# algebra precalculus – False proof that \$frac{13}{6}=0\$

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 $$t=1$$ 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, by transitivity, $$frac{a_3}{a_1}=0$$.

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

Where is the mistake?