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Chapter 1: Problem 40

If \(m, n, n^{\prime} \in \mathbb{Z}\) are such that \(m\) and \(n\) are relatively prime and \(m \mid n n^{\prime}\), then show that \(m \mid n^{\prime}\). Deduce that if \(p\) is a prime (which means that \(p\) is an integer \(>1\) and the only positive integers that divide \(p\) are 1 and \(p\) ) and if \(p\) divides a product of two integers, then it divides one of them.

Short Answer

Expert verified
In summary, if \(m, n, n' \in \mathbb{Z}\) are such that \(m\) and \(n\) are relatively prime and \(m \mid nn'\), then \(m \mid n'\). We first show that \(m\) must be a prime number, and then apply Euclid's Lemma to deduce that \(m\) must divide one of the integers involved in the product, in this case, \(n'\). This directly leads us to the result that if \(p\) is a prime number and \(p\) divides a product of two integers, then it must divide at least one of the integers involved in the product.

Step by step solution

01

Write down the given information

We are given the following information: 1. m, n, n' 鈭� 鈩� (That is to say, they are integers) 2. The greatest common divisor of m and n is 1. (i.e., gcd(m, n) = 1) 3. m 鈭� nn' (m divides the product of n and n') Our goal is to show that m 鈭� n' (m divides n').
02

Use Euclid's Lemma

Euclid's Lemma states that if a prime number p divides the product of two numbers, say a and b, then p must divide either a or b (or both). We will use this lemma after proving that m must be a prime number.
03

Prove that m is a prime number

We know that m and n are relatively prime numbers, which means their gcd is 1. Now let's assume that m is not a prime number. Then m can be written as the product of two integers a and b, where 1 < a, b < m. But this would mean that the gcd(m, n) is not necessarily 1, which is not possible since m and n are relatively prime. Therefore, m must be a prime number.
04

Apply Euclid's Lemma to our case

Now that we know m is a prime number, we can apply Euclid's Lemma to our problem: Since m 鈭� nn', by Euclid's Lemma, m has to divide either n or n' (or both). But we know that gcd(m, n) = 1 which means m cannot divide n. This implies that m must divide n', so m 鈭� n'.
05

Deduce the result for prime numbers

We are now asked to deduce that if p is a prime number and p 鈭� ab (where a and b are integers), then p must divide either a or b (or both). This is a direct result of our work in Steps 3 and 4. Since we showed that if two numbers are relatively prime and their product is divisible by a prime number, then the prime number must divide at least one of them, we can directly apply this result to our case here. Therefore, if p is a prime number and p 鈭� ab, then p must divide either a or b (or both).

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Key Concepts

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

Euclid's Lemma
Understanding the relationship between prime numbers and divisibility is crucial for grasping many concepts in number theory, and one key theorem that helps in this understanding is Euclid's Lemma. In essence, this lemma states that if a prime number divides the product of two numbers, then it must divide at least one of those numbers. For example, if a prime number p divides the product ab, then p must divide a or b (or both).

Here's why this is so powerful: the definition of a prime number already tells us that it has no divisors other than 1 and itself. When a prime number divides a product, it cannot 'share' factors with both numbers unless it is a factor of one of them. This property is not necessarily true for non-prime numbers, which makes Euclid's Lemma particularly unique for prime numbers.

In the context of our exercise, we are asked to use Euclid鈥檚 Lemma to show that if m divides nn' and is also prime, it must divide n or n'. Given that m and n are relatively prime, we can safely say that m cannot divide n, hence it must divide n'.
Greatest Common Divisor (gcd)
The greatest common divisor (gcd) of two numbers is the largest number that divides both of them without leaving a remainder. It's a fundamental concept when examining the relationships between different numbers. Often, finding the gcd of two numbers is the first step in simplifying fractions, solving Diophantine equations, or analyzing number divisibility.

For instance, if we consider the numbers 8 and 12, their divisors are 1, 2, 4, and 8 for the former, and 1, 2, 3, 4, 6, and 12 for the latter. The largest common divisor they share is 4, so gcd(8, 12) = 4.

The exercise provided tests our understanding of gcd in the context where the gcd of two numbers, m and n, is 1. This special condition implies that m and n are coprime or relatively prime. It is this relationship that allows us to infer the divisibility of m over n' when m divides their product, nn'.
Relatively Primes
Two integers are considered relatively prime or coprime if they have no common positive integer factors other than 1. In other words, their gcd is exactly 1. An easy mistake to make is to think that relative primes must be prime numbers, which isn't always true. For example, 8 and 9 are relatively prime, but 8 is not a prime number.

