Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 3: Problem 4

Show that \(\operatorname{gcd}(a, b)=1\) and \(\operatorname{gcd}(a, c)=1\) imply that \(\operatorname{gcd}(a, b c)=1\).

Short Answer

Expert verified
\(\operatorname{gcd}(a, b c) = 1\) because \(a\) is coprime to both \(b\) and \(c\).

Step by step solution

01

Understand the Concepts of GCD and Coprime Numbers

The greatest common divisor (gcd) of two numbers is the largest positive integer that divides both numbers without a remainder. Two numbers are said to be coprime (or relatively prime) if their gcd is 1. Here, it is given that \(\operatorname{gcd}(a, b)=1\) and \(\operatorname{gcd}(a, c)=1\). This means that \(a\) is coprime to both \(b\) and \(c\).
02

Apply the Properties of GCD

One important property of gcd is that \(\operatorname{gcd}(x, yz) = \operatorname{gcd}(\operatorname{gcd}(x, y), z)\). We apply this concept by recognizing that \(\operatorname{gcd}(a, bc)\) can be compared stepwise through the factors of \(bc\).
03

Use the Prime Factorization Concept

Since \(\operatorname{gcd}(a, b)=1\), \(a\) and \(b\) share no common prime factors. Similarly, \(a\) and \(c\) also share no common prime factors because \(\operatorname{gcd}(a, c)=1\). This implies that any common factor of \(a\) and \(bc\) must divide both \(b\) and \(c\), which cannot happen since they are both coprime with \(a\).
04

Conclude with GCD of Combined Product

Given that \(a\) is coprime to both \(b\) and \(c\), any combination product \(bc\) will have no prime factor in common with \(a\). Thus, \(\operatorname{gcd}(a, bc) = 1\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coprime Numbers
Coprime numbers, also known as relatively prime numbers, are two numbers with no common prime factors other than 1. This implies that the greatest common divisor (GCD) of two coprime numbers is automatically 1. Let's consider the expressions \( \operatorname{gcd}(a, b) = 1 \) and \( \operatorname{gcd}(a, c) = 1 \). These statements indicate that the number \( a \) does not share any prime factors with either \( b \) or \( c \). Consequently, \( a \) and \( b \) are coprime, and \( a \) and \( c \) are also coprime.

A useful tip for identifying coprime numbers: if you can verify that two numbers have no common divisors other than 1, they are indeed coprime. In our specific exercise, since \( a \) is coprime to both \( b \) and \( c \), it follows the logical deduction that it is also coprime to their product \( bc \). This understanding is pivotal in comprehending the properties of the GCD.
Prime Factorization
Prime factorization is the process of breaking down a number into its basic building blocks: prime numbers. For example, the number 12 can be expressed as \( 2^2 \times 3 \), which illustrates that 2 and 3 are the prime factors of 12.

Applying this concept to our exercise, if \( \operatorname{gcd}(a, b) = 1 \), then \( a \) and \( b \) do not share prime factors. The same logic applies to \( a \) and \( c \) with \( \operatorname{gcd}(a, c) = 1 \). This tells us that there is no common prime factor between each pair in these expressions.

When we consider \( bc \) — the product of \( b \) and \( c \) — any primes in \( bc \) are also in either \( b \) or \( c \). Because \( a \) is coprime with both individually, it cannot share any common prime factor with the collective factors in \( bc \). This realization assists in understanding why \( \operatorname{gcd}(a, bc) = 1 \).
Properties of GCD
The greatest common divisor, aside from determining coprimeness, also has several noteworthy properties that can simplify complex problems. One such property is that if two numbers are coprime, any linear combination of them will also be coprime with those numbers. Furthermore, the GCD is associative, meaning it can be broken down interactively among various terms: \( \operatorname{gcd}(x, yz) = \operatorname{gcd}(\operatorname{gcd}(x, y), z) \).

In our problem, this property allows us to treat the product \( bc \) as a combined number when calculating the GCD with \( a \). Since both \( \operatorname{gcd}(a, b) \) and \( \operatorname{gcd}(a, c) \) equate to 1, using the properties of GCD ensures that \( \operatorname{gcd}(a, bc) \) will also be 1.

