Nerding out

This morning, I had some nerdiness busting out. I was thinking, and I figured out a way to determine how many decimal places come from a fraction. The first step is to reduce it to the simplest form--1/2 instead of 4/8 or 5/6 instead of 25/30. Then you can ignore the numerator. The denominator is all that matters. The next step is to attempt to reduce the denominator to the form (2^x)(5^y) where x and y are whole, non-negative numbers (0, 1, 2, etc.). If the denominator can't be reduced to this form, the decimal version goes on infinitely. If it can be reduced to this form, compare x and y. The higher of the two  numbers is how many decimal places your answer will have. For example, a denominator of 40 would reduce to (2³5¹), giving 3 as the higher of x and y. This is how many decimal places will always be in a number with 40 as the denominator (unless it can be reduced to a smaller denominator). This stems from the fact that decimals come from exact powers of 10, which can be written as (2^x5^x) and anything with those numbers the same (or lower) fits evenly.

Now is there a way to figure out what denominator(s) you should pick to have a decimal that is a certain length? Yes. Let's imagine that you are a math teacher, and you want a problem to have a decimal with 3 places. So you take your (2^x)(5^y) form and set x or y equal to 3. With x, that gives 2³. Then vary y from 0 to 3, giving you (2³)(5⁰)=8, (2³)(5¹)=40, (2³)(5²)=200, and (2³)(5³)=1000. Then starting from this last, vary x down to 0 to get (2²)(5³)=500, (2¹)(5³)=250, and (2⁰)(5⁵)=125. Then you can always multiply the numerator and denominators by whatever number you want to make the problem seem harder. Just be careful to not make it possible to reduce the fraction further than you wanted.

I ran through all of this in about 15 minutes this morning--for no real purpose. Yeah, I'm a nerd. And proud of it.