But you usage the hash vital just before 0.8bar3and additionally 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 political parties by 10

#" "90x=75#

Divide both political parties by 90

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

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

Here"s another means you can transform decimals to fractions if you have a calculator to hand.

You are watching: What is 8.3 repeating as a fraction

We usage the calculator to discover the end continued portion expansion because that the offered number, then unwrap it come a regular fraction.

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

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

#color(blue)(0) + #

Take the reciprocal of the given number to get a an outcome something like: #1.2000000048#. We can ignore the rolling digits #48# as they are just a rounding error. So through our new result #1.2# note that the number before the decimal allude is #1#. Compose that under as the next coefficient in the ongoing fraction:

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

then subtract the to get #0.2#. Take it the reciprocal, obtaining the an outcome #5.0#. This has actually the number #5# prior to the decimal suggest and no remainder. So add that come 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 do the method a little clearer, let us think about a more complicated example:

Given:

#3.82857142857#

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

#1.20689655173#

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

#4.99999999525#

Let"s speak to that #color(blue)(5)# and stop.

Taking the number we have actually 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))#