What is the intuition behind the fact that:
$$ 0.9999999\ldots = 1 $$
My Teacher's Proof:
We know that any number which belongs to the interval $[0,1)$ can be uniquely written as decimal expansion:
$$ x=0.a_1a_2a_3 \ldots = a_1×10^-1+a_2×10^-2+a_3×10^-3 + \ldots $$
where the digits $a_1,a_2,a_3 \ldots$ satisfy the following:
Each digit $a_i$ is an integer in the 10-element set ${0,1,2,....,9}$
So, $$ 0.9999999 \ldots = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} \ldots = 9 [$\frac{1}{10} + \frac{1}{100} + \frac{1}{1000} \ldots$)] = 9 × \frac{1}{9} = 1 $$
My doubt:
Since
$$ [\frac{1}{10} + \frac{1}{100} + \frac{1}{1000} \ldots)] ≈ \frac{1}{9} $$
So $0.9999999 \ldots ≈ 1$ not exactly equal to $1$.
Am I correct?
Suppose some people jump towards the right with velocity $u$ relative to a cart. Let the mass of each person be $m$ and that of cart be $M$. If there are $N$ people then, $Nm=M$
Case 2: All people jump simultaneously.
$$ \Rightarrow Nm(u-v_B)=M(v_B) $$
Hence, $v_B=\frac{u}{2}$
Case 1: People jump one after other.
After the first man jumps off the velocity $v_1$, acquired by the cart to the left,
$$ m(u-v_1)=(M+Nm-m)v_1 $$
Hence, $v_1 = \frac{mu}{M+Nm}$
After the second person jumps off, the velocity $v_2$ acquired by cart to the left,
$$ m(u-v_2)+(M+Nm-m)v_1=(M+Nm-2m)v_2 $$
Hence, $v_2=v_1+\frac{mu}{M+Nm-m}$
Similarly, after the $n^{th}$ person jumps off, the velocity $v_n$ acquired by the cart to the left $(2≤n≤N)$.
So, $v_n=v_{n-1}+\frac{mu}{M+Nm-nm}$
Therefore the velocity $(v_A)$ after all people have jumped off,
$$ v_A=\frac{mu}{M+Nm}+\frac{mu}{M+(N-1)m}+\cdots+\frac{mu}{M+m} $$
How do I proceed from here?
Answer:
$$\frac{v_A}{v_B}=2ln2 $$
