How do you convert 0.83 (3 repeating) to a fraction?

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Tony B Share
Aug 23, 2017

Answer:

#" so "x= 0.8bar3 = 5/6#

Explanation:

To format this question type you write: 0.8bar3

But you use the hash key just before 0.8bar3
and also at the end. So you end up with

#" "0.8bar3#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let #x=0.8bar3#

Then #10x = 8.bar3#

So #(10x-x)=" " 8.3333bar3#
#color(white)("bbnnn.nnnnnnnbb")underline(0.8333bar3-# Subtracting
#color(white)("bbbbb.bbbbbbbbb")7.5#

#" "9x = 7.5#

Multiply both sides by 10

#" "90x=75#

Divide both sides by 90

#" "x=75/90 = 5/6#

#" so "x= 0.8bar3 = 5/6#

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Write your answer here...
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Then teach the underlying concepts
Don't copy without citing sources
preview
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Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

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13
Aug 23, 2017

Answer:

Here's a method using a calculator to help...

Explanation:

Here's another way you can convert decimals to fractions if you have a calculator to hand.

We use the calculator to find the terminating continued fraction expansion for the given number, then unwrap it to a regular fraction.

For our example, type #0.83333333# into your calculator.

Note that the portion before the decimal point is #0#, so write that down:

#color(blue)(0) + #

Take the reciprocal of the given number to get a result something like: #1.2000000048#. We can ignore the trailing digits #48# as they are just a rounding error. So with our new result #1.2# note that the number before the decimal point is #1#. Write that down as the next coefficient in the continued fraction:

#color(blue)(0) + 1/color(blue)(1)#

then subtract it to get #0.2#. Take the reciprocal, getting the result #5.0#. This has the number #5# before the decimal point and no remainder. So add that to our continued fraction as the next reciprocal to get:

#color(blue)(0) + 1/(color(blue)(1)+1/color(blue)(5)) = 0+1/(6/5) = 5/6#

#color(white)()#
Another example

Just to make the method a little clearer, let us consider a more complex example:

Given:

#3.82857142857#

Note the #color(blue)(3)#, subtract it and take the reciprocal to get:

#1.20689655173#

Note the #color(blue)(1)#, subtract it and take the reciprocal to get:

#4.83333333320" "color(lightgrey)"Note the rounding error"#

Note the #color(blue)(4)#, subtract it and take the reciprocal to get:

#1.20000000019#

Note the #color(blue)(1)#, subtract it and take the reciprocal to get:

#4.99999999525#

Let's call that #color(blue)(5)# and stop.

Taking the numbers we have found, we have:

#3.82857142857 = color(blue)(3) + 1/(color(blue)(1)+1/(color(blue)(4)+1/(color(blue)(1)+1/color(blue)(5))))#

#color(white)(3.82857142857) = color(blue)(3) + 1/(color(blue)(1)+1/(color(blue)(4)+5/6))#

#color(white)(3.82857142857) = color(blue)(3) + 1/(color(blue)(1)+6/29)#

#color(white)(3.82857142857) = color(blue)(3) + 29/35#

#color(white)(3.82857142857) = 134/35#

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