Irrational Numbers Worksheets - 15 Worksheets Library - Free Printable

To solve the problem of expressing non-terminating recurring decimals in the form \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \neq 0 \), we will use algebraic methods to convert each recurring decimal into a fraction. Let's go through each problem step by step.

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1. \( 0.\overline{3} \)

Let \( x = 0.\overline{3} \).

Multiply both sides by 10 (since there is one digit in the repeating block):
\[ 10x = 3.\overline{3} \]

Subtract the original equation \( x = 0.\overline{3} \) from this:
\[ 10x - x = 3.\overline{3} - 0.\overline{3} \]
\[ 9x = 3 \]
\[ x = \frac{3}{9} = \frac{1}{3} \]

So, \( 0.\overline{3} = \frac{1}{3} \).

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2. \( 1.\overline{4} \)

Let \( x = 1.\overline{4} \).

Multiply both sides by 10 (since there is one digit in the repeating block):
\[ 10x = 14.\overline{4} \]

Subtract the original equation \( x = 1.\overline{4} \) from this:
\[ 10x - x = 14.\overline{4} - 1.\overline{4} \]
\[ 9x = 13 \]
\[ x = \frac{13}{9} \]

So, \( 1.\overline{4} = \frac{13}{9} \).

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3. \( 0.\overline{57} \)

Let \( x = 0.\overline{57} \).

Multiply both sides by 100 (since there are two digits in the repeating block):
\[ 100x = 57.\overline{57} \]

Subtract the original equation \( x = 0.\overline{57} \) from this:
\[ 100x - x = 57.\overline{57} - 0.\overline{57} \]
\[ 99x = 57 \]
\[ x = \frac{57}{99} = \frac{19}{33} \]

So, \( 0.\overline{57} = \frac{19}{33} \).

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4. \( 0.0\overline{01} \)

Let \( x = 0.0\overline{01} \).

Multiply both sides by 100 (to move the decimal point past the non-repeating part):
\[ 100x = 0.\overline{01} \]

Now, let \( y = 0.\overline{01} \). Multiply both sides by 100 (since there are two digits in the repeating block):
\[ 100y = 1.\overline{01} \]

Subtract the original equation \( y = 0.\overline{01} \) from this:
\[ 100y - y = 1.\overline{01} - 0.\overline{01} \]
\[ 99y = 1 \]
\[ y = \frac{1}{99} \]

Thus, \( 100x = \frac{1}{99} \), so:
\[ x = \frac{1}{9900} \]

So, \( 0.0\overline{01} = \frac{1}{9900} \).

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5. \( 0.\overline{53} \)

Let \( x = 0.\overline{53} \).

Multiply both sides by 100 (since there are two digits in the repeating block):
\[ 100x = 53.\overline{53} \]

Subtract the original equation \( x = 0.\overline{53} \) from this:
\[ 100x - x = 53.\overline{53} - 0.\overline{53} \]
\[ 99x = 53 \]
\[ x = \frac{53}{99} \]

So, \( 0.\overline{53} = \frac{53}{99} \).

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6. \( 2.\overline{93} \)

Let \( x = 2.\overline{93} \).

Multiply both sides by 100 (since there are two digits in the repeating block):
\[ 100x = 293.\overline{93} \]

Subtract the original equation \( x = 2.\overline{93} \) from this:
\[ 100x - x = 293.\overline{93} - 2.\overline{93} \]
\[ 99x = 291 \]
\[ x = \frac{291}{99} = \frac{97}{33} \]

So, \( 2.\overline{93} = \frac{97}{33} \).

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7. \( 18.\overline{48} \)

Let \( x = 18.\overline{48} \).

Multiply both sides by 100 (since there are two digits in the repeating block):
\[ 100x = 1848.\overline{48} \]

Subtract the original equation \( x = 18.\overline{48} \) from this:
\[ 100x - x = 1848.\overline{48} - 18.\overline{48} \]
\[ 99x = 1830 \]
\[ x = \frac{1830}{99} = \frac{610}{33} \]

So, \( 18.\overline{48} = \frac{610}{33} \).

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8. \( 0.2\overline{35} \)

Let \( x = 0.2\overline{35} \).

Multiply both sides by 10 (to move the decimal point past the non-repeating part):
\[ 10x = 2.\overline{35} \]

Now, let \( y = 2.\overline{35} \). Multiply both sides by 100 (since there are two digits in the repeating block):
\[ 100y = 235.\overline{35} \]

Subtract the original equation \( y = 2.\overline{35} \) from this:
\[ 100y - y = 235.\overline{35} - 2.\overline{35} \]
\[ 99y = 233 \]
\[ y = \frac{233}{99} \]

Thus, \( 10x = \frac{233}{99} \), so:
\[ x = \frac{233}{990} \]

So, \( 0.2\overline{35} = \frac{233}{990} \).

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9. \( 0.00\overline{32} \)

Let \( x = 0.00\overline{32} \).

Multiply both sides by 100 (to move the decimal point past the non-repeating part):
\[ 100x = 0.\overline{32} \]

Now, let \( y = 0.\overline{32} \). Multiply both sides by 100 (since there are two digits in the repeating block):
\[ 100y = 32.\overline{32} \]

Subtract the original equation \( y = 0.\overline{32} \) from this:
\[ 100y - y = 32.\overline{32} - 0.\overline{32} \]
\[ 99y = 32 \]
\[ y = \frac{32}{99} \]

Thus, \( 100x = \frac{32}{99} \), so:
\[ x = \frac{32}{9900} \]

So, \( 0.00\overline{32} = \frac{32}{9900} \).

