False proof that $\frac{13}{6}=0$

Question

At start, the length of a line segment is $a_0=0$. When $3$ hours have elapsed since start, its length is $a_3$. When $1$ hour has elapsed since start, its length increased by $\frac{a_3}{2}$ with respect to $a_0$ (call the new length $a_1$). When $2$ hours have elapsed since start, its length increased by $\frac{a_3}{3}$ (call the new length $a_2$) with respect to $a_1$. When $3$ hours have elapsed since start, its length increased by $\frac{a_3}{4}$ with respect to $a_2$. What is the value of $\frac{a_3}{a_1}$?

At $t=0$, the length is $a_0=0$. At $t=1$ hour, the length is $a_1=\frac{a_3}{2}$. At $t=2$ hours, the length is $a_2=\frac{a_3}{2}+\frac{a_3}{3}$. At $t=3$ hours, the length is $\color{red}{a_3=\frac{a_3}{2}+\frac{a_3}{3}+\frac{a_3}{4}=\frac{13}{12}a_3}$. So $$\frac{a_3}{a_1}=\frac{\frac{13}{12}a_3}{\frac{a_3}{2}}=\frac{13}{6}$$ is the answer. But notice that the red equation enables solving for $a_3$ which gives $a_3=0$. Therefore $\frac{a_3}{a_1}=0$ and, by transitivity, we have:

Conclusion: $\frac{13}{6}=0$

Where is the mistake?

Answer 1

I wrote my answer in spoiler blocks so you can decide when to see it.

It is defined that for $t = 3$, $l = a_{3}$.

For $t = 0$, $l = 0$. The remaining length should be $a_{3}$.

For $t = 1$, $l = \frac{a_{3}}{2}$. The remaining length should be $\frac{a_{3}}{2}$.

For $t = 2$, $l = \frac{a_{3}}{2} + \frac{a_{3}}{3}$. The remaining length should be $\frac{a_{3}}{6}$.

For $t = 3$, $l = \frac{a_{3}}{2} + \frac{a_{3}}{3} + \frac{a_{3}}{4}$. This is where the problem arises, as the added length is greater than the remaining $(\frac{a_{3}}{4} > \frac{a_{3}}{6})$. This is the mistake.

Answer 2

The red equation gives $a_3 = 0$. So dividing by $a_1 = \dfrac{a_3}{2}$ is the same as dividing by $0$.

Answer 3

If $\color{red}{a_3=\frac{a_3}{2}+\frac{a_3}{3}+\frac{a_3}{4}=\frac{13}{12}a_3}$ then $a_3$$=0$

So $a_1$ = $a_{3}/2$ $=0 .$ But you can’t divide by $0$ .