# Recurring Proof Note $5$

I'll prove that

$y = 0.x_1\overline{x_2} =$$\frac{x_1x_2 - x_1}{90}$

Let:

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

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

$10y - y = 9y$

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

$9y = x_1.x_2 - x_1$

$\frac{9y}{9}$ $=$ $\frac{x_1.x_2 - x_1}{9}$

$\frac{90y}{90}$ $=$ $\frac{x_1x_2 - x_1}{90}$

$y =$$\frac{x_1x_2 - x_1}{90}$

Therefore, $y = 0.x_1\overline{x_2} =$$\frac{x_1x_2 - x_1}{90}$

Note by Yajat Shamji
1 year, 2 months ago

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