Being relatively prime has interesting implications in number theory, especially in divisibility. As our exercise demonstrates, if two numbers are relatively prime and one of them divides the product of the other and a third number, then it must also divide the third number.

Essentially, being relatively prime is a statement about independence in divisibility. When a number is relatively prime to another, there are no shared prime factors to consider when analyzing divisibility by other numbers. This concept is central to understanding the Euclidean algorithm for computing gcd, the foundation of many encryption algorithms, and the fundamental theorem of arithmetic.

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Most popular questions from this chapter

Let \(I\) be an interval and let \(f: I \rightarrow \mathbb{R}\) be convex on \(I .\) Given any \(r \in \mathbb{R}\), show that \(r f\) is a convex function on \(I\) if \(r \geq 0\) and a concave function on \(I\) if \(r<0\)

Using only the algebraic properties A1-A5 on page 2, prove the following. (i) 0 is the unique real number such that \(a+0=a\) for all \(a \in \mathbb{R}\). In other words, if some \(z \in \mathbb{R}\) is such that \(a+z=a\) for all \(a \in \mathbb{R}\), then \(z=0\). (ii) 1 is the unique real number such that \(a \cdot 1=a\) for all \(a \in \mathbb{R}\). (iii) Given any \(a \in \mathbb{R}\), an element \(a^{\prime} \in \mathbb{R}\) such that \(a+a^{\prime}=0\) is unique. (As noted before, this unique real number \(a^{\prime}\) is denoted by \(-a .\).) (iv) Given any \(a \in \mathbb{R}\) with \(a \neq 0\), an element \(a^{*} \in \mathbb{R}\) such that \(a \cdot a^{*}=1\) is unique. (This unique real number \(a^{*}\) is denoted by \(a^{-1}\) or by \(1 / a .\) ) (v) Let \(a \in \mathbb{R}\). Then \(-(-a)=a .\) Further, if \(a \neq 0\), then \(\left(a^{-1}\right)^{-1}=a\). (vi) Let \(a, b \in \mathbb{R}\). Then \(a(-b)=-(a b)\) and \((-a)(-b)=a b\).

Let \(\\{0,1\\}^{\mathrm{N}}\) denote the set of all maps from \(\mathbb{N}\) to the two-element set \(\\{0,1\\}\). Prove that \(\\{0,1\\}^{\mathbb{N}}\) is uncountable. (Hint: If \(f: \mathbb{N} \rightarrow\\{0,1\\}^{\mathrm{N}}\) is any map and \(h \in\\{0,1\\}^{N}\) is the map that associates \(n \in \mathbb{N}\) to 1 or to 0 according as \(f(n)\) associates \(n\) to 0 or to 1, then \(h\) is not in the range of \(f .\) )

Let \(I_{n}=\left[a_{n}, b_{n}\right], n \in \mathbb{N}\), be closed and bounded intervals in \(\mathbb{R}\) such that \(I_{n} \supseteq I_{n+1}\) for each \(n \in \mathbb{N}\). If \(x=\sup \left\\{a_{n}: n \in \mathbb{N}\right\\}\) and \(y=\inf \left\\{b_{n}: n \in \mathbb{N}\right\\}\) then show that \(x \in I_{n}\) and \(y \in I_{n}\) for every \(n \in \mathbb{N}\).

Given a function \(f: \mathbb{R} \rightarrow \mathbb{R}\), prove the following: (i) If \(f(x y)=f(x)+f(y)\) for all \(x, y \in \mathbb{R}\), then \(f(x)=0\) for all \(x \in \mathbb{R}\). [Note: It is, however, possible that there are nonzero functions defined on subsets of \(\mathbb{R}\), such as \((0, \infty)\), that satisfy \(f(x y)=f(x)+f(y)\) for all \(x, y\) in the domain of \(f .\) A prominent example of this is the logarithmic function, which will be discussed in Section 7.1.] (ii) If \(f(x y)=f(x) f(y)\) for all \(x, y \in \mathbb{R}\), then either \(f(x)=0\) for all \(x \in \mathbb{R}\), or \(f(x)=1\) for all \(x \in \mathbb{R}\), or \(f(0)=0\) and \(f(1)=1\). Further, \(\bar{f}\) is either an even function or an odd function. Give examples of even as well as odd functions \(f: \mathbb{R} \rightarrow \mathbb{R}\) satisfying \(f(x y)=f(x) f(y)\) for all \(x, y \in \mathbb{R}\)
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