But you usage the hash vital just prior to 0.8bar3and also at the end. Therefore you finish 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 political parties by 10

#" "90x=75#

Divide both sides by 90

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

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

Here"s another method you can convert decimals to fractions if you have actually a calculator to hand.

You are watching: .083 repeating as a fraction

We use the calculator to uncover the end continued portion expansion because that the provided number, climate unwrap it come a continual fraction.

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

See more: Number: Prime Numbers Between 1 And 50 Are (1) 18 (2) 12 (3) 15 (4) 20

Note the the part before the decimal suggest is #0#, so compose that down:

#color(blue)(0) + #

Take the reciprocal of the given number to acquire a an outcome something like: #1.2000000048#. We deserve to ignore the trailing digits #48# as they are simply a round off error. So v our new result #1.2# keep in mind that the number prior to the decimal allude is #1#. Compose that down as the following coefficient in the ongoing fraction:

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

then subtract that to acquire #0.2#. Take the reciprocal, obtaining the an outcome #5.0#. This has the number #5# prior to the decimal allude and no remainder. So add that to our continued fraction as the following reciprocal come 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, allow us consider a more complicated example:

Given:

#3.82857142857#

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

#1.20689655173#

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

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

Note the #color(blue)(4)#, subtract it and also 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 also stop.

Taking the numbers we have found, us 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))#