Addition of Fractions Misconception
Which of the following is equal to the sum \(\frac {4}{11} + \frac {3}{8}?\) \[\]
\[\begin{array} &(A)~ \frac {4+3}{11+8} = \frac {7}{19} &&&(B)~ \frac {4 \times 8 + 3 \times 11}{11 \times 8} = \frac {65}{88} &&&\end{array}\]
(B) is the correct answer.
Proof:
Remember that fraction is just another way to show division. The expression in the question can be rewritten as \[\dfrac {4}{11}+\dfrac{3}{8} = 4 \div 11 + 3 \div 8.\] By Order of Operations we must do division first, so we cannot add the numerators (4 and 3) and the denominators (11 and 8) to get the answer. There is an easy way to solve the problem at hand. Note that \(\frac {a}{b} = \frac {ac}{bc}\). Now, we want to convert both fractions so they both have like (equal) denominators, as per the method shown in this wiki. Since the product of two numbers is divisible by both of those numbers, we have \[\dfrac {4}{11} + \dfrac {3}{8} = \dfrac {4 \times 8 }{11 \times 8}+\dfrac {3 \times 11 }{11 \times 8}.\] Simplifying this expression gives \[\dfrac {32}{88} + \dfrac {33}{88}.\] Now we can add the numerators together (never add the denominators), thus getting \(\dfrac {65}{88}.\) Therefore, the answer is \((B).\) \(_\square\)
Rebuttal: But adding is just going across the page! Same applies here!
Reply: No, remember that fractions are division, which means we must follow order of operations. \(_\square\)