How do you prove:
Q1) Why is every rational number (say m/n, where m and n are both positive integers) either a terminating or a repeating decimal? Q2) Why is every repeating decimal (or terminating decimal) a rational number?
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$\begingroup$ See this answer: math.stackexchange.com/questions/295195/… (for Q2 at least) $\endgroup$– Milo BrandtCommented Oct 5, 2014 at 20:11
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$\begingroup$ See en.wikipedia.org/wiki/… $\endgroup$– HenryCommented Oct 5, 2014 at 20:15
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2 Answers
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For Q1 do long division. Either you reach a remainder $0$ and the division terminates, or you don't. If you don't, the number of possible remainders is finite and you must repeat.
For Q2, if the repeating decimal has $n$ digits, multiply by $10^n$ and subtract the original. If it terminates with $n$ digits after the decimal point, just multiply by $10^n$
I'm fairly sure this is a duplicate, and that answers already exist. But you need to work it through for yourself at least once. I've sketched the key points, but left you to work them out in detail.
answered Oct 5, 2014 at 20:16
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Every rational number not having a finite number of decimals is a repeating decimal because of the following reason. When you make the division of $m$ by $n$ you get, at most, $n$ different remainders. So, if the division doesn't end, after, at most, $n+1$ steps, you get the same remainder you have got in a previous step. From this, you get a periodic repetition of remainders.
As said by @Meelo the other part of the question has been answered in H0w t0 prove that periodic decimal numbers are rational? $a_1...a_k(b_1b_2..b_l)={m \over n}$
