Problem 6
Question
Let \(\operatorname{gcd}(a, b)=c .\) Write \(a=c a^{\prime}\) and \(b=c b^{\prime} .\) Then $$ \operatorname{gcd}\left(a^{\prime}, b^{\prime}\right)=\operatorname{gcd}\left(a, b^{\prime}\right)=\operatorname{gcd}\left(a^{\prime}, b\right)=1 $$
Step-by-Step Solution
Verified Answer
All expressions simplify by proving shared divisibility conflicts with the maximal gcd assumption, confirming 1 is the greatest common divisor for each.
1Step 1: Understanding the Exercise
We start with the given situation: the greatest common divisor (gcd) of two numbers \(a\) and \(b\) is \(c\). Thus, both \(a\) and \(b\) can be rewritten in terms of \(c\) and new variables \(a'\) and \(b'\), such that \(a = c \cdot a'\) and \(b = c \cdot b'\). The goal is to show that certain gcd equations involving \(a'\) and \(b'\) equal 1.
2Step 2: Expressing in Terms of New Variables
Since \(a = c \cdot a'\) and \(b = c \cdot b'\), \(c\) divides both \(a\) and \(b\). The fact that \(c\) is the gcd of \(a\) and \(b\) means \(c\) is the largest number dividing both. Consequently, any factor common to \(a'\) and \(b'\) must not affect \(a\) or \(b\) beyond \(c\), because any common factor of \(a'\) and \(b'\) would contravene the gcd condition of \(a\) and \(b\).
3Step 3: Proving \(\operatorname{gcd}(a', b') = 1\)
Since \(a = c a'\) and \(b = c b'\), any common factor of \(a'\) and \(b'\) would make \(c\) not maximal if it were greater than 1, contradicting \(c = \operatorname{gcd}(a, b)\). Thus, the only possible common divisor of \(a'\) and \(b'\) is 1: \(\operatorname{gcd}(a', b') = 1\).
4Step 4: Proving \(\operatorname{gcd}(a, b') = 1\) and \(\operatorname{gcd}(a', b) = 1\)
To show \(\operatorname{gcd}(a, b') = 1\), assume a common divisor \(d\) exists for \(a = c a'\) and \(b'\). If so, then \(d\) divides \(c a'\) and thus \(c\) and \(a'\). Since \(c\) already divides both \(a\) and \(b\), and \(b'\) is free of additional factor beyond \(c\), it follows \(d\) must be 1.Similarly, for \(\operatorname{gcd}(a', b) = 1\), a common divisor \(d\) of \(a'\) and \(b = c b'\) implies \(d\) divides \(ca'\) and \(b\), restricting \(d\) to involvement only through \(c\). Hence, \(d\) must be 1.
Key Concepts
Integer FactorizationCoprime IntegersNumber Theory Concepts
Integer Factorization
Integer factorization involves breaking down a composite number into a product of smaller integers or prime factors. When given two numbers, say \(a\) and \(b\), if their greatest common divisor (GCD) is \(c\), then we can express these numbers like this:
This breakdown is useful because it helps isolate the shared component of \(a\) and \(b\) (which is \(c\)), allowing us to consider what's left (\(a'\) and \(b'\)) independently. By understanding this process, it becomes easier to work with these numbers and apply further number theory analyses, such as determining coprime relationships.
- \(a = c \cdot a'\)
- \(b = c \cdot b'\)
This breakdown is useful because it helps isolate the shared component of \(a\) and \(b\) (which is \(c\)), allowing us to consider what's left (\(a'\) and \(b'\)) independently. By understanding this process, it becomes easier to work with these numbers and apply further number theory analyses, such as determining coprime relationships.
Coprime Integers
Coprime integers, or relatively prime integers, are pairs of numbers whose greatest common divisor is 1. In this exercise, the pair \((a', b')\) are elements that arise from integer factorization.
- Since \(a = c \cdot a'\) and \(b = c \cdot b'\) with \(c\) as \(\operatorname{gcd}(a, b)\), any common factor between \(a'\) and \(b'\) would indicate a higher GCD than \(c\), which cannot be true.
- Therefore, \(\operatorname{gcd}(a', b') = 1\).
Number Theory Concepts
Number theory is an area of mathematics dedicated to studying integers and their properties.
Number theory concepts such as the GCD, integer factorization, and coprimeness offer insight into the structure and relations of numbers. Deep understanding comes with the application of:
- GCD: Utilizing the greatest common divisor to simplify and solve complex integer problems.
- Factorization: Decomposing integers to manage and analyze complex expressions.
- Coprimeness: Playing a crucial role in developing further mathematical properties, easing calculations with relatively prime numbers.
Other exercises in this chapter
Problem 6
If \(\operatorname{gcd}(a, b)=1\), then \(\operatorname{lcm}(a, b)=a b\).
View solution Problem 6
Suppose \(\operatorname{gcd}(a, b)=1\) and \(c \mid a b\). Then there exist integers \(r\) and \(s\) such that \(c=r s\) \(a, s \mid b\), and \(\operatorname{gc
View solution Problem 6
If \(a>0\) and \(b>0\) and \(a \mid b\), then \(a \leq b\).
View solution Problem 7
If \(\operatorname{lcm}(a, b)=a b\), then \(\operatorname{gcd}(a, b)=1\)
View solution