Understanding this property is vital as it simplifies identifying whether products of numbers maintain coprimality with others. It broadens our understanding by illustrating that regardless of how complex a product of coprime numbers becomes, their essential mutual relationship remains unchanged.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use the Extended Euclidean Algorithm to find \(\operatorname{ged}(f, g)\), for \(f, g \in \mathbb{Z}_{p}[x]\) in each of the following examples (arithmetic in \(\mathrm{Z}_{p}=\\{0, \ldots, p-1\\}\) is done modulo \(p\) ). In each case compute the corresponding polynomials \(s\) and \(t\) such that \(\operatorname{gcd}(f, g)=s f+t \mathrm{~g}\). (i) \(f=x^{3}+x+1, g=x^{2}+x+1\) for \(p=2\) and \(p=3\). (ii) \(f=x^{4}+x^{3}+x+1, g=x^{3}+x^{2}+x+1\) for \(p=2\) and \(p=3\). (iii) \(f=x^{5}+x^{4}+x^{3}+x+1, g=x^{4}+x^{3}+x^{2}+x+1\) for \(p=5\). (iv) \(f=x^{5}+x^{4}+x^{3}-x^{2}-x+1, g=x^{3}+x^{2}+x+1\) for \(p=3\) and \(p=5\).

We consider the following property of a Euclidean function on an integral domain \(R\) : $$ d(a b) \geq d(b) \text { for all } a, b \in R \backslash\\{0\\} \text {. } $$ Our two familiar examples, the degree on \(F[x]\) for a field \(F\) and the absolute value on Z, both fulfill this property. This exercise shows that every Euclidean domain has such a Euclidean function. (i) Show that \(\delta: \mathbb{Z} \longrightarrow \mathbb{N}\) with \(\delta(3)=2\) and \(\delta(a)=|a|\) if \(a \neq 3\) is a Euclidean function on \(\mathbb{Z}\) violating (9). (ii) Suppose that \(R\) is a Euclidean domain and \(D=\\{\delta: \delta\) is a Euclidean function on \(R\\}\). Then \(D\) is nonempty, and we may define a function \(d: R \rightarrow N \cup\\{-\infty\\}\) by \(d(a)=\min \\{\delta(a): \delta \in D\\}\). Show that \(d\) is a Euclidean function on \(R\) (called the minimal Euclidean function). (iii) Let \(\delta\) be a Euclidean function on \(R\) such that \(\delta(a b)<\delta(b)\) for some \(a, b \in R \backslash\\{0\\}\). Find another Euclidean function \(\delta^{*}\) that is smaller than \(\delta\). Conclude that the minimal Euclidean function \(d\) satisfies (9). (iv) Show that for all \(a, b \in R \backslash\\{0\\}\) and a Euclidean function \(d\) satisfying \((9)\), we have \(d(0)

(i) Show that for polynomials \(f, g \in F[x]\) of degrees \(n \geq m\), where \(F\) is a field, computing all entries \(s_{i}\) in the traditional Extended Euclidean Algorithm from the quotients \(q_{1}\) takes at most \(2 m^{2}+2 m\) additions and multiplications in \(F\). Hint: Exhibit the bound for the normal case and prove that this is the worst case. (ii) Prove that the corresponding estimate for the Extended Euclidean Algorithm is \(2 m^{2}+3 m+1\).

For each of the following pairs of integers, find their greatest common divisor using the Euclidean Algorithm: (i) 34,21 : (ii) 136,51 : (iii) 481,325 ; (iv) 8771,3206 .

Let \(R\) be a Euclidean domain, with a Euclidean function \(d: R \rightarrow N \cup\\{-\infty\\}\) that has the additional properties o \(d(a b)=d(a)+d(b) .\) o \(d(a+b) \leq \max \\{d(a), d(b)\\}\), with equality if \(d(a) \neq d(b)\). o \(d\) is surjective. for all \(a, b \in R\). Prove that \(R\) is a polynomial ring with \(d\) as degree function. Proceed as follows: (i) Prove that \(d(a)=-\infty\) if and only if \(a=0\). (ii) Show that \(F=\\{a \in R: d(a) \leq 0\\}\) is a subfield of \(R\). (iii) Let \(x \in R\) be such that \(d(x)=1\), and prove that every nonzero \(a \in R\) has a unique representation $$ a=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0} $$ where \(n=d(a), a_{0} \ldots, a_{n} \in F_{\text {, and }} a_{n} \neq 0\).
See all solutions

Recommended explanations on Computer Science Textbooks

Data Structures

Read Explanation

Big Data

Read Explanation

Computer Systems

Read Explanation

Cloud Services

Read Explanation

Fintech

Read Explanation

Game Design in Computer Science

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free