I don't know how to convert infinite periodic decimal number x=3,1(42)=3.1424242…to a fraction aba,b are integers. Need to find a,b

Equationx=3.142―first multiply by 10 we get10x=31.42―now multiply by 100 we have1000x=3142.42―from third equation subtract second we get990x=3111x=3111990=1037330

Any repeating decimal can be written as a convergent geometric series. Here we have3.1424242…=3.1+42103+42105+42107+⋯=3.1+42103(1+1100+11002+11003+⋯)=3.1+4210311−1100=3.1+42990

Equationx=3.142―first multiply by 10 we get10x=31.42―now multiply by 100 we have1000x=3142.42―from third equation subtract second we get990x=3111x=3111990=1037330

Any repeating decimal can be written as a convergent geometric series. Here we have3.1424242…=3.1+42103+42105+42107+⋯=3.1+42103(1+1100+11002+11003+⋯)=3.1+4210311−1100=3.1+42990

Best solutions to - "I don't know how to convert infinite periodic..."

Rowan Callahan

Answered question

2022-04-01

I don't know how to convert infinite periodic decimal number

x = 3, 1 (42) = 3.1424242 \dots

to a fraction

\frac{a}{b}

a,b are integers. Need to find a,b

Answer & Explanation

anghoelv1lw

Beginner2022-04-02Added 19 answers

Equation

x = 3.1 \overset{―}{42}

first multiply by 10 we get

10 x = 31 . \overset{―}{42}

now multiply by 100 we have

1000 x = 3142 . \overset{―}{42}

from third equation subtract second we get

990 x = 3111

x = \frac{3111}{990} = \frac{1037}{330}

Carter Lin

Beginner2022-04-03Added 13 answers

Any repeating decimal can be written as a convergent geometric series. Here we have

3.1424242 \dots = 3.1 + \frac{42}{10^{3}} + \frac{42}{10^{5}} + \frac{42}{10^{7}} + \dots

= 3.1 + \frac{42}{10^{3}} (1 + \frac{1}{100} + \frac{1}{100^{2}} + \frac{1}{100^{3}} + \dots)

= 3.1 + \frac{42}{10^{3}} \frac{1}{1 - \frac{1}{100}} = 3.1 + \frac{42}{990}

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Pre-Algebra

Didn't find what you were looking for?