The Cool Thing about Repeating Decimals

Get out your calculator and type in 331/999 and press enter. What do you get? Something along the lines of 0.331331331…? Now type in 21/9999 and press enter. What do you get? 0.002100210021…? This cool trick works with any fraction less than 1 with all nines in the denominator, which are always of the form 10^A – 1, for some number A. Here’s a proof:

• Say x = 0.A₁A₂A₃…A_n…… (Where x is a repeating decimal, and A₁A₂A₃…A_n is the string of numbers that is repeated in the decimal, and n is the length of the string)
• If we multiply both sides by 10ⁿ, then we would get: x10ⁿ = A₁A₂A₃…A_n.A₁A₂A₃…A_n…, because the decimal point is moved n times to the right
• x10ⁿ = A₁A₂A₃…A_n.A₁A₂A₃…A_n… is the same thing as A₁A₂A₃…A_n + 0.A₁A₂A₃…A_n, and since 0.A₁A₂A₃…A_n = x, then x10ⁿ = A₁A₂A₃…A_n + x
• By subtracting both sides of the equation by x, you would get x10ⁿ – x = A₁A₂A₃…A_n, or x(10ⁿ – 1) = A₁A₂A₃…A_n
• By dividing both sides by 10ⁿ – 1, you would get… x = (A₁A₂A₃…A_n)/(10ⁿ – 1)
• Therefore, by substituting 0.A₁A₂A₃…A_n… for x, you would get…
• A₁A₂A₃…A_n… = (A₁A₂A₃…A_n)/(10ⁿ – 1), where A₁A₂A₃…A_n is a string of numbers of length n, and 10ⁿ – 1 is a string of n 9s, for example 10³ – 1 = 1000 – 1 = 999

This proof is a generalization of a trick you learn in middle school. Say you want to convert 0.21… into a fraction. You would set that equal to x to get x = 0.21… and multiply both sides by 100 to get 100x = 21.21…, which is 100x = 21 + 0.21…, or 100x = 21 + x. Next you would subtract x from both sides to get 99x = 21, and x = 21/99. Don’t forget to simplify J.

Note: If you want to do something like 7/99999, add 0s in front of the 7 to make the numerator and denominator have the same number of digits to get 00007/99999, and the decimal of that would be .000070000700007…