Math 220.906
Writing Assignment #3
October 1, 2013
Due Thursday, October 10.
Instructions: Answer the questions in the boxes.
Problem B: Throughout Z represents the set of integers, and Q represents the set of rational
numbers. First a little terminology:
• For b, a ∈ Z, with b 6= 0, we say that b divides a if ab ∈ Z. That is, b divides a if there
is no remainder when a is divided by b. The common abbreviation for “b divides a” is
“b | a”, and if b | a, then b is called a divisor of a.
• An integer p ≥ 2 is called a prime number if the only positive integers that divide it
are 1 and p. Note that 1 is not a prime number.
The following is a fact about integers that you are probably familiar with. We may prove it
later in the semester, but for now we will accept it as a fact.
Unique Factorization of Integers: Given an integer n ≥ 2, there are prime numbers p1 ,
p2 , . . . , pr so that n = p1 p2 · · · pr . This expression of n as the product of primes is unique up
to reordering the factors.
I. Given two positive integers a and b, we call d the greatest common divisor if
• d | a and d | b (d is a common divisor of a and b);
• if e is any other positive integer such that e | a and e | b, then e ≤ d.
Using the fact that every integer can be factored uniquely into the product of prime
numbers, prove that every pair of positive integers a and b has a greatest common divisor.
A couple of comments: (1) We often abbreviate “the greatest common divisor of a and b”
with gcd(a, b). (2) The greatest common divisor is unique (hence why it is the “greatest”).
Why is that? (3) Using unique factorization is not the most efficient way to compute the
greatest common divisor (far from it); in fact one often uses the gcd to factor large integers,
rather than the other way around.
II. Let Z+ denote the set of all positive integers, and let Q+ denote the set of all positive
rational numbers, so
na
o
Q+ =
: a, b ∈ Z+ , b 6= 0 .
b
a
+
We say that b ∈ Q is written in lowest terms if gcd(a, b) = 1.
Suppose we try to define a function f : Q+ → Z+ by f ab = 2a 3b . Does this
• define a legitimate function from Q+ to Z+ ? Why or why not? If not, how
can we change the definition of f so that we obtain an actual function?
Let S = {2m 3n : m, n ∈ Z+ }. Prove that the function g : S → Q+ defined
• by g(2m 3n ) = m
is a legitimate function. (On what properties of the integers
n
does this depend?)