)k+1 xk +
k=1 2n+1

+ 1.

(2) The study of Fermat number is important; for the details the reader can see the book, An Introduction To The Theory OfNumbers by Loo-Keng Hua, pp 15. (Chinese Version)

1.5 The Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, ... are defined by the recursion formula xn+1 = xn + xn−1 , with x1 = x2 = 1. Prove that (xn , xn+1 ) = 1 and that xn = (an − bn ) / (a − b) , where a and b are the roots of the quadratic equation x2 − x − 1 = 0. Proof : Let d = g.c.d. (xn , xn+1 ) , then d |xn and d |xn+1 = xn + xn−1 . So, d |xn−1. Continue the process, we finally have d |1 . So, d = 1 since d is positive. Observe that xn+1 = xn + xn−1 , and thus we consider xn+1 = xn + xn−1 , i.e., consider x2 = x + 1 with two roots, a and b. If we let Fn = (an − bn ) / (a − b) , 3

then it is clear that F1 = 1, F2 = 1, and Fn+1 = Fn + Fn−1 for n > 1. So, Fn = xn for all n. Remark: The study of the Fibonacci numbers is important; thereader can see the book, Fibonacci and Lucas Numbers with Applications by Koshy and Thomas.

1.6 Prove that every nonempty set of positive integers contains a smallest member. This is called the well–ordering Principle. Proof : Given (φ =) S (⊆ N ) , we prove that if S contains an integer k, then S contains the smallest member. We prove it by Mathematical Induction of second form as follows. As k= 1, it trivially holds. Assume that as k = 1, 2, ..., m holds, consider as k = m + 1 as follows. In order to show it, we consider two cases. (1) If there is a member s ∈ S such that s < m + 1, then by Induction hypothesis, we have proved it. (2) If every s ∈ S, s ≥ m + 1, then m + 1 is the smallest member. Hence, by Mathematical Induction, we complete it. Remark: We give a fundamental result to...
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