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Chapter 1: Problem 38

Given any integers \(m\) and \(n\), not both zero, a positive integer \(d\) satisfying (i) \(d \mid m\) and \(d \mid n \quad\) and (ii) \(e \in \mathbb{Z}, e \mid m\) and \(e|n \Longrightarrow e| d\) is called a greatest common divisor, or simply a GCD, of \(m\) and \(n .\) If \(m=n=0\), we set the \(\mathrm{GCD}\) of \(m\) and \(n\) to be \(0 .\) Given any \(m, n \in \mathbb{Z}\), show that a GCD of \(m\) and \(n\) exists and is unique; it is denoted by \(\mathrm{GCD}(m, n)\). Also show that \(\mathrm{GCD}(m, n)=u m+v n\) for some \(u, v \in \mathbb{Z}\). (Hint: Consider the least element of \(\\{u m+v n: u, v \in \mathbb{Z}\) with \(u m+v n>0\\} .\) )

Short Answer

Expert verified
We proved the existence and uniqueness of the GCD for any integers m and n, not both zero, by considering the least element d of the set {um + vn: u, v 鈭� Z with um + vn > 0} and showing that d is the GCD of m and n. We also demonstrated that GCD(m, n) = d = um + vn, which means it can be expressed as a linear combination of m and n.

Step by step solution

01

Consider the least element of the set

Let's consider the least element of the set S = {um + vn: u, v 鈭� Z with um + vn > 0}. We define d as the smallest positive integer that can be represented as a linear combination of m and n (meaning d = um + vn for some integers u and v).
02

Prove d divides m and n

We will show that d divides both m and n. Apply the Division Algorithm to m and n by d: m = qd + r, where 0 鈮� r < d Rearranging: r = m - qd. Since d = um + vn, substitute it into the equation: r = m - q(um + vn) = (1 - qu)m + (-qv)n This shows r can also be expressed as a linear combination of m and n. However, d is the smallest positive integer that can be represented as a linear combination of m and n. Since 0 鈮� r < d, the only possibility is r = 0. Thus, d divides m (which means d 鈭� m). Similarly, we can show that d divides n (d 鈭� n).
03

Prove d is the greatest common divisor of m and n

Now we need to prove d is the greatest common divisor of m and n, which means that if any integer e divides both m and n, then e must also divide d. So, suppose e is an integer such that e 鈭� m and e 鈭� n. As e divides m and n, m = ke and n = le for some integers k and l. From step 1, we have d = um + vn. Substitute m and n with ke and le, respectively: d = u(ke) + v(le) = e(uk + vl) This shows that e divides d (e 鈭� d). Since any integer e that divides m and n also divides d, we conclude that d is the GCD of m and n.
04

Prove the uniqueness of the GCD

Let's show that the GCD of m and n is unique. Suppose d鈧� and d鈧� are both GCDs of m and n. Following the definition of the GCD, d鈧� 鈭� m and d鈧� 鈭� n, so we have d鈧� 鈭� d鈧� (since d鈧� is also a GCD). Analogously, d鈧� divisibility to m and n implies d鈧� 鈭� d鈧�. The fact that d鈧� 鈭� d鈧� and d鈧� 鈭� d鈧� means that d鈧� and d鈧� are associate numbers, they only differ by a unit factor (卤1). With d鈧� and d鈧� both being positive, their only possible difference is a sign. In this case, d鈧� = d鈧� or d鈧� = -d鈧�. However, since d鈧� and d鈧� are positive by definition, it must be that d鈧� = d鈧�. The GCD of m and n is unique, and we can denote it as GCD(m, n).
05

Show GCD(m, n) = um + vn

In step 1, we defined d as the smallest positive integer that can be represented as a linear combination of m and n (d = um + vn for some integers u and v). In steps 2 and 3, we showed that d is the GCD of m and n. So, we have GCD(m, n) = d = um + vn. We proved the existence and uniqueness of the GCD for any integers m and n, and that GCD(m, n) can be expressed as a linear combination of m and n.

