Problem 44 - Discrete Mathematics with Applications

Question

Evaluate each sum and product, where \(p\) is a prime and \(I=\\{1,2,3,5\\}.\) $$\begin{aligned} &\prod i^{j}\\\ &i, j \in I\\\ &i \leq j \end{aligned}$$

Step-by-Step Solution

Answer

The value of the expression \(\prod_{i, j \in I} i^{j}\), where \(I=\{1, 2, 3, 5\}\), is 36279705600000.

Step 1: List the Elements of I

We can start by listing the set of elements, which are: \(I = \{1, 2, 3, 5\}\).

Step 2: Write Out All Possible Combinations of i and j in I

Next, list all possible combinations of i and j in I: $$ 1^{1}, 1^{2}, 1^{3}, 1^{5}, 2^{1}, 2^{2}, 2^{3}, 2^{5}, 3^{1}, 3^{2}, 3^{3}, 3^{5}, 5^{1}, 5^{2}, 5^{3}, 5^{5} $$

Step 3: Calculate Each Element

Calculate each of the given combinations: $$ 1, 1, 1, 1, 2, 4, 8, 32, 3, 9, 27, 243, 5, 25, 125, 3125 $$

Step 4: Calculate the Product of All Elements

Finally, find the product of all the calculated terms: $$ \prod_{i, j \in I} i^{j} = 1 \cdot 1 \cdot 1 \cdot 1 \cdot 2 \cdot 4 \cdot 8 \cdot 32 \cdot 3 \cdot 9 \cdot 27 \cdot 243 \cdot 5 \cdot 25 \cdot 125 \cdot 3125 \\ = 36279705600000 $$ So the value of the given expression \(\prod_{i, j \in I} i^{j}\) is 36279705600000.

Key Concepts

Prime Numbers

Prime numbers are fundamental elements in the world of mathematics. They are numbers greater than 1 that have no divisors other than 1 and themselves. In other words, a prime number can only be divided evenly by 1 or by the number itself without leaving a remainder. This simplicity is what makes them a unique cornerstone for various mathematical concepts, including cryptography and number theory.
For example:

• The number 2 is a prime number because its only divisors are 1 and 2.
• The number 3 is also a prime number, as it can only be divided evenly by 1 and 3.
• Conversely, 4 is not a prime number because it can also be divided by 2 without a remainder.

In the context of determining products or sums involving prime numbers, understanding their properties ensures correct application and calculation in formulas or algorithms. Identifying primes within a set can help optimize calculations since their unique divisors streamline multiplicative processes.

Exponents

Exponents are a shorthand way to express repeated multiplication of the same number. When you see an expression like \(a^b\), it means you multiply \(a\) by itself \(b\) times. This concept is crucial for simplifying large calculations and expressing numbers compactly.
Here’s how exponents work:

• \(2^3\) means \(2 \times 2 \times 2\), which equals 8.
• \(5^2\) is \(5 \times 5\), giving us 25.
• The base is the number being multiplied, and the exponent indicates how many times the multiplication is repeated.

Using exponents allows us to manage large numbers quickly. For instance, instead of writing \(5 \times 5 \times 5 \times 5 \times 5\) for \(5^5\), we neatly express it using an exponent, simplifying both writing and calculation.
In our exercise, calculating terms like \(i^j\) with values from set \(I\) involves using exponents to resolve and compute each term efficiently.

Product Notation

Product notation, symbolized by the Greek letter \(\prod\), is a systematic way to express the product of a sequence of numbers. Just like summation notation (\(\Sigma\)) is used for summing numbers, product notation provides a compact method for multiplying a series of factors.
How it works:

• \(\prod_{i=1}^{n} a_i\) signifies the product of all elements \(a_i\) for each \(i\) from 1 to \(n\).
• Each term inside the product is multiplied together consecutively.
• It is widely used in calculus, algebra, and other areas of mathematics to simplify expressions involving multiple multiplication processes.

Using the product notation in mathematical operations can vastly improve clarity and efficiency, especially when handling sequences involving many terms.
In the problem provided, we use product notation to evaluate \(\prod_{i,j \in I} i^j\), allowing us to compute the multiplication of all possible combinations of elements in the set \(I\). This significantly reduces the complexity of the task by organizing calculations into a straightforward notation, leading to the final product of 36,279,705,600,000.