Using the notation that a bar over a digit or string of digits denotes them repeating infinitely, let x = 0.bar(3)x=0.¯3
=> 10x = 3.bar(3)⇒10x=3.¯3
=> 10x - x = 3.bar(3)-0.bar(3)⇒10x−x=3.¯3−0.¯3
=> 9x = 3⇒9x=3
:. x = 3/9 = 1/3
This technique will actually work for converting any repeating decimal into a fraction. Suppose we have a repeating decimal x=0.bar(a_1a_2...a_n) (that is, with n repeating digits). We multiply x by 10^n, subtract x, and then divide by 10^n-1 to obtain the fractional form:
10^nx = a_1a_2...a_n.bar(a_1a_2...a_n)
=> 10^nx - x = a_1a_2...a_n.bar(a_2a_2...a_n)-0.bar(a_1a_2...a_n)
=> x(10^n-1) = a_1a_2...a_n
:. x = (a_1a_2...a_n)/(10^n-1)
Note that once we know the technique, we can skip to the end as a shortcut: 0.bar(a_1a_2...a_n) = (a_1a_2...a_n)/(10^n-1). In the given problem, only a single digit, 3, is repeating, so the shortcut gives us: 0.bar(3) = 3/(10^1-1) = 3/9 = 1/3.
We can also modify the technique to handle cases where the repeating portion does not begin until after a finite number of non-repeating digits, such as numbers like 0.1232323...=0.1bar(23). The idea is still to multiply by an appropriate power of 10 and then subtracting away the repeating portion. As an exercise, a student can try finding the fractional form of such a number.
