How I do it in my head is that I start from the fraction #6/6# and subtract #1/6#.

You are watching: What is 5/6 in decimal form

#6/6# is just a ratio that is equal to #1#. Any number divided by itself is equal to #1#. So, we know that #5/6# is less than #1.bar(00)#.

Next, to find #1/6#, what you can do is find what numbers get close to #100# upon multiplying by #6#, then look at the remainder and divide by #6#. So, you can:

Divide #100# by #6#.Divide by #100# again.

Since #6xx16 = 96#, we have:

#100/6 = stackrel("quotient")(overbrace(16)) + stackrel("remainder")(overbrace(4/6)#

#= 16 + 2/3#

#= 16.bar(66)#

Therefore:

#((100)/6)/100 = 1/6 = 0.1bar(66)#

#color(blue)(5/6) = 6/6 - 1/6#

#= 1 - 0.1bar(66)#

#= color(blue)(0.8bar(33))# EZ as pi
Jul 9, 2016

#5/6 = 0.833# or shown as #0.8dot3#

Explanation:

6 cannot be multiplied by any number to give an exact power of 10, so it is not possible to write an equivalent fraction to get a decimal.

Divide 5 by 6:

#5÷6 = 0.8333333...# The answer is a recurring decimal.

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It is usually rounded to 2 or 3 decimal places. Tony B
Jul 9, 2016

#0.83bar3#

Explanation:

#color(blue)("This method is "Long Division" in disguise")#

Think of 5 as #5+ 0/10+0/100+0/1000+....#

Observe that #5 ( 5 is less than 6 )

But 5 can be written as #50/10# so we have

#50/10-:6" " ->" " (50-:6)/10" " =" " 8/10+" remainder of " 2/10#"~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~So now we have #5-:6 = 8/10+(2-:6)/10#

But #2/10 #can be written as #20/100# giving

#8/10+(20-:6)/100" "->" "8/10+3/100+ "remainder of "2/100#"~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~So now we have:

#5-:6=8/10+3/100+(2-:6)/100#

But #2/100# can be written as #20/1000# giving

#8/10+3/100+(20-:6)/1000# giving #8/10+3/100+3/1000+(2-:6)/1000#"~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~This process repeats indefinitely so we can write this in decimal form as:

#0.8333.."repeating "3#

Mathematically this is written as

#0.83bar3# or #0.83dot3# depending on your preference.

Notice the bar over the last 3 in #0.83bar3# and the dot over the last 3 in #0.83dot3# 