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Chapter 2: Problem 5

(The Euclidean algorithm) a) Let \(m, n \in \mathbb{N}\) and \(m>n .\) Their greatest common divisor \((\operatorname{gcd}(m, n)=d \in \mathbb{N}\) ) can be found in a finite number of steps using the following algorithm of Euclid involving successive divisions with remainder. $$ \begin{aligned} m &=q_{1} n+r_{1} \quad\left(r_{1}

Short Answer

Expert verified
Question: Use the Euclidean algorithm to find the greatest common divisor (gcd) of two natural numbers \(m\) and \(n\). Show that there exist integers \(p\) and \(q\) such that \(pm+qn=d\), where \(d\) is their gcd. If \(m\) and \(n\) are relatively prime, show that \(pm+qn=1\).

Step by step solution

01

Understand the Euclidean algorithm

The Euclidean algorithm is an efficient method for finding the gcd of two numbers. We'll perform successive divisions with remainders, in which we divide the larger number by the smaller number, then replace the larger number with the smaller number, and the smaller number with the remainder of this division. This process continues until the remainder is zero.
02

Carry out the Euclidean Algorithm

Given two natural numbers \(m\) and \(n\) with \(m>n\), we will perform the Euclidean algorithm as shown: $$ \begin{aligned} m &= q_1n + r_1 \quad (r_1 < n) \\ n &= q_2r_1 + r_2 \quad (r_2 < r_1) \\ r_1 &= q_3r_2 + r_3 \quad (r_3 < r_2) \\ & \ \vdots \\ r_{k-1} &= q_{k+1}r_k + 0 \end{aligned} $$ The gcd, \(d\), is the last nonzero remainder, which is \(r_k\) in this case.
03

Find integers \(p\) and \(q\) such that \(pm + qn = d\).

To find integers \(p\) and \(q\), we will work our way up in the algorithm's steps, starting from the second-to-last equation (\(r_{k-1} = q_{k}r_k + r_{k+1}\), where \(r_{k+1} = 0\)) and using back-substitution to express \(r_k\) (which is our gcd \(d\)) in terms of \(m\) and \(n\). First, express \(r_k\) in terms of \(r_{k-1}\) and \(r_{k-2}\): $$r_k = r_{k-1} - q_kr_{k-2}$$ Now, for each back-substitution step, replace the remainders with the corresponding expressions involving the previous remainders (replacing \(r_i\) with \(r_{i-1} - q_ir_{i-2}\)) until the expression involves only \(m\) and \(n\). Finally, we'll obtain the desired relationship \(pm + qn = d\).
04

If \(m\) and \(n\) are relatively prime, \(pm + qn = 1\)

When \(m\) and \(n\) are relatively prime, it means their gcd is 1. In this case, we will have \(d = 1\) and consequently, we will find \(p\) and \(q\) such that \(pm + qn = 1\). This demonstrates that the relationship holds when \(m\) and \(n\) are relatively prime.

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Key Concepts

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

Greatest Common Divisor (GCD)
The greatest common divisor (GCD), also known as the greatest common factor or the highest common divisor, is the largest number that can divide two integers without leaving a remainder. For any two positive integers, the GCD is of significant importance in various fields such as number theory, algebra, and cryptography.

Consider two numbers, say 48 and 18. The divisors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48, while the divisors of 18 are 1, 2, 3, 6, 9, and 18. The common divisors are 1, 2, 3, and 6, but the greatest one is 6, so the GCD of 48 and 18 is 6. There are several ways to calculate the GCD, but one of the most efficient and ancient methods is the Euclidean algorithm.
Successive Divisions with Remainder
The Euclidean algorithm employs successive divisions with remainders to find the GCD of two integers. It's based on the principle that the GCD of two numbers also divides their difference. To grasp the procedure, let's focus on how the algorithm works through an example using the numbers 1071 and 462.

Implementation Steps:

  • Divide 1071 by 462 to get a quotient of 2 and a remainder of 147: \( 1071 = 2 \times 462 + 147 \).
  • Now, divide the previous divisor (462) by the remainder (147): \( 462 = 3 \times 147 + 21 \).
  • Continue the process by dividing 147 by 21: \( 147 = 7 \times 21 + 0 \).
The last nonzero remainder, 21 in this example, is the GCD of 1071 and 462. This algorithm simplifies the problem step by step, reducing the numbers involved until the remainder becomes zero, indicating that the previous remainder is the GCD.
Relatively Prime Integers
Relatively prime integers, also known as coprime or mutually prime numbers, are a pair of numbers that have no common divisors other than 1. In other words, their GCD is 1. For example, 9 and 28 are relatively prime because they share no divisors other than 1.

