Recurring Proof Note 11

I'll prove that:

y=0.x1x2=y = 0.\overline{x_1}x_2 =x19\frac{x_1}{9}


y=0.x1x2y = 0.\overline{x_1}x_2

10y=x1.x1x210y = x_1.\overline{x_1}x_2

10yy=9y10y - y = 9y

x1.x1x20.x1x2=x1x_1.\overline{x_1}x_2 - 0.\overline{x_1}x_2 = x_1

9y=x19y = x_1

9y9\frac{9y}{9} == x19\frac{x_1}{9}

y=y = x19\frac{x_1}{9}

Therefore, y=0.x1x2=y = 0.\overline{x_1}x_2 =x19\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?

Yajat Shamji - 1 year ago

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What first proof?

Zakir Husain - 1 year ago

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The one above?

Yajat Shamji - 1 year ago

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A result for general case: Consider a number l=0.x1x2x3...xnl=0.\overline{x_1x_2x_3...x_n} 10nl=x1x2x3...xn.x1x2x3...xn10^nl=x_1x_2x_3...x_n\large{.}\overline{x_1x_2x_3...x_n} 10nll=x1x2x3...xn\Rightarrow 10^nl-l={x_1x_2x_3...x_n} (10n1)l=x1x2x3...xn(10^n-1)l=x_1x_2x_3...x_n l=x1x2x3...xn10n1\boxed{l=\dfrac{x_1x_2x_3...x_n}{10^n-1}} 0.x1x2x3...xn=x1x2x3...xn10n1\boxed{0.\overline{x_1x_2x_3...x_n}=\dfrac{x_1x_2x_3...x_n}{10^n-1}}

Note :

  • x1x2x_1x_2 act as number with digits x1,x2x_1,x_2 for example if x1=5x_1=5 and x2=8x1x2=58x_2=8\Rightarrow x_1x_2=58 dont confuse (x1x2=x1×x2x_1x_2\cancel{=}x_1\times x_2)

  • 0.a=0.aaaaa...0.\overline{a}=0.aaaaa...

Zakir Husain - 1 year ago

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@Yajat Shamji - Here?

Derivation - v2u2=2aSv^{2} - u^{2} = 2aS

We know that vut=a\dfrac{v - u}{t} = a because vuv-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+12at2S = 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.

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

S=ut+12at2S=uvua+12avua)2S=uvu2a+v22uv+u22aS=2uv2u22a+v22uv+u22a2aS=2uv2u2+v22uv+u22aS=v2u2v2u2=2aS\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 - 7 months, 4 weeks ago

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@Percy Jackson ABads=ABdvdtds=ABdvv\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} =ABvxdvx+ABvydvy+ABvzdvz = { {}} \int_{A}^{B} v_x dv_x + {{{ }}} \int_{A}^{B} v_y dv_y + { { }}\int_{A}^{B} v_z dv_z =12(vx,B2vx,A2+vy,B2vy,A2+vz,B2vz,A2)=12(vBvBvAvA)=\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}}) 2ABads=vBvBvAvA\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 a\vec{a} 2a(sBsA)=vBvBvAvA2\vec{a}(\vec{s}_B - \vec{s}_A) = \vec{v_{B}} \cdot \vec{v_{B}} - \vec{v_{A}} \cdot \vec{v_{A}}

Zakir Husain - 7 months, 4 weeks ago

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That's fine.

Yajat Shamji - 7 months, 3 weeks ago

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