How I perform it in mine head is that I begin from the portion #6/6# and also subtract #1/6#.

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

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

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

Divide #100# by #6#.Divide through #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)#

Finally, ours answer is:

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

#= 1 - 0.1bar(66)#

#= color(blue)(0.8bar(33))#


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EZ together pi
Jul 9, 2016

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


Explanation:

6 can not be multiplied by any type of number come give specific power that 10, so that is not possible to create an equivalent fraction to obtain a decimal.

Divide 5 by 6:

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

See more: About How Many Songs Can 16Gb Iphone Hold ? How Many Songs Is 16 Gb

It is usually rounded come 2 or 3 decimal places.


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Tony B
Jul 9, 2016

#0.83bar3#


Explanation:

#color(blue)("This an approach 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 deserve to be written as #50/10# so us have

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

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

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

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

But #2/100# deserve to be written as #20/1000# giving

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

#0.8333.."repeating "3#

Mathematically this is composed as

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

Notice the bar over the critical 3 in #0.83bar3# and the period over the critical 3 in #0.83dot3#


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