Express 0-001 in the form p-q- where p and q are integers and q - 0-

0.001= 0.001001... Let nbsp;x nbsp;= 0.001001... ⇒ 1000x nbsp;= 1.001001… nbsp;[Since, there are three repeating decimal digits, so multiply x with 100 ...

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Standard VII

Mathematics

Addition of Integers

Express 0.001...

Question

Express $0. ¯ ¯¯¯¯¯¯ ¯ 001$ in the form $\frac{p}{q}$ , where p and q are integers and $q \neq 0$ .

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Solution

$0. ¯ ¯¯¯¯¯¯ ¯ 001$ = 0.001001...
Let x = 0.001001...
$\Rightarrow$ 1000x = 1.001001… [Since, there are three repeating decimal digits, so multiply x with 1000.]
$\Rightarrow$ 1000x = 1+ 0.001001
$\Rightarrow$ 1000x = 1 + x
$\Rightarrow$ 999x = 1
$\Rightarrow x = \frac{1}{999}$

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Express 0.¯¯¯¯¯¯¯¯001 in the form pq, where p and q are integers and q≠0.

0.¯¯¯¯¯¯¯¯001= 0.001001... Let x = 0.001001... ⇒ 1000x = 1.001001… [Since, there are three repeating decimal digits, so multiply x with 1000.] ⇒ 1000x = 1+ 0.001001 ⇒ 1000x = 1 + x ⇒ 999x = 1 ⇒x=1999

Express $0. ¯ ¯¯¯¯¯¯ ¯ 001$ in the form $\frac{p}{q}$ , where p and q are integers and $q \neq 0$ .

$0. ¯ ¯¯¯¯¯¯ ¯ 001$ = 0.001001...
Let x = 0.001001...
$\Rightarrow$ 1000x = 1.001001… [Since, there are three repeating decimal digits, so multiply x with 1000.]
$\Rightarrow$ 1000x = 1+ 0.001001
$\Rightarrow$ 1000x = 1 + x
$\Rightarrow$ 999x = 1
$\Rightarrow x = \frac{1}{999}$