Come si converte 0.916 (6 ripetizioni) in una frazione?

In case you have not encountered it, you can indicate a repeating sequence of digits in a decimal expansion by placing a bar over it.

Così:

#0.91666... = 0.91bar(6)#

#color(white)()#
Metodo 1

Moltiplicato per #100(10-1) = 1000-100# per ottenere un numero intero:

#(1000-100) 0.91bar(6) = 916.bar(6) - 91.bar(6) = 825#

Dividi entrambe le estremità per #1000-100# trovare:

#0.91bar(6) = 825/(1000-100) = 825/900 = (color(red)(cancel(color(black)(75)))*11)/(color(red)(cancel(color(black)(75)))*12) = 11/12#

Perché #100(10-1)# ?

The factor #100# shifts the given number two places left, leaving the repeating section starting just after the decimal point. The factor #(10-1)# shifts the number a further #1# place - the length of the repeating pattern - then subtracts the original to cancel out the repeating tail.

#color(white)()#
Metodo 2

Dato:

#0.91bar(6)#

Recognise the repeating #6# tail as the result of dividing by #3#, so multiply by #color(blue)(3)# trovare:

#color(blue)(3) * 0.91bar(6) = 2.75#

Notare che #2.75# ends with a #5#, so we can attempt to simplify the decimal by multiplying by #color(blue)(2)#:

#color(blue)(2) * 2.75 = 5.5#

Notare che #5.5# ends with a #5#, so we can attempt to simplify by multiplying by #color(blue)(2)# ancora una volta:

#color(blue)(2) * 5.5 = 11#

Having arrived at an integer, we can divide by the numbers we multiplied by to get a fraction:

#0.91bar(6) = 11/(2*2*3) = 11/12#