Chapter 6: Problem 9
If \(n\) is a square-free integer, prove that \(\tau(n)=2^{r}\), where \(r\) is the number of prime divisors of \(n\).
Short Answer
Expert verified
\(\tau(n) = 2^r\) because each prime factor increases divisors by one additional factor.
Step by step solution
01
Understand the Problem
We are given that \(n\) is a square-free integer, which means it has no repeated prime factors. Our goal is to show that the number of divisors of \(n\), denoted as \(\tau(n)\), is equal to \(2^r\), where \(r\) is the number of distinct prime divisors of \(n\).
02
Express \(n\) as a product of primes
Since \(n\) is square-free, its prime factorization can be expressed as \(n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}\) where \(a_i = 1\) for all \(i\) because \(n\) is square-free, and \(p_1, p_2, \ldots, p_k\) are distinct primes. Thus, \(n = p_1 p_2 \cdots p_k\).
03
Use the divisor formula
The formula for the number of divisors \(\tau(n)\) of a number \(n\) in terms of its prime factorization \(n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}\) is \(\tau(n) = (a_1 + 1)(a_2 + 1) \cdots (a_k + 1)\). Since for a square-free \(n\), each \(a_i = 1\), we have \(\tau(n) = (1+1)(1+1) \cdots (1+1)\) for \(k\) terms.
04
Simplify the expression
This simplifies to \(\tau(n) = 2 \times 2 \times \cdots \times 2 = 2^k\), where \(k\) is the number of distinct prime factors of \(n\). In this context, \(k\) equals \(r\), the number of prime divisors.
05
Conclusion
Since we have shown that \(\tau(n) = 2^k\) and \(k = r\), it follows that \(\tau(n) = 2^r\) for any square-free integer \(n\). Thus, the theorem is proved.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Square-Free Integer
A square-free integer is a fascinating concept in number theory. It refers to a number that is not divisible by the square of any prime number. In simpler terms, a square-free number has no repeated prime factors. For example, the number 30 is square-free because its prime factorization is \(2 \times 3 \times 5\), and none of these primes are repeated.
On the other hand, a number like 18 is not square-free because its prime factorization \(2 \times 3^2\) includes the repeated prime factor 3. Understanding square-free integers is crucial because it determines the structure of a number and influences how many divisors the number has.
On the other hand, a number like 18 is not square-free because its prime factorization \(2 \times 3^2\) includes the repeated prime factor 3. Understanding square-free integers is crucial because it determines the structure of a number and influences how many divisors the number has.
- Key Property: Each prime appears only once in the factorization.
- Examples: 10, 21, and 33 are square-free; 9, 12, and 18 are not.
Prime Factorization
Prime factorization is the process of expressing a number as a product of prime numbers. Every integer greater than 1 can be uniquely written as a product of primes. This is a foundational idea in number theory.
For square-free integers, the prime factorization is especially straightforward: each prime number is used only once. Suppose we have a number \(n\) that is square-free. Its prime factorization might look like \(n = p_1 \times p_2 \times \cdots \times p_k\). Here, each \(p_i\) is a distinct prime number, and each exponent is 1.
For square-free integers, the prime factorization is especially straightforward: each prime number is used only once. Suppose we have a number \(n\) that is square-free. Its prime factorization might look like \(n = p_1 \times p_2 \times \cdots \times p_k\). Here, each \(p_i\) is a distinct prime number, and each exponent is 1.
- This is significant because it simplifies the calculation of the number of divisors.
- It also helps in proving other properties related to divisors and their behaviors.
Number of Divisors
The number of divisors of a number \(n\), denoted \(\tau(n)\), is a crucial topic when studying integers and their properties. The formula for finding the number of divisors, based on the prime factorization of a number, is beautiful in its simplicity:
If \(n = p_1^{a_1} \times p_2^{a_2} \times \cdots \times p_k^{a_k}\), then \(\tau(n) = (a_1 + 1)\times (a_2 + 1) \times \cdots \times (a_k + 1)\). In square-free numbers, where each \(a_i = 1\), this simplifies nicely to \((1+1)(1+1) \cdots (1+1)\) for \(k\) terms.
This equals \(2^k\), where \(k\) is the number of distinct prime factors of \(n\). It shows that each prime factor gives rise to two possibilities: the factor itself and 1.
If \(n = p_1^{a_1} \times p_2^{a_2} \times \cdots \times p_k^{a_k}\), then \(\tau(n) = (a_1 + 1)\times (a_2 + 1) \times \cdots \times (a_k + 1)\). In square-free numbers, where each \(a_i = 1\), this simplifies nicely to \((1+1)(1+1) \cdots (1+1)\) for \(k\) terms.
This equals \(2^k\), where \(k\) is the number of distinct prime factors of \(n\). It shows that each prime factor gives rise to two possibilities: the factor itself and 1.
- This property helps in understanding why square-free numbers have a power of 2 as their divisor count.
- It also reflects the simple structure of square-free numbers' prime factorizations.
Mathematical Proof
In mathematics, proving statements like \(\tau(n) = 2^r\), where \(n\) is a square-free integer and \(r\) is the number of prime divisors, is pivotal. It involves clear logical steps.
We begin by understanding the definitions and properties. For a square-free number \(n = p_1 \times p_2 \times \cdots \times p_k\), primes are distinct, meaning no prime is repeated. The number of divisors \(\tau(n)\) then follows the formula \((1+1)(1+1) \cdots (1+1)\), simplifying to \(2^k\).
We begin by understanding the definitions and properties. For a square-free number \(n = p_1 \times p_2 \times \cdots \times p_k\), primes are distinct, meaning no prime is repeated. The number of divisors \(\tau(n)\) then follows the formula \((1+1)(1+1) \cdots (1+1)\), simplifying to \(2^k\).
- Knowing \(k=r\), the number of distinct primes, implies \(\tau(n) = 2^r\).
- This chain of logical reasoning supports the statement completely.