Question #aee35

As it happens, any real number whose decimal representation has a portion which repeats indefinitely or terminates* can be expressed as a ratio of integers, and thus is rational.

We can find the fraction using a bit of algebra.

Suppose #0.bar(a_1a_2...a_n)# is the decimal representation of a number (note that the bar denotes a repeating portion).

Let #x = 0.bar(a_1a_2...a_n)#

#=> 10^nx = a_1a_2...a_n.bar(a_1a_2...a_n)#

#=> 10^nx - x = a_1a_2...a_n.bar(a_1a_2...a_n) - 0.bar(a_1a_2...a_n)#

#=> (10^n-1)x = a_1a_2...a_n#

#=> x = (a_1a_2...a_n)/(10^n-1)#

So, in our given example, we have

#0.bar(5) = 5/(10^1-1) = 5/9#

Note that this process can be generalized to handle cases in which the repeating portion starts after nonrepeating digits as well.

*A terminating decimal can be converted to a fraction by multiplying and dividing by an appropriate power of #10#. E.g.

#0.1234567 = 0.1234567*10^7/10^7 = 1234567/10^7#

#color(brown)("Note")#
If the use of the #x# confuses you do the following: every time you see #x# think of it as saying "the unknown fraction value"

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Let #x=0.555bar5" " larr" "# the 5's go on for ever

So #10x=5.555bar5#

So #10x-x# is:

#10x=5.555bar5#
#ul(color(white)(10)x=0.555bar5) larr" Subtract"#
#color(white)(1)9x=5#

Divide both sides by 9

#x=5/9#