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Key Concepts

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

Division Algorithm
The Division Algorithm provides a systematic way to divide two integers. Given integers \( a \) and \( b \) with \( b > 0 \), it states that there exist unique integers \( q \) (quotient) and \( r \) (remainder) such that:
  • \( a = bq + r \)
  • \( 0 \leq r < b \)
This simple formula is a cornerstone for more advanced mathematical concepts like the Greatest Common Divisor (GCD).
In the context of finding the GCD, the Division Algorithm helps determine the remainder when dividing two numbers. For instance, if \( m \) and \( n \) are divided by the potential GCD \( d \), the algorithm confirms that if the remainder is zero, then \( d \) divides the number perfectly.
Through this process, it establishes that the minimal positive integer obtained from expressions like \( um + vn \) is indeed a divisor of those integers.
Linear Combination
A linear combination involves creating a new value by multiplying integers and summing the results. Specifically, with integers \( m \) and \( n \), a linear combination takes the form \( um + vn \), where \( u \) and \( v \) are also integers.
The significance of linear combinations in finding the GCD is profound. The smallest positive integer that can be constructed using a linear combination of \( m \) and \( n \) is considered their GCD. This is due to the fact that any integer that divides both \( m \) and \( n \) must also divide any such linear combination.
By selecting appropriate values for \( u \) and \( v \), linear combinations provide a method to demonstrate how the GCD itself can be represented in this form. Hence, the equation \( \text{GCD}(m, n) = um + vn \) encapsulates this beautifully.
Unique Existence
The unique existence of the GCD states that every pair of integers \( m \) and \( n \) has a single greatest common divisor. This uniqueness ensures that although different methods or expressions might lead to the same conclusion, they all result in identifying one specific GCD.
To prove uniqueness, suppose two numbers can both be seen as the GCD of \( m \) and \( n \). If both are to divide the same integers perfectly so that each is a divisor of the other, then these numbers can only differ by a sign. Upon reassessing these as positive by definition, they must indeed be identical.
Thus, this characteristic confirms there is one GCD for any set of two integers, making it consistent and reliable when applied in problems and theories. It prevents ambiguity and establishes a universal standard for computations.
Associative Property
The associative property is an essential principle in mathematics, facilitating the manipulation of expressions without altering their results. It states that in operations like addition and multiplication, the grouping of numbers does not matter:
  • For addition: \((a + b) + c = a + (b + c)\)
  • For multiplication: \((a \times b) \times c = a \times (b \times c)\)
In the context of finding the GCD, although the associative property might not be directly employed, it underlies the hidden operations within any linear combination or algorithmic process. By guaranteeing that integer addition and multiplication are order-agnostic, the associative property ensures calculations remain accurate regardless of how components are grouped.
This property supports intuitive thinking and simpler manipulation of algebraic expressions, enhancing clarity and efficiency in proofs and problem-solving.

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Most popular questions from this chapter

Use Exercise 53 to deduce the following: (i) [Basic Inequality for Rational Powers] Given any \(a, b \in \mathbb{R}\) and \(r \in\) \(\mathbb{Q}\) such that \(a \geq 0, b \geq 0\) and \(r \geq 1\), we have \(\left|a^{r}-b^{r}\right| \leq r M^{r-1}|a-b|\), where \(M=\max \\{|a|,|b|\\}\) (ii) [Inequality for Rational Roots] Given any \(a, b \in \mathbb{R}\) and \(r \in \mathbb{Q}\) such that \(a \geq 0, b \geq 0\) and \(0a / 2\), then \(a^{r}-b^{r} \leq r b^{r-1}(a-b) \leq r(a-b)^{r}\) Note that \(\max \left\\{r, 2^{r}\right\\} \leq 2 .\) ) (iii) [Binomial Inequality for Rational Powers] Given any \(a \in \mathbb{R}\) and \(r \in \mathbb{Q}\) such that \(1+a \geq 0\) and \(r \geq 1\), we have \((1+a)^{r} \geq 1+r a\) Further, if \(r>1\), then \((1+a)^{r}>1+r a\)

Let \(f:(0, \infty) \rightarrow \mathbb{R}\) be the function defined by the following: (i) \(f(x)=\sqrt{x}\), (ii) \(f(x)=x^{3 / 2}\), (iii) \(f(x)=\frac{1}{|x|}\), (iv) \(f(x)=\frac{1}{x^{2}}\). Sketch the graph of \(f\) and determine the points at which \(f\) has local extrema as well as the points of inflection of \(f\), if any, in each case.

Given any \(p(x) \in \mathbb{R}[x]\) and \(\alpha \in \mathbb{R}\), show that there is a unique polynomial \(q(x) \in \mathbb{R}[x]\) such that \(p(x)=(x-\alpha) q(x)+p(\alpha)\). Deduce that \(\alpha\) is a root of \(p(x)\) if and only if the polynomial \((x-\alpha)\) divides \(p(x)\).

Let \(m, n \in \mathbb{N}\) and let \(x \in \mathbb{R}\) be such that \(x \geq 0\) and \(x \neq 1\). Show that \(\frac{x^{m}-1}{m}>\frac{x^{n}-1}{n} \quad\) if \(m>n \quad\) and \(\quad \frac{x^{m}-1}{m}<\frac{x^{n}-1}{n} \quad\) if \(mn\). Write \(n\left(x^{m}-1\right)-m\left(x^{m}-1\right)\) as \((x-1)\left[n\left(x^{n}+x^{n+1}+\cdots+x^{m-1}\right)-(m-n)\left(1+x+\cdots+x^{n-1}\right)\right]\). Compare the \(m-n\) elements \(x^{n}, x^{n+1}, \ldots, x^{m-1}\) as well as the \(n\) elements \(1, x, \ldots, x^{n-1}\) with \(x^{n}\), when \(x<1\) and \(x>1 .\) )

Given any \(\ell, m \in \mathbb{Z}\) with \(\ell \neq 0\), prove that there are unique integers \(q\) and \(r\) such that \(m=\ell q+r\) and \(0 \leq r<|\ell| .\) (Hint: Consider the least element of the subset \(\\{m-\ell n: n \in \mathbb{Z}\) with \(m-\ell n \geq 0\\}\) of \(\mathbb{Z} .\) )
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