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10. \( 0.\overline{9} \)

Let \( x = 0.\overline{9} \).

Multiply both sides by 10 (since there is one digit in the repeating block):
\[ 10x = 9.\overline{9} \]

Subtract the original equation \( x = 0.\overline{9} \) from this:
\[ 10x - x = 9.\overline{9} - 0.\overline{9} \]
\[ 9x = 9 \]
\[ x = 1 \]

So, \( 0.\overline{9} = 1 \).

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11. \( 2.4\overline{178} \)

Let \( x = 2.4\overline{178} \).

Multiply both sides by 10 (to move the decimal point past the non-repeating part):
\[ 10x = 24.\overline{178} \]

Now, let \( y = 24.\overline{178} \). Multiply both sides by 1000 (since there are three digits in the repeating block):
\[ 1000y = 24178.\overline{178} \]

Subtract the original equation \( y = 24.\overline{178} \) from this:
\[ 1000y - y = 24178.\overline{178} - 24.\overline{178} \]
\[ 999y = 24154 \]
\[ y = \frac{24154}{999} \]

Thus, \( 10x = \frac{24154}{999} \), so:
\[ x = \frac{24154}{9990} = \frac{12077}{4995} \]

So, \( 2.4\overline{178} = \frac{12077}{4995} \).

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12. \( 2.\overline{36} \)

Let \( x = 2.\overline{36} \).

Multiply both sides by 100 (since there are two digits in the repeating block):
\[ 100x = 236.\overline{36} \]

Subtract the original equation \( x = 2.\overline{36} \) from this:
\[ 100x - x = 236.\overline{36} - 2.\overline{36} \]
\[ 99x = 234 \]
\[ x = \frac{234}{99} = \frac{26}{11} \]

So, \( 2.\overline{36} = \frac{26}{11} \).

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13. \( 1.\overline{323} \)

Let \( x = 1.\overline{323} \).

Multiply both sides by 1000 (since there are three digits in the repeating block):
\[ 1000x = 1323.\overline{323} \]

Subtract the original equation \( x = 1.\overline{323} \) from this:
\[ 1000x - x = 1323.\overline{323} - 1.\overline{323} \]
\[ 999x = 1322 \]
\[ x = \frac{1322}{999} \]

So, \( 1.\overline{323} = \frac{1322}{999} \).

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14. \( 0.\overline{3178} \)

Let \( x = 0.\overline{3178} \).

Multiply both sides by 10000 (since there are four digits in the repeating block):
\[ 10000x = 3178.\overline{3178} \]

Subtract the original equation \( x = 0.\overline{3178} \) from this:
\[ 10000x - x = 3178.\overline{3178} - 0.\overline{3178} \]
\[ 9999x = 3178 \]
\[ x = \frac{3178}{9999} \]

So, \( 0.\overline{3178} = \frac{3178}{9999} \).

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15. \( 0.4\overline{07} \)

Let \( x = 0.4\overline{07} \).

Multiply both sides by 10 (to move the decimal point past the non-repeating part):
\[ 10x = 4.\overline{07} \]

Now, let \( y = 4.\overline{07} \). Multiply both sides by 100 (since there are two digits in the repeating block):
\[ 100y = 407.\overline{07} \]

Subtract the original equation \( y = 4.\overline{07} \) from this:
\[ 100y - y = 407.\overline{07} - 4.\overline{07} \]
\[ 99y = 403 \]
\[ y = \frac{403}{99} \]

Thus, \( 10x = \frac{403}{99} \), so:
\[ x = \frac{403}{990} \]

So, \( 0.4\overline{07} = \frac{403}{990} \).

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16. \( 32.1\overline{235} \)

Let \( x = 32.1\overline{235} \).

Multiply both sides by 10 (to move the decimal point past the non-repeating part):
\[ 10x = 321.\overline{235} \]

Now, let \( y = 321.\overline{235} \). Multiply both sides by 1000 (since there are three digits in the repeating block):
\[ 1000y = 321235.\overline{235} \]

Subtract the original equation \( y = 321.\overline{235} \) from this:
\[ 1000y - y = 321235.\overline{235} - 321.\overline{235} \]
\[ 999y = 320914 \]
\[ y = \frac{320914}{999} \]

Thus, \( 10x = \frac{320914}{999} \), so:
\[ x = \frac{320914}{9990} = \frac{160457}{4995} \]

So, \( 32.1\overline{235} = \frac{160457}{4995} \).

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Final Answers:

\[
\boxed{
\begin{array}{ll}
1. & \frac{1}{3} \\
2. & \frac{13}{9} \\
3. & \frac{19}{33} \\
4. & \frac{1}{9900} \\
5. & \frac{53}{99} \\
6. & \frac{97}{33} \\
7. & \frac{610}{33} \\
8. & \frac{233}{990} \\
9. & \frac{32}{9900} \\
10. & 1 \\
11. & \frac{12077}{4995} \\
12. & \frac{26}{11} \\
13. & \frac{1322}{999} \\
14. & \frac{3178}{9999} \\
15. & \frac{403}{990} \\
16. & \frac{160457}{4995} \\
\end{array}
}
\]