But you usage the hash crucial just prior to 0.8bar3and likewise at the end. Therefore 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 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 fountain if you have a calculator come hand.

You are watching: .83 repeating as a fraction

We use the calculator to find the end continued fraction expansion for the provided number, climate unwrap it come a constant fraction.

For our example, form #0.83333333# right into your calculator.

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Note the the portion before the decimal allude is #0#, so compose that down:

#color(blue)(0) + #

Take the mutual of the offered number to obtain a result something like: #1.2000000048#. We deserve to ignore the trailing digits #48# as they are just a round off error. So through our brand-new result #1.2# note that the number before the decimal suggest is #1#. Create that under as the next coefficient in the ongoing fraction:

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

then subtract the to acquire #0.2#. Take it the reciprocal, acquiring the an outcome #5.0#. This has the number #5# before the decimal point and no remainder. So include that to our continued portion 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 do the an approach a little clearer, permit us consider a more complicated example:

Given:

#3.82857142857#

Note the #color(blue)(3)#, subtract it and also 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 round off error"#

Note the #color(blue)(4)#, subtract it and take the mutual 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 number we have actually 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))#