Why Are Repeating Decimals Less Than Denominator?

[uranium_235]

If the repeating decimal is equal to a/b, why is the number of repeating digits less than the denominator b?
This was a question on my homework, and I could not find mathematical proof for the conclusion that number repeating digits is always less than b.

 

[Hurkyl]

The key lies in doing long division (at least that's how I discovered it way back when); run through a few dozen examples, maybe you'll pick up on it.

 

[uranium_235]

I don’t wish to seem a bit impatient, but is it possible one could post the mathematical proof? We will most likely be going over this in class tomorrow, and I do not wish to be behind in the event I am not able to find the answer before then.

 

[Tide]

Okay - imagine performing long division of an integer by an integer. In each step you have a remainder. There are only N possible remainders (from 0 up to N-1). If one of them is 0 then you're done. Otherwise, of the other possible remainders (from 1 up to N-1) you can only go through N-1 steps before one of those very same remainders occurs once again - it's inevitable! At that point, each step repeats the previous ones and therefore your quotient must repeat over and over!

 

[uranium_235]

Thank you for your help, Tide and Hurkyl.