# Recurring Proof Note $1$

I'll prove that:

$y = 0.\overline{x_1}x_2 =$$\frac{x_1}{9}$

Let:

$y = 0.\overline{x_1}x_2$

$10y = x_1.\overline{x_1}x_2$

$10y - y = 9y$

$x_1.\overline{x_1}x_2 - 0.\overline{x_1}x_2 = x_1$

$9y = x_1$

$\frac{9y}{9}$ $=$ $\frac{x_1}{9}$

$y =$$\frac{x_1}{9}$

Therefore, $y = 0.\overline{x_1}x_2 =$$\frac{x_1}{9}$

Note by Yajat Shamji
1 year ago

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@Zakir Husain, what do you think of my first proof?

- 1 year ago

What first proof?

- 1 year ago

The one above?

- 1 year ago

A result for general case: Consider a number $l=0.\overline{x_1x_2x_3...x_n}$ $10^nl=x_1x_2x_3...x_n\large{.}\overline{x_1x_2x_3...x_n}$ $\Rightarrow 10^nl-l={x_1x_2x_3...x_n}$ $(10^n-1)l=x_1x_2x_3...x_n$ $\boxed{l=\dfrac{x_1x_2x_3...x_n}{10^n-1}}$ $\boxed{0.\overline{x_1x_2x_3...x_n}=\dfrac{x_1x_2x_3...x_n}{10^n-1}}$

Note :

• $x_1x_2$ act as number with digits $x_1,x_2$ for example if $x_1=5$ and $x_2=8\Rightarrow x_1x_2=58$ dont confuse ($x_1x_2\cancel{=}x_1\times x_2$)

• $0.\overline{a}=0.aaaaa...$

- 1 year ago

@Yajat Shamji - Here?

Derivation - $v^{2} - u^{2} = 2aS$

We know that $\dfrac{v - u}{t} = a$ because $v-u$ is the change in velocity, and dividing by time we just get change in velocity over time, which is nothing but acceleration.

We also know that $S = ut + \dfrac{1}{2}at^{2}$ as it has been proved in the preface of a kinematics problem. Here is the image of the proof -

Now we can use these equations to prove the one we used to solve the problem.

In the equation, we don't have time, so let's find the value of time, and then substitute.

$\dfrac{v - u}{t} = a$ can be changed to $\dfrac{v - u}{a} = t$ by some transposition. Let's substitute this in the other equation now.

\begin{aligned} S &= ut + \dfrac{1}{2}at^{2} \\ S &= u\dfrac{v-u}{a} + \dfrac{1}{2}a\dfrac{v-u}{a})^{2} \\ S &= \dfrac{uv-u^{2}}{a} + \dfrac{v^{2}-2uv + u^{2}}{2a} \\ S &= \dfrac{2uv-2u^{2}}{2a} + \dfrac{v^{2}-2uv + u^{2}}{2a} \\ 2aS &= 2uv-2u^{2} + v^{2}-2uv + u^{2} \\ 2aS &= v^{2} - u^{2} \\ v^{2} - u^{2} &= 2aS \end{aligned}

Hence Proved :)

- 7 months, 4 weeks ago

@Percy Jackson $\int_{A}^{B} \vec{a} \cdot d\vec{s} = \int_{A}^{B} \frac{d\vec{v}}{dt} \cdot d\vec{s} = \int_{A}^{B} d\vec{v} \cdot \vec{v}$ $= { {}} \int_{A}^{B} v_x dv_x + {{{ }}} \int_{A}^{B} v_y dv_y + { { }}\int_{A}^{B} v_z dv_z$ $=\frac{1}{2}(v^2_{x,B} - v^2_{x,A} + v^2_{y,B} - v^2_{y,A} + v^2_{z,B} - v^2_{z,A}) = \frac{1}{2}(\vec{v_{B}} \cdot \vec{v_{B}} - \vec{v_{A}} \cdot \vec{v_{A}})$ $\Rightarrow 2 \int_{A}^{B} \vec{a} \cdot d\vec{s} = \vec{v_{B}} \cdot \vec{v_{B}} - \vec{v_{A}} \cdot \vec{v_{A}}$ For constant $\vec{a}$ $2\vec{a}(\vec{s}_B - \vec{s}_A) = \vec{v_{B}} \cdot \vec{v_{B}} - \vec{v_{A}} \cdot \vec{v_{A}}$

- 7 months, 4 weeks ago

That's fine.

- 7 months, 3 weeks ago