Problem 32 - Discrete Mathematics with Applications

Question

Let \(p, q,\) and \(r\) be prime numbers, and \(i, j,\) and \(k\) whole numbers. Find the sum of the positive divisors of each. $$p^{i} q^{j}$$

Step-by-Step Solution

Answer

The sum of the positive divisors of \(p^i q^j\) is given by the expression \(\sigma(p^i q^j) = (p^{i + 1} - 1)(q^{j + 1 }- 1)\).

Step 1: Understand the divisor function formula

To find the sum of the positive divisors of an integer, we can use the divisor function formula. For a given number of the form \(n = p_1^{e_1} p_2^{e_2} \dots p_k^{e_k}\), where \(p_1, p_2, \dots, p_k\) are prime factors and \(e_1, e_2, \dots, e_k\) are their respective exponents, we can find the sum using the following formula: $$ \sigma(n) = (\sigma_1^{e_1 + 1} - 1)(\sigma_2^{e_2 + 1} - 1)\dots(\sigma_k^{e_k + 1} - 1) $$ where \(\sigma_i\) denotes the ith prime factor. The sum of divisors of \(p^i q^j\) can be found using this formula.

Step 2: Apply the divisor function formula

Now, we can apply the formula to find the sum of the divisors of \(p^i q^j\). In this case, \(\sigma_1 = p\) and \(\sigma_2 = q\), while \(e_1 = i\) and \(e_2 = j\). Plugging these values into the formula yields: $$ \sigma(p^i q^j) = (p^{i + 1} - 1)(q^{j + 1} - 1) $$

Step 3: Write the final solution

The sum of the positive divisors of \(p^i q^j\) is given by the following expression: $$ \sigma(p^i q^j) = (p^{i + 1} - 1)(q^{j + 1 }- 1) $$

Key Concepts

Divisor Function

The divisor function, often denoted by \( \sigma(n) \), is a mathematical tool used to calculate the sum of all positive divisors of a given integer \( n \). This includes 1 and the number itself. Understanding the divisor function is crucial for students who study number theory or need to solve problems related to divisors.

When the number \( n \), is expressed as a product of its prime factors with their respective exponents, the formula for the divisor function looks like this:\[\sigma(n) = \prod_{k=1}^{m} \frac{p_k^{e_k + 1} - 1}{p_k - 1}\]Where \( p_k \) are the prime factors of \( n \) and \( e_k \) are their exponents. The product runs over all prime factors of the number. The beauty of this formula is that it simplifies the process of finding the sum of all divisors by breaking the problem into smaller parts related to each prime factor.

To illustrate, if we have \( p^i \) as part of our integer, the sum of divisors that include this prime factor to the power of \( i \) is \( (p^{i+1} - 1) / (p - 1) \) because we consider all powers of \( p \) from 0 to \( i \) as potential divisors. When combined for multiple prime factors, we multiply these sums for each to obtain the total sum of divisors.

Prime Factorization

Prime factorization is the process of breaking down a composite number into its constituent prime numbers, which, when multiplied together, give back the original number. It is a fundamental concept in understanding the nature of numbers and plays a key role in various areas of mathematics, including the calculation of divisors.

In the case of \( p^i q^j \) from our exercise, \( p \) and \( q \) are already prime numbers, and \( i \) and \( j \) are their exponents. To find the prime factorization of this expression, you write it as \( p \) multiplied by itself \( i \) times and \( q \) multiplied by itself \( j \) times. Thus, \( p^i q^j \) is already in its prime factored form.

Using prime factorization is critical when applying formulas such as the divisor function because you need to know the prime factors and their exponents to plug into the formula. Moreover, prime factorization aids in simplifying fractions, finding the greatest common divisors, and lowest common multiples, making it a versatile tool in problem-solving.

Exponents in Mathematics

Exponents in mathematics represent a shorthand way of expressing repeated multiplication of the same number. The exponent tells us how many times to multiply the base number by itself. For example, \( 3^4 \) means \( 3 \) multiplied by itself 4 times: \( 3 \times 3 \times 3 \times 3 \).

In the context of our exercise, we're dealing with prime numbers raised to certain exponents, such as \( p^i \) or \( q^j \). Understanding exponents is vital in calculating the sum of positive divisors because the exponents determine the potential divisors of the number. For instance, if we take the prime number \( p \) raised to the exponent \( i \) (i.e., \( p^i \) ), we can think of every exponent from 0 up to \( i \) as providing a new divisor for the number \( p^i \).

Being able to manipulate exponents efficiently is also important when simplifying expressions and solving equations, especially when working with the divisor function formula. Remember that in the formula, the exponent applied is one greater than the original exponent of the prime factor (\( i + 1 \) for \( p^i \) in our case), and understanding how to calculate powers quickly can save time and reduce errors in computations.