I would recommend that you interpret "terminating decimal" as a "repeating decimal" where the repeating part happens to be 0. That way, you can simply show that any rational number is a "repeating" decimal.

let m and n be two integers. Thenmeans"m divided by n". If m> n, then there exist q and r ("quotient" and "remainder") such that m= qn+ r with r< n. That is, so it is sufficient to prove this for where r is less than n. When we "long divide" n into r, we will have, at each decimal place, a "remainder" that is less than n and larger than or equal to 0, then bring down a "0" and continue. The critical point is that there can only be ndifferentremainders, 0 to n-1 as I said. After at most n steps, a remainder must repeat, we bring down a "0" again so we are dividing n into exactly the same number as before and everything repeats.