Question
Jonathan says that when you add 7,2/3
to 2,2/3
, you must add the whole numbers together first ( 7+2=9
). Then you just add the numerators together ( 2+2=4
), and you add the denominators together ( 3+3=6
). So the answer is 9,4/6
.
Unfortunately, Jonathan's strategy is incorrect. Solve this problem, and then briefly explain the flaw in Jonathan's reasoning.
(4 points)
to 2,2/3
, you must add the whole numbers together first ( 7+2=9
). Then you just add the numerators together ( 2+2=4
), and you add the denominators together ( 3+3=6
). So the answer is 9,4/6
.
Unfortunately, Jonathan's strategy is incorrect. Solve this problem, and then briefly explain the flaw in Jonathan's reasoning.
(4 points)
Answers
GPT-4o mini
Let's solve the problem of adding \(7 \frac{2}{3}\) and \(2 \frac{2}{3}\) correctly.
1. **Convert mixed numbers to improper fractions**:
- For \(7 \frac{2}{3}\):
\[
7 \frac{2}{3} = \frac{7 \times 3 + 2}{3} = \frac{21 + 2}{3} = \frac{23}{3}
\]
- For \(2 \frac{2}{3}\):
\[
2 \frac{2}{3} = \frac{2 \times 3 + 2}{3} = \frac{6 + 2}{3} = \frac{8}{3}
\]
2. **Add the improper fractions**:
\[
\frac{23}{3} + \frac{8}{3} = \frac{23 + 8}{3} = \frac{31}{3}
\]
3. **Convert back to a mixed number** (if needed):
\[
\frac{31}{3} = 10 \frac{1}{3} \quad \text{(since \(31 \div 3 = 10\) remainder \(1\))}
\]
So, \(7 \frac{2}{3} + 2 \frac{2}{3} = 10 \frac{1}{3}\).
### Flaw in Jonathan's Reasoning:
Jonathan's reasoning contains a fundamental misunderstanding of how to operate with fractions. When he adds the fractions, he misapplies the operations on the numerators and denominators.
1. **Adding Whole Numbers**: While it is correct to add the whole numbers together (7 and 2), the way he handles the fractions is incorrect.
2. **Adding Numerators and Denominators**: You cannot simply add the numerators (2 + 2) and the denominators (3 + 3) together as if they were separate fractions. Instead, you must first have a common denominator to perform the addition of fractions correctly.
3. **Final Result**: This leads to an invalid answer of \(9 \frac{4}{6}\). The correct solution follows from combining like terms with a common denominator, which gives us a total of \(10 \frac{1}{3}\).
In summary, the flaw lies in treating fractions as if their parts can be added independently rather than as components of a single collective fraction.
1. **Convert mixed numbers to improper fractions**:
- For \(7 \frac{2}{3}\):
\[
7 \frac{2}{3} = \frac{7 \times 3 + 2}{3} = \frac{21 + 2}{3} = \frac{23}{3}
\]
- For \(2 \frac{2}{3}\):
\[
2 \frac{2}{3} = \frac{2 \times 3 + 2}{3} = \frac{6 + 2}{3} = \frac{8}{3}
\]
2. **Add the improper fractions**:
\[
\frac{23}{3} + \frac{8}{3} = \frac{23 + 8}{3} = \frac{31}{3}
\]
3. **Convert back to a mixed number** (if needed):
\[
\frac{31}{3} = 10 \frac{1}{3} \quad \text{(since \(31 \div 3 = 10\) remainder \(1\))}
\]
So, \(7 \frac{2}{3} + 2 \frac{2}{3} = 10 \frac{1}{3}\).
### Flaw in Jonathan's Reasoning:
Jonathan's reasoning contains a fundamental misunderstanding of how to operate with fractions. When he adds the fractions, he misapplies the operations on the numerators and denominators.
1. **Adding Whole Numbers**: While it is correct to add the whole numbers together (7 and 2), the way he handles the fractions is incorrect.
2. **Adding Numerators and Denominators**: You cannot simply add the numerators (2 + 2) and the denominators (3 + 3) together as if they were separate fractions. Instead, you must first have a common denominator to perform the addition of fractions correctly.
3. **Final Result**: This leads to an invalid answer of \(9 \frac{4}{6}\). The correct solution follows from combining like terms with a common denominator, which gives us a total of \(10 \frac{1}{3}\).
In summary, the flaw lies in treating fractions as if their parts can be added independently rather than as components of a single collective fraction.