Problem 81
Question
If \(m=\) number of distinct rational numbers \(\frac{p}{q} \in(0,1)\) such that \(p, q \in\\{1,2,3,4,5\\}\) and \(n=\) number of mappings from \(\\{1,2,3\\}\) onto \(\\{1,2\\}\), then \(m-n\) is (A) 1 (B) \(-1\) (C) 0 (D) None of these
Step-by-Step Solution
Verified Answer
The answer is (D) None of these, because \( m-n = 2 \).
1Step 1: Identify Rational Numbers
First, we need to find the number of distinct rational numbers in the form \( \frac{p}{q} \) where \( p \) and \( q \) belong to the set \( \{1,2,3,4,5\} \) and the condition \( 0 < \frac{p}{q} < 1 \) holds. This means \( p < q \).
2Step 2: List Valid Fractions
List the possible fractions: When \( q = 2 \), \( p \) can be 1. When \( q = 3 \), \( p \) can be 1, 2. When \( q = 4 \), \( p \) can be 1, 2, 3. When \( q = 5 \), \( p \) can be 1, 2, 3, 4. This results in the set of fractions \( \{ \frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5} \} \). After simplifying, the distinct fractions are \( \{ \frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5} \} \).
3Step 3: Count Distinct Fractions
These distinct simplified fractions are \( \frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5} \). So, there are 8 distinct fractions. Therefore, \( m = 8 \).
4Step 4: Calculate Onto Mappings
Next, calculate the number of onto mappings from \( \{1,2,3\} \) to \( \{1,2\} \). An onto function from \( A \) to \( B \) means every element of \( B \) is mapped to by at least one element from \( A \). For this, use the formula \( n^k - \binom{n}{1}(n-1)^k \), where \( n \) is the cardinality of \( B \) and \( k \) is the cardinality of \( A \). For our sets, \( k = 3 \) and \( n = 2 \), giving \( 2^3 - \binom{2}{1}(1)^3 = 8 - 2 = 6 \). Thus, \( n = 6 \).
5Step 5: Compute m-n
Now compute \( m - n \). We have \( m = 8 \) and \( n = 6 \), so \( m - n = 8 - 6 = 2 \).
Key Concepts
Distinct FractionsOnto MappingsSet Theory
Distinct Fractions
Distinct fractions refer to unique fractions that cannot be simplified to the same value as any other fraction in a given set. In the context of the problem we're looking at, we seek to identify fractions of the form \( \frac{p}{q} \) where \( p \) and \( q \) are integers chosen from a specific set, such as \( \{1,2,3,4,5\} \). These fractions must meet specific criteria to be considered 'distinct.' For example:
- The condition \( 0 < \frac{p}{q} < 1 \) implies that \( p \) must be less than \( q \), which helps limit the possible values for \( p \) and \( q \) in constructing these fractions.
- After generating potential fractions, simplification is key. Simplifying fractions such as \( \frac{2}{4} = \frac{1}{2} \) is crucial to ensuring each fraction in the list is distinct.
Onto Mappings
Onto mappings, or surjective functions, are a type of function where every element in the target set (codomain) is mapped to by at least one element in the source set (domain). The task of determining the number of onto mappings between two sets involves understanding combinations and assignments between elements:
- An onto map from set \( A \) to set \( B \) requires that for every \( b \in B \), there exists at least one \( a \in A \) such that the map \( f: A \to B \) satisfies \( f(a) = b \).
- To calculate the number of such mappings from a set \( \{1,2,3\} \) (3 elements) onto a set \( \{1,2\} \) (2 elements), we utilize the formula for the number of onto functions: \[ n^k - \binom{n}{1}(n-1)^k \], where \( n \) is the size of the codomain, and \( k \) is the size of the domain.
Set Theory
Set theory underpins the mathematical framework of the problem by defining and utilizing sets and their elements to solve problems effectively. Key aspects include:
- Understanding set notation, such as \( \{1,2,3,4,5\} \), which denotes a collection of distinct elements. Elements are important as they are the building blocks for identifying all possible fractions and for determining mappings.
- Applying set operations — the difference between a function being onto or not relates back to how elements are being utilized across two sets (the function's domain and codomain).
- Utilizing combinations and permutations when considering how multiple mappings can occur. Without the structure that set theory provides, calculating the function mappings between two sets would be a complex task.
Other exercises in this chapter
Problem 79
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Given 5 different green dyes, 4 different blue dyes and 3 different red dyes, the number of combinations of dyes that can be chosen by taking at least one green
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