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)