Dan observes that $$\frac{6}{10}\div \frac{2}{10} = 6 \div 2$$ He says,
I think that if we are dividing a fraction by a fraction with the same denominator, then we can just divide the numerators.Is Dan’s conjecture true for all fractions? Explain how you know.
Yes, Dan's rule is correct for all fractions.
Explaining with analogies: One way of explaining why the rule is correct is to bear in mind that \(6\over 10\) refers to six "items", where the "item" is \(1\over 10\). $${6\over 10}=6\,\Bigl({1\over 10}\Bigr),$$ so the original division problem can be rephrased as $${6\,\Bigl({1\over 10}\Bigr)}\div{2\,\Bigl({1\over 10}\Bigr)}.$$ The computation $${6\,\Bigl({1\over 10}\Bigr)}\div{2\,\Bigl({1\over 10}\Bigr)} = 6\div 2 =3$$ follows the same logic as: $$6\,{\rm apples}\div 2\,{\rm apples} = 6\div 2 = 3.$$ We would probably not write it as \(6\,{\rm apples}\div 2\,{\rm apples}\); instead we might have a situation where we have 6 apples and want to know how many groups of 2 apples we can make. Since we can make 3 groups with 2 apples in each group, the answer is 3. There was nothing special about the apples, the same reasoning applies to any "objects":
How many groups of 2 peaches can I make if I have 6 peaches?
How many groups of $2 dollars can I make if I have $6?
How many groups of 2 tenths can I make if I have 6 tenths?
When we divide a quantity consisting of \(m\) units divided into groups of size \(n\) of the same units, then the result does not depend on what the units are. The answer is found by dividing the number \(m\) by the number \(n\).
In summary, Dan's rule is true not only for dividing fractions with the same denominator, but also for any division of one quantity (number of units) by another quantity with the same units. In all cases, we find the answer by dividing the numbers, and the kind of unit does not matter.
Explaining with symbols: Dan might have made his conjecture based on using the "invert and multiply" rule: $${6\over 10}\div{2\over 10} = {6\over 10}\times{10\over 2}= {6\times 10\over 2\times10} = {6\over 2} = 6\div 2.$$ This works just as well for any denominator \(d\not=0\). If \(m\) and \(n\) are any integers and \(n\not=0\), then: $${m\over d}\div{n\over d} = {m\over d}\times{d\over n} = {m\times d\over n\times d} = {m\over n} = m\div n.$$
Submitted by J. Madden. This solution was developed by the 12 middle- and secondary-school teachers in the "LaMSTI On-Ramp Course" at LSU, with major contributions from N. Revaula.
Commentary
The purpose of this task is to help students explore the meaning of fraction division and to connect it to what they know about whole-number division.