Proof: Suppose #\frac{a}{b}# is a rational number, in which #a# and #b# are integers and #b\gt0#. If we perform the long division by #b# on the decimal development #a.000\cdots#, then, at each step after the decimal point of #a.000\cdots#, a new decimal is added. We keep adding a #0# to the decimal development of #a#, before we divide again in order to determine the next decimal digit of the quotient. At each step, a remainder results: a number less than #b#. There are no more than #b+1# possibilities for that remainder. This means that after at most #b+1# steps the remainder is equal to what it was once before. After this occurrence, we go through the same steps as before to determine the next decimal digit, and therefore get the same result. We conclude that the decimal development a repetitive tail.

Conversely, imagine #a# is a real number with a #w#-repetitive tail for a given series #w# consisting of #n# numbers. We have to show that #a# is a rational number. If #a\le0#, we replace #a# by #-a#. We can do this without loss of generality, since #-a# is rational if and only if #-a# is rational. If the #w#-repetitive tail of #a# starts at the #l#-th decimal, this development has the form #m.a_1a_2\cdots a_{l-1} www\cdots#, for an integer #m# and digits #a_1,a_2,\ldots,a_{l-1}#. This means that \[a=r+10^{-l}(w+\frac{w}{10^n}+\frac{w}{10^{2n}}+\cdots)\] where #r=m+\frac{a_1a_2\cdots a_{l-1}}{10^{l-1}}# is a rational number.

It is therefore sufficient to prove that the contribution of the tail #10^{-l}\left(w+\frac{w}{10^n}+\frac{w}{10^2}+\cdots\right)# is rational. We can write this number as #\frac{w}{10^l}\left(1+10^{-n}+{10^{-2n}}+\cdots\right)#. Thanks to the theory of geometrical series, we know that #1+10^{-n}+{10^{-2n}}+\cdots = \frac{1}{1-10^{-n}}= \frac{10^n}{10^n-1}#. This is a rational number. This establishes that #a=m+\frac{w\cdot 10^{nl}}{10^n-1}# is rational.