# Repeating Decimals As Fraction

• August 13th 2010, 04:21 AM
fdrhs1984
Repeating Decimals As Fraction
Express the given repeating decimals as a fraction.(1) 0.929292...(2) 52.31272727...
• August 13th 2010, 05:28 AM
Sudharaka
Quote:

Originally Posted by fdrhs1984
Express the given repeating decimals as a fraction.(1) 0.929292...(2) 52.31272727...

Dear fdrhs1984,

I shall show you an example similar to the ones that you have given. Suppose you have to express 0.62626262....as a fraction.

$0.626262.....=0.62+0.0062+0.000062+......=\frac{62 }{100}+\frac{62}{10000}+\frac{62}{1000000}+.......$

$0.62626262.....=\frac{62}{10^2}+\frac{62}{10^4}+\f rac{62}{10^6}+....=\frac{62}{10^2}\left(1+\frac{1} {10^2}+\frac{1}{10^4}+.......\right)$

$0.626262......=\frac{62}{10^2}\left(\displaystyle{ \lim_{n\to\infty}\frac{1-\frac{1}{100^n}}{1-\frac{1}{100}}}\right)=\frac{62}{100}\times{\frac{ 100}{99}}=\frac{62}{99}$

Hopefully using the same method you will be able to do your two problems.
• August 13th 2010, 05:29 AM
mr fantastic
Quote:

Originally Posted by fdrhs1984
Express the given repeating decimals as a fraction.(1) 0.929292...(2) 52.31272727...

Here is the slacker's way:

S = 0.929292..... eqn (1)

100S = 92.9292.... eqn (2)

Eqn (2) - eqn (1): 99S = 92 => S = 92/99.

--------------------------------------------------------------------------------------

S = 52.31272727.... eqn (1)

100S = 5231.272727.... eqn (2)

10000S = 523127.2727..... eqn (3)

Eqn (3) - eqn (2): etc.
• August 13th 2010, 05:43 AM
earboth
Quote:

Originally Posted by fdrhs1984
Express the given repeating decimals as a fraction.(1) 0.929292...(2) 52.31272727...

1. Let n denote the number of the repeated decimals. Then you can use the following 2 equations:

$\begin{array}{rcr}x&=&0.929292... \\ 10^2 x&=&92.929292...\end{array}$

The exponent of the base 10 corresponds to n.

3. Now subtract the first equation from the second cloumnwise and you'll get:

$99x = 92$

Solve for x and simplify if possible: $x = \dfrac{92}{99}$

4. With your 2nd example you have to multiply first to get an immediately repeating decimal before you can use the described method:

$\begin{array}{rcr}x&=&52.3127272727... \\ 100x&=&5231.27272727... \\ 10000x&=&523127.272727...\end{array}$

Subtract the last two equations:

$9900x = 517896$

I'll leave the rest for you.

EDIT: ... too late - again.
• August 13th 2010, 06:32 AM
fdrhs1984
Quote:

Originally Posted by Sudharaka
Dear fdrhs1984,

I shall show you an example similar to the ones that you have given. Suppose you have to express 0.62626262....as a fraction.

$0.626262.....=0.62+0.0062+0.000062+......=\frac{62 }{100}+\frac{62}{10000}+\frac{62}{1000000}+.......$

$0.62626262.....=\frac{62}{10^2}+\frac{62}{10^4}+\f rac{62}{10^6}+....=\frac{62}{10^2}\left(1+\frac{1} {10^2}+\frac{1}{10^4}+.......\right)$

$0.626262......=\frac{62}{10^2}\left(\displaystyle{ \lim_{n\to\infty}\frac{1-\frac{1}{100^n}}{1-\frac{1}{100}}}\right)=\frac{62}{100}\times{\frac{ 100}{99}}=\frac{62}{99}$

Hopefully using the same method you will be able to do your two problems.

This method is a bit too advanced since we have not covered limits in class.

Quote:

Originally Posted by mr fantastic
Here is the slacker's way:

S = 0.929292..... eqn (1)

100S = 92.9292.... eqn (2)

Eqn (2) - eqn (1): 99S = 92 => S = 92/99.

--------------------------------------------------------------------------------------

S = 52.31272727.... eqn (1)

100S = 5231.272727.... eqn (2)

10000S = 523127.2727..... eqn (3)

Eqn (3) - eqn (2): etc.

This method is slightly better than the first reply involving limits.
• August 13th 2010, 06:39 AM
undefined
Quote:

Originally Posted by fdrhs1984
This method is slightly better than the first reply involving limits.

Simpler maybe, but I don't think it's fair to call it better. I think they're equally good.

Quote:

Originally Posted by fdrhs1984
This method is a bit too advanced since we have not covered limits in class.

• August 13th 2010, 09:58 AM
fdrhs1984
Quote:

Originally Posted by earboth
1. Let n denote the number of the repeated decimals. Then you can use the following 2 equations:

$\begin{array}{rcr}x&=&0.929292... \\ 10^2 x&=&92.929292...\end{array}$

The exponent of the base 10 corresponds to n.

3. Now subtract the first equation from the second cloumnwise and you'll get:

$99x = 92$

Solve for x and simplify if possible: $x = \dfrac{92}{99}$

4. With your 2nd example you have to multiply first to get an immediately repeating decimal before you can use the described method:

$\begin{array}{rcr}x&=&52.3127272727... \\ 100x&=&5231.27272727... \\ 10000x&=&523127.272727...\end{array}$

Subtract the last two equations:

$9900x = 517896$

I'll leave the rest for you.

EDIT: ... too late - again.

Nice break down of the questions. I'll do the rest.

Quote:

Originally Posted by undefined
Simpler maybe, but I don't think it's fair to call it better. I think they're equally good.