Consider the set of all fractions equal to $$ \frac{1}{2} $$. If you take any 10 of those fractions, add their numerators to obtain the numerator of a new fraction and add their denominators to obtain the denominator of a new fraction. How does the new fraction relate to $$ \frac{1}{2} $$ ? Generalize what was found and explain. Choose the correct generalization below. The sum of the mumerators gives $$ a+r_{1}+a \cdot r_{2}+\ldots+a \cdot r_{n}=a\left(r_{1}+r_{2}+\ldots+r_{n}\right) $$. Adding the denominators gives A. $$ b \cdot r_{1}+b \cdot r_{2}+\ldots+b \cdot r_{n}=b\left(r_{1}+r_{2}+\ldots+r_{n}\right) $$. The new fraction is $$ \frac{a\left(r_{1}+r_{2}+\ldots+r_{n}\right)}{b\left(r_{1}+r_{2}+\ldots+r_{n}\right)} $$ and it is equal to $$ \frac{a}{b} $$. The sum of the numerators gives $$ a \cdot r_{1}+a \cdot r_{2}+\ldots+a \cdot r_{n}=a\left(r_{1}+r_{2}+\ldots+r_{n}\right) $$. Adding the denominators gives B. $$ b+r_{1}+b \cdot r_{2}+\ldots+b \cdot r_{n}=b\left(r_{1}+r_{2}+\ldots+r_{n}\right) $$. The new fraction is $$ \frac{a\left(r_{1}+r_{2}+\ldots+r_{n}\right)}{b\left(r_{1}+r_{2}+\ldots+r_{n}\right)} $$ and it is leess than $$ \frac{a}{b} $$ The sum of the numerators gives $$ a \cdot r_{1}+a \cdot r_{2}+\ldots+a \cdot r_{n}=a\left(r_{1}+r_{2}+\ldots+r_{n}\right) $$. Adding the denominators gives C. $$ b \cdot r_{1}+b * r_{2}+\ldots+b * r_{n}=b\left(r_{1}+r_{2}+\ldots+r_{n}\right) $$. The new fraction is $$ \frac{a\left(r_{1}+r_{2}+\ldots+r_{n}\right)}{b\left(r_{1}+r_{2}+\ldots+r_{n}\right)} $$ and it is greater then $$ \frac{a}{b} $$

Question

UpStudy AI · Accepted Answer

The problem involves fractions that are all equal to $$ \frac{1}{2} $$. Let's break it down step by step and then examine the generalizations provided.

### Understanding the Problem

1. **Set of Fractions:** All fractions are equal to $$ \frac{1}{2} $$, meaning each fraction can be written as $$ \frac{a}{b} = \frac{1}{2} $$. This implies that $$ a = \frac{b}{2} $$.

2. **Summing Numerators and Denominators:**
   - If you take any $$ n $$ fractions $$ \frac{a}{b} $$, their numerators sum to $$ n 	imes a $$.
   - Their denominators sum to $$ n 	imes b $$.

3. **New Fraction Formation:**
   - The new fraction formed by these sums is $$ \frac{n 	imes a}{n 	imes b} = \frac{a}{b} $$.

Since $$ \frac{a}{b} = \frac{1}{2} $$, the new fraction will also equal $$ \frac{1}{2} $$.

### Evaluating the Generalizations

Let's analyze each option provided:

**Option A:**
- **Numerator Sum:** $$ a \cdot r_1 + a \cdot r_2 + \ldots + a \cdot r_n = a(r_1 + r_2 + \ldots + r_n) $$.
- **Denominator Sum:** $$ b \cdot r_1 + b \cdot r_2 + \ldots + b \cdot r_n = b(r_1 + r_2 + \ldots + r_n) $$.
- **New Fraction:** $$ \frac{a(r_1 + r_2 + \ldots + r_n)}{b(r_1 + r_2 + \ldots + r_n)} = \frac{a}{b} $$.

**Option B:**
- **Denominator Sum Error:** It incorrectly states the denominator as $$ b + r_1 + b \cdot r_2 + \ldots + b \cdot r_n $$, which doesn't align with how denominators should be summed.

**Option C:**
- **Incorrect Inequality:** It suggests the new fraction is greater than $$ \frac{a}{b} $$, which contradicts the calculation.

### Conclusion

**Option A** correctly generalizes the process, showing that the new fraction remains equal to $$ \frac{a}{b} $$, which in this specific problem is $$ \frac{1}{2} $$.

**Answer:**  
**A.** The sum of the numerators gives $$ a\left(r_{1}+r_{2}+\ldots+r_{n}ight) $$. Adding the denominators gives  
$$ b\left(r_{1}+r_{2}+\ldots+r_{n}ight) $$. The new fraction is $$ \frac{a\left(r_{1}+r_{2}+\ldots+r_{n}ight)}{b\left(r_{1}+r_{2}+\ldots+r_{n}ight)} $$ and it is equal to $$ \frac{a}{b} $$.
The new fraction remains equal to $$ \frac{1}{2} $$.

Upstudy AI Solution

Answer

Solution

Beyond the Answer

Related Questions

Latest Algebra Questions