0

While doing math the other day, I ran across this strange equation: $$\frac{20}{71} \cdot \frac{5}{4} = \frac{20 + 5}{71}$$

At first I tried to represent it as $$\frac ab \cdot \frac{\frac ad}{d} = \frac{a + \frac ad}{b}$$ But this turned out to not hold true for all $a$ or $d$ (though it did for all $b$), and it also turned out to not be algebraically equivalent anyway.

My partner did point out the following explanation: $$\frac{20}{71} \cdot \frac{5}{4} = \frac{20 \cdot 5}{71 \cdot 4} = \frac{5 \cdot 5}{71} = \frac{25}{71}$$ However, if possible I'd like to generalize to some relationship of the structure $$\frac ab \cdot \frac cd = \frac{a + c}{b}$$

What's going on here?

Drake P
  • 119

1 Answers1

1

Looking at the cancellation you did on the third line, it's clear to see that the equality on the fourth line will always hold so long as $$ca=d(c+a)$$ This equation always has the integer solution $a=c=2d$, so you can vary $d\in\mathbb{Z}$ to generate solutions to the expression, such as $d=6$ yielding: $$\frac{12}{b}\cdot\frac{12}{6} = \frac{12}{b}\cdot 2 = \frac{12}{b}+\frac{12}{b}=\frac{12 + 12}{b}=\frac{24}{b}$$ Moreover, we may want to ask: if we choose any $a,c\in\mathbb{Z}$, can we always find a satisfying $d\in\mathbb{Z}$? I can hypothesize that we can, but I'll come back to rigorously proving it later. Maybe consider giving it a try on your own. Because we're seeking integer answers (I imagine), I recommend taking a modular algebra approach to solving $ca=d(c+a)$ for $d$.

dsillman2000
  • 568