### 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)
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