Find a counterexample to show that the following statement is false. Adding the same number to both the numerator and the denominator (top and bottom) of a fraction does not change the fraction's value. Choose the correct answer below. $$ \frac{1+0}{2+0}=\frac{1}{2} $$; the fraction $$ \frac{1}{2} $$ is equal to $$ \frac{1}{2} $$. $$ \frac{3+4}{3+4}=\frac{7}{7} $$; the fraction $$ \frac{7}{7} $$ is not equal to $$ \frac{3}{3} $$. $$ \frac{1 \cdot 3}{2+1}=\frac{2}{3} $$; the fraction $$ \frac{2}{3} $$ is not equal to $$ \frac{1}{2} $$; the fraction $$ \frac{3}{6} $$ is equal to $$ \frac{1}{2} $$.

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UpStudy AI · Accepted Answer

To demonstrate that adding the same number to both the numerator and the denominator of a fraction can **change** the fraction's value, let's examine the provided options:

1. **Option 1:**
   $$
   \frac{1+0}{2+0} = \frac{1}{2}
   $$
   - **Analysis:** Here, adding 0 to both the numerator and the denominator leaves the fraction unchanged:
     $$
     \frac{1}{2} = \frac{1}{2}
     $$
   - **Conclusion:** This does **not** serve as a counterexample since the fraction's value remains the same.

2. **Option 2:**
   $$
   \frac{3+4}{3+4} = \frac{7}{7} = 1
   $$
   - **Analysis:** Initially, consider the original fraction before adding:
     $$
     \frac{3}{3} = 1
     $$
     After adding 4 to both:
     $$
     \frac{7}{7} = 1
     $$
   - **Conclusion:** The fraction's value remains unchanged. Additionally, the statement claims $$\frac{7}{7}$$ is not equal to $$\frac{3}{3}$$, which is incorrect since both equal 1. Thus, this is **not** a valid counterexample.

3. **Option 3:**
   $$
   \frac{1 + 1}{2 + 1} = \frac{2}{3}
   $$
   - **Analysis:** Start with the original fraction:
     $$
     \frac{1}{2}
     $$
     After adding 1 to both numerator and denominator:
     $$
     \frac{2}{3}
     $$
     Comparing the two:
     $$
     \frac{2}{3} \neq \frac{1}{2}
     $$
     Additionally, note that:
     $$
     \frac{3}{6} = \frac{1}{2}
     $$
   - **Conclusion:** Adding the same number (in this case, 1) to both the numerator and the denominator **changes** the value of the fraction from $$\frac{1}{2}$$ to $$\frac{2}{3}$$. This serves as a valid **counterexample** to the original statement.

**Final Answer:**

**Option 3:**  
$$
\frac{1+1}{2+1} = \frac{2}{3}
$$
The fraction $$\frac{2}{3}$$ is not equal to $$\frac{1}{2}$$; thus, adding the same number changes the fraction's value.
Adding the same number to both the numerator and denominator changes the fraction's value. For example, $$\frac{1+1}{2+1} = \frac{2}{3}$$, which is not equal to $$\frac{1}{2}$$.

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