Problem 13 Use Mathematical Induction to prove that if a set S has n elements with n > 2, then has n(n-1) subsets containing exactly two elements.
Problem 13 Use Mathematical Induction to prove that if a set S has n elements with n > 2, then has n(n-1) subsets containing exactly two elements.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Problem 13
**Objective:** Use Mathematical Induction to prove that if a set \( S \) has \( n \) elements with \( n > 2 \), then it has \( \frac{n(n-1)}{2} \) subsets containing exactly two elements.
**Explanation of the Problem:**
This problem aims to demonstrate the principle of mathematical induction to prove a combinatorial identity. Specifically, you need to show that for any set \( S \) with \( n \) elements, the number of 2-element subsets (also known as 2-combinations) is given by the formula \( \frac{n(n-1)}{2} \).
### Steps for the Proof Using Mathematical Induction:
1. **Base Case:**
Prove that the formula holds for the initial value \( n = 2 \):
- For \( n = 2 \), the set \( S \) can be written as \( \{a, b\} \).
- The number of 2-element subsets of \( S \) is exactly 1, which can be calculated as follows:
\[
\frac{2(2-1)}{2} = \frac{2 \times 1}{2} = 1
\]
- Thus, the base case holds true.
2. **Inductive Step:**
Assume that the formula holds for some integer \( k \geq 2 \), i.e., suppose that a set with \( k \) elements has \( \frac{k(k-1)}{2} \) subsets of 2 elements.
- Now, consider a set with \( k + 1 \) elements. Let this set be \( S = \{a_1, a_2, \dots, a_k, a_{k+1}\} \).
- The number of 2-element subsets in this case can be divided into two groups:
1. Subsets that do not include \( a_{k+1} \). According to the inductive hypothesis, the number of such subsets is \( \frac{k(k-1)}{2} \).
2. Subsets that include \( a_{k+1} \). Each of the \( k \) other elements can pair with \( a_{k+1} \), contributing an additional \( k \) subsets.
- Therefore, the total](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe872339c-719c-4a81-bc7c-37653a99cb41%2Fae97aad7-c24f-4bb4-97b6-f2d4d7b0cd73%2Fnvw5swe_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 13
**Objective:** Use Mathematical Induction to prove that if a set \( S \) has \( n \) elements with \( n > 2 \), then it has \( \frac{n(n-1)}{2} \) subsets containing exactly two elements.
**Explanation of the Problem:**
This problem aims to demonstrate the principle of mathematical induction to prove a combinatorial identity. Specifically, you need to show that for any set \( S \) with \( n \) elements, the number of 2-element subsets (also known as 2-combinations) is given by the formula \( \frac{n(n-1)}{2} \).
### Steps for the Proof Using Mathematical Induction:
1. **Base Case:**
Prove that the formula holds for the initial value \( n = 2 \):
- For \( n = 2 \), the set \( S \) can be written as \( \{a, b\} \).
- The number of 2-element subsets of \( S \) is exactly 1, which can be calculated as follows:
\[
\frac{2(2-1)}{2} = \frac{2 \times 1}{2} = 1
\]
- Thus, the base case holds true.
2. **Inductive Step:**
Assume that the formula holds for some integer \( k \geq 2 \), i.e., suppose that a set with \( k \) elements has \( \frac{k(k-1)}{2} \) subsets of 2 elements.
- Now, consider a set with \( k + 1 \) elements. Let this set be \( S = \{a_1, a_2, \dots, a_k, a_{k+1}\} \).
- The number of 2-element subsets in this case can be divided into two groups:
1. Subsets that do not include \( a_{k+1} \). According to the inductive hypothesis, the number of such subsets is \( \frac{k(k-1)}{2} \).
2. Subsets that include \( a_{k+1} \). Each of the \( k \) other elements can pair with \( a_{k+1} \), contributing an additional \( k \) subsets.
- Therefore, the total
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