When two numbers are relatively prime, it has interesting consequences in number theory and applications in cryptography. Using the Euclidean algorithm, if we find that the GCD of two numbers is 1, we can also find integers \(p\) and \(q\) such that \(pm + qn = 1\), illustrating their coprimality. This relationship is particularly important in solving linear Diophantine equations and understanding the multiplicative inverse in modular arithmetic.

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Most popular questions from this chapter

Let \(n \in \mathbb{N}\) and \(n>1\). In the set \(E_{n}=\\{0,1, \ldots, n-1\\}\) we define the sum and product of two elements as the remainders when the usual sum and product in \(\mathbb{R}\) are divided by \(n\). With these operations defined on it, the set \(E_{n}\) is denoted \(\mathbb{Z}_{n}\). a) Show that if \(n\) is not a prime number, then there are nonzero numbers \(m, k\) in \(\mathbb{Z}_{n}\) such that \(m \cdot k=0\). (Such numbers are called zero divisors.) This means that in \(\mathbb{Z}_{n}\) the equation \(a \cdot b=c \cdot b\) does not imply that \(a=c\), even when \(b \neq 0\). b) Show that if \(p\) is prime, then there are no zero divisors in \(\mathbb{Z}_{p}\) and \(\mathbb{Z}_{p}\) is a field. c) Show that, no matter what the prime \(p, \mathbb{Z}_{p}\) cannot be ordered in a way consistent with the arithmetic operations on it.

Show that if \(\mathbb{R}\) and \(\mathbb{R}^{\prime}\) are two models of the set of real numbers and \(f: \mathbb{R} \rightarrow \mathbb{R}^{\prime}\) is a mapping such that \(f(x+y)=f(x)+f(y)\) and \(f(x \cdot y)=f(x) \cdot f(y)\) for any \(x, y \in \mathbb{R}\), then a) \(f(0)=0^{\prime}\); b) \(f(1)=1^{\prime}\) if \(f(x) \not \equiv 0^{\prime}\), which we shall henceforth assume; c) \(f(m)=m^{\prime}\) where \(m \in \mathbb{Z}\) and \(m^{\prime} \in \mathbb{Z}^{\prime}\), and the mapping \(f: \mathbb{Z} \rightarrow \mathbb{Z}^{\prime}\) is injective and preserves the order. d) \(f\left(\frac{m}{n}\right)=\frac{m^{\prime}}{n^{\prime}}\), where \(m, n \in \mathbb{Z}, n \neq 0, m^{\prime}, n^{\prime} \in \mathbb{Z}^{\prime}, n^{\prime} \neq 0^{\prime}, f(m)=m^{\prime}\), \(f(n)=n^{\prime} .\) Thus \(f: \mathbb{Q} \rightarrow \mathbb{Q}^{\prime}\) is a bijection that preserves order. e) \(f: \mathbb{R} \rightarrow \mathbb{R}^{\prime}\) is a bijective mapping that preserves order.

Using the principle of induction, show that a) the sum \(x_{1}+\cdots+x_{n}\) of real numbers is defined independently of the insertion of parentheses to specify the order of addition; b) the same is true of the product \(x_{1} \cdots x_{n}\) c) \(\left|x_{1}+\cdots+x_{n}\right| \leq\left|x_{1}\right|+\cdots+\left|x_{n}\right|\) d) \(\left|x_{1} \cdots x_{n}\right|=\left|x_{1}\right| \cdots\left|x_{n}\right|\) e) \(((m, n \in \mathbb{N}) \wedge(m-1\) and \(n \in \mathbb{N}\), equality holding only when \(n=1\) or \(x=0\) (Bernoulli's inequality); g) \((a+b)^{n}=a^{n}+\frac{n}{1 !} a^{n-1} b+\frac{n(n-1)}{2 !} a^{n-2} b^{2}+\cdots+\frac{n(n-1) \cdots 2}{(n-1) !} a b^{n-1}+b^{n}(\) Newton's binomial formula).

Show that if we take only the set \(\mathbb{Q}\) of rational numbers instead of the complete set \(\mathbb{R}\) of real numbers, taking a closed interval, open interval, and neighborhood of a point \(r \in \mathbb{Q}\) to mean respectively the corresponding subsets of \(\mathbb{Q}\), then none of the three lemmas proved above remains true.

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