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

#Algebra

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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Comments

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

Yajat Shamji - 11 months, 2 weeks ago

What first proof?

Zakir Husain - 11 months, 2 weeks ago

The one above?

Yajat Shamji - 11 months, 2 weeks 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 ( $x_1x_2\cancel{=}x_1\times x_2$ )
$0.\overline{a}=0.aaaaa...$

Zakir Husain - 11 months, 2 weeks 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 :)

A Former Brilliant Member - 6 months, 2 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 $\vec{a}$ $2\vec{a}(\vec{s}_B - \vec{s}_A) = \vec{v_{B}} \cdot \vec{v_{B}} - \vec{v_{A}} \cdot \vec{v_{A}}$

Zakir Husain - 6 months, 2 weeks ago

That's fine.

Yajat Shamji - 6 months, 2 weeks ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Recurring Proof Note 111

Comments

Recurring Proof Note $1$