The value of 1.999... in the form frac{p}{q}, | vedclass

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(B) Let $x = 1.999... = 1.\overline{9} \quad \dots(1)$Multiply both sides by $10$:$10x = 19.999... = 19.\overline{9} \quad \dots(2)$Subtract equation

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$2$

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(B) Let $x = 1.999... = 1.\overline{9} \quad \dots(1)$Multiply both sides by $10$:$10x = 19.999... = 19.\overline{9} \quad \dots(2)$Subtract equation $(1)$ from equation $(2)$:$10x - x = 19.999... - 1.999...$$9x = 18$Divide by $9$:$x = \frac{18}{9} = 2$Since $2$ can be written as $\frac{2}{1},$ where $p=2$ and $q=1$ are integers and $q 
eq 0,$ the value is $2.$Hence,option $(b)$ is the correct answer.

The value of $1.999...$ in the form $\frac{p}{q},$ where $p$ and $q$ are integers and $q \neq 0,$ is

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The value of $1.999...$ in the form $\frac{p}{q},$ where $p$ and $q$ are integers and $q \neq 0,$ is

Similar Questions

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