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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)

1 answer

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.