In response, Ms. Lawrence poses another problem, this time involving unlike denominators: “How would we find the sum of these two?” she asks. Stepping back, she gives the students a chance to think. She then asks whether the sum would be less than or greater than 1. Several students raised their hands, eager to respond. Ms. Lawrence calls on Susan, who explains that the sum would be less than 1 because is less than and equals exactly 1.
Ms. Lawrence then asks how you could find the exact sum. Jim raises his hand and offers and as equivalent fractions with a common denominator. Ms. Lawrence writes on the chalkboard as Jim dictates:
She asks Jim why he chose 12 as the common denominator. “Twelve is the smallest number that both 3 and 4 go into,” replies Jim. “How did you come up with that?” Ms. Lawrence asks. “By multiplying 3 and 4,” he answers.
Ms. Lawrence turns to the class. “Let’s take a closer look. Jim got the equivalent fractions by multiplying the numerator and denominator of each fraction by the denominator of the other fraction. So if we show all the steps, it looks like this.” She then reworks the problem to make her point, justifying each step by giving a property of the rational numbers:
Ms. Lawrence stops and looks at the students. “How do we know that what Jim did makes sense? How do we know that he is adding the same fractions as in the original problem: and This is really important. Maybe he has just added two other fractions.”