Why **Excursion length** is less than **Denominator**? Can anyone prove that the Excursion length of any fraction's decimal representation which is cyclic, is less than its Denominator?

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Why **Excursion length** is less than **Denominator**? Can anyone prove that the Excursion length of any fraction&#039;s decimal representation which is cyclic, is less than its Denominator?

Olive Avila · Accepted Answer

Let us consider       1    7   as an example. Successive digits of this fraction are determined by division with remainder. Since 7 does not fit into 1, we start with  0.with a remainder of 1. Then for the second digit, we multiply this remainder by 10, giving 10, and repeating: 7 fits into 10 once, with remainder 3. We obtain  0.1and we repeat. 7 fits into 30 four times, with remainder 2; into 20 twice, with remainder 6; into 60 eight times, with remainder 4; into 40 five times, with remainder 5; and into 50 seven times with remainder 1. This gives  0.142857and now we are back to the start of remainder 1. Thus the number starts repeating. The repeating part could never have been longer than 7, because there are only 7 remainders to cycle through: 0,1,2,3,4,5,6. And in fact, the cycle always had to be less than 7, because if the remainder 0 ever occurs, then the fraction just terminates (which gives cycle length 1), so the remainders can only really cycle through 1,2,3,4,5,6.Now, the same works for       m    n   for any   m  &amp;gt;  0  ,  n  &amp;gt;  1

Why Excursion length is less than Denominator? Can anyone prove that the Excursion length of any fraction's decimal representation which is cyclic, is less than its Denominator?

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