Question

Prove that the power set \(\mathrm{P}(\mathrm{A})\) of any set \(\mathrm{A}\) of \(\mathrm{n}\) elements contains exactly \(2^{\mathrm{n}}\) elements.

Short answer

For a set A with n elements, the total number of subsets in the power set P(A) is equal to the sum of all possible combinations with different k values, which can be represented as \( \sum_{k=0}^{n} C(n, k) \). According to the binomial theorem, this sum is equal to \( 2^n \). Therefore, the power set P(A) of any set A of n elements contains exactly 2^n elements.

Step-by-step solution

Understanding the power set and binomial coefficients

The power set P(A) is the set of all subsets of a given set A. If A has n elements, then the elements in P(A) are all possible subsets of A with zero elements, one element, two elements, ..., and n elements. For example, let A = {1, 2, 3} with n=3 elements, the power set P(A) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}. To count the number of subsets of A with a specific number of elements, we can use binomial coefficients, which are denoted by \( C(n, k) \) (also written as \( \binom{n}{k} \)). \( C(n, k) \) represents the number of ways to choose k elements from a set of n elements without considering the order of those elements.

Counting the number of subsets with k elements

For a given set A with n elements, the number of subsets with k elements can be calculated using the binomial coefficient as \( C(n, k) \) = \( \binom{n}{k} \) = \( \frac{n!}{k!(n-k)!} \), where k ranges from 0 to n. Now, let's count the total number of subsets in P(A), considering subsets with any number of elements (from 0 to n), which is the sum of all possible combinations with different k values: Number of subsets in P(A) = \( \sum_{k=0}^{n} C(n, k) \)

Understanding the binomial theorem

The binomial theorem states that for any non-negative integer n and any real numbers a and b, \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \] Setting a = b = 1, the binomial theorem simplifies to: \[ 2^n = \sum_{k=0}^{n} \binom{n}{k} \]

Connecting the counting of subsets to the binomial theorem

Comparing the expression obtained in Step 2 and the simplified binomial theorem in Step 3, we can see that they are exactly the same: \[ 2^n = \sum_{k=0}^{n} C(n, k) \] This means that for a set A with n elements, the total number of subsets in the power set P(A) is equal to 2^n. Therefore, the power set P(A) of any set A of n elements contains exactly 2^n elements.

Key concepts

Binomial Coefficients

Binomial coefficients are a fundamental concept in combinatorics and play a crucial role in counting subsets. If you have a set with \( n \) elements, binomial coefficients help determine how many ways you can select \( k \) items from that set. This is denoted as \( C(n, k) \) or \( \binom{n}{k} \). It is calculated using the formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] Here, \( n! \) indicates factorial, which is the product of all positive integers up to \( n \). Understanding binomial coefficients allows you to comprehend how many combinations of elements exist in a subset of a certain size. For instance, selecting 2 elements from a set of 3 elements is computed as \( \binom{3}{2} = 3 \), indicating there are three different subsets of size 2.

Subset Counting

Counting subsets is a key activity when working with power sets. A power set \( P(A) \) of a set \( A \) includes every possible subset of \( A \) from the empty set to the full set itself. To find the total number of subsets, we sum the binomial coefficients for all possible subset sizes, i.e., \( k \), ranging from 0 to \( n \). This sum is expressed as: \[ \sum_{k=0}^{n} \binom{n}{k} \] Here's a quick breakdown:

• \( \binom{n}{0} \): Number of subsets with 0 elements (just the empty set).
• \( \binom{n}{1} \): Number of subsets with 1 element.
• Continue adding these up to \( \binom{n}{n} \), which is just the set itself.

This process demonstrates that all possible ways to select elements from your set \( A \) are accounted for.

Binomial Theorem

The binomial theorem is a powerful equation involving binomial coefficients: \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k \] When working with power sets, we simplify this expression by setting \( a = 1 \) and \( b = 1 \). This changes the expression to: \[ 2^n = \sum_{k=0}^{n} \binom{n}{k} \] This is fascinating because it connects directly to our previous exercise of counting subsets. It tells us that the total number of subsets (the size of the power set) of a set with \( n \) elements is exactly \( 2^n \). This explains why power sets are termed power sets—they relate directly to powers of 2.

Combinatorics

Combinatorics is the mathematical study of counting and arrangement possibilities. Within this field, many concepts like permutations, combinations, and subsets explore how elements can be combined. In our context, the problem of determining the size of the power set is a classic combinatorial exercise. If we grasp how to utilize binomial coefficients and the binomial theorem, we unlock the ability to count intricate element combinations. Combinatorics provides tools and methods to methodically and efficiently count sets, ensuring we're not just guessing but using mathematical certainty to achieve accurate results. This makes it not only a chapter in mathematics but an essential tool in problem-solving across various fields.