Sunday, July 19, 2009

Solutions manual for NTB - 1.3 Some consequences of unique factorization

Let's recall some notations and facts before we begin. For each prime , we may define the function , mapping non-zero integers to non-negative integers, as follows: for every integer , if , where , then .

We may then write the factorization of into primes as:

,
where the product is over all primes p; although syntactically this is an infinite product, all but finitely many of its terms are equal to 1, and so this expression makes sense.

Observe that if and are non-zero integers, then

for all primes ,

and

for all primes .

From this, it is clear that:



.

The greatest common divisor of is denoted and is the unique non-negative integer satisfying

for all primes .

The least common multiple of is denoted and is the unique non-negative integer satisfying

for all primes .



Exercise 1.19. Let n be an integer. Generalizing Exercise 1.11, show that if is a pairwise relatively prime family of integers, where each divides , then their product also divides .
Proof

We have:



Since is a pairwise relatively prime family of integers, for all primes , we have: , for some . (1)

Since
, hence for all primes . (2)

From (1) and (2), we see that
must divides .

(q.e.d)



Exercise 1.20. Show that for all integers , , , we have:

(a) ,

(b) ,

(c) ,

(d) .

Proof

a, b, c are trivial and can be easily derived from the definition of least command multiple. We prove only d here:




(q.e.d)




Exercise 1.21. Show that for all integers , , we have:

(a) ;

(b) .

Proof

(a) Observe that:



for all primes
.

(b) This directly follows from (a).

(q.e.d)



Exercise 1.22. Let with . Show that:

(a) ;

(b) .

Proof

(Intentionally left blank)

(q.e.d)



Exercise 1.23. Let with . Show that ; in particular, there exist integers such that .

Proof

We can prove this by generalizing Theorem 1.7 to many integers.

Observe that each is an ideal of , and so is their sum .

Let is an ideal of such that . According to Theorem 1.6, there exists an unique integer such that . We show that is indeed the greatest common divisor of . Note that .

Since , we see that for . So we see that is a common divisor of .

Since , there exist integers such that .

Now suppose
is another common divisor of with . Then the equation implies that , which says that .

Thus, any common divisor of divides . That proves that is a greatest common divisor of .

(q.e.d)



Exercise 1.24. Show that if is a pairwise relatively prime family of integers, then .

Proof

Since is a pairwise relative prime family of integers, we see that for all primes , for some , and for all .

Hence we see that for all primes ,



for some .

Therefore,



(q.e.d)



Exercise 1.25. Show that every non-zero can be expressed as:

,

where the ’s are distinct primes and the ’s are non-zero integers, and that this expression in unique up to a reordering of the primes.

Proof

We know that we can represent every non-zero as a fraction of the form where and are relatively prime; moreover, the values of and are uniquely determined up to sign.

Applying Theorem 1.3 (the fundamental theorem of arithmetic) and the fact that and are relatively prime, we see that:



where are distinct primes and are positive integers. Moreover, this expression is unique, up to a reordering of the primes.

(q.e.d)



Exercise 1.26. Let and be positive integers, and suppose such that for some . Show that . In other words, is either an integer or is irrational.

Proof

Using Exercise 1.25's result, we have:

,

where the ’s are distinct primes and the ’s are non-zero integers, and that this expression in unique up to a reordering of the primes. To prove , we must show that for all .

Since for some positive integers and , we see that:

.

According to Theorem 1.3, we see that are positive integers. Since , that means for all , which proves our claim.

(q.e.d)


2 comments:

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Sam Li said...

I was wondering if he has the complete solution for the book. I only see section 1.1 - 1.3