Because $10 \div \dfrac{4}{3}$ isn't an integer, [I changed the numbers in this Reddit post.](https://old.reddit.com/r/learnmath/comments/ckv75r/why_multiplying_by_the_reciprocal_works_while/evurf0g/) My $X = 6, A = 2, B = 3$. Undoubtedly, I know $6 \div \dfrac{2}{3} = 6 \times \dfrac{3}{2} = 9$, but don't use $\dfrac{a}{b} \div \dfrac{c}{d} \equiv \dfrac{a}{b} \times \dfrac{d}{c}$ here to explain.
> But you could replace 6, 3, and 4 with any numbers X, A, and B. If you do that, the same sort of logic holds. $X \times \dfrac{A}{B}$ means taking a full group of X items and splitting it apart into B equal parts but only taking A of these parts as your final amount that you have. **Meanwhile, $X \div \dfrac{A}{B}$ means starting out with a total of X items, splitting it up into B equal "partial groups" (where a full group is actually A of these partial groups), and then rounding up A partial groups to get a full group.** Again, the actions you do are the same in both cases - you start with X, split it into B equal groups, and take A of these groups as your final answer.
I don't understand the bolded sentence. Can you please picture all this?
1. What exactly are my B (= 3) equal "partial groups" here?
2. What exactly is my full group that's "a full group is actually A [= 2] of these partial groups"?
3. How do I round "up A [= 2] partial groups to get a full group"?
Chgg Clou
·
2021-02-18T06:38:08Z (almost 4 years ago)