Periodicity of Decimal Expansion

[Gear300]

I am asked to prove that a real number is rational if and only if it has a periodic decimal expansion.

I have shown that any periodic decimal expansion has an integer p such that multiplication returns an integer. For the case of showing that all rational numbers have a periodic decimal expansion, I have shown that the expansion can eventually become periodic (repeated 9's being a trivial case)...but I'm not sure if this is what is being. If it isn't, then is it true that any rational number has an entirely periodic decimal expansion (I can't really come up with an example for some numbers)?

 

[uart]

If it isn't, then is it true that any rational number has an entirely periodic decimal expansion (I can't really come up with an example for some numbers)?

Yes it is true (assuming that by "periodic" you mean either recurring or terminating).

I'm not sure of the best way to formalize this into a proof, but the standard way of demonstrating it is by long division (of "n" by "d"). Here it is obvious that with each division by "d" there can be at most "d" different remainders (including a zero remainder). So eventually (that is, after you've gone through all the digits of the dividend "n" and are working on the ".0000..." part at the end) if you ever get a zero remainder then the decimal expansion terminates right there. Now if no zero remainder occurs but the same remainder ever reoccurs then the decimal expansion must repeat from that point.

This demonstrates that the decimal expansion of the fraction "n/d" either terminates or it repeats with a maximum cycle length of "d-1" digits (corresponding the the "d-1" maximum number of possible different non zero remainders that you can get when you divide by "d").

Do the long hand division of 1 divide 7 (7 into 1.00000000...) and you'll soon see how this works in practice. You keep getting a different non-zero remainder for the first 6 divisions, but obviously that situation can't continue for ever and of course you eventually get a remainder that's you've already had, and at that point the decimal expansion repeats.

 

[HallsofIvy]

I would just put the specific calculation uart talks about in general terms:

Saying that a number has "periodic" decimal expansion means it is of the form "$N.a_1a_2...a_nb_1b_2...b_mb_1b_2...b_m...$" where "N" is the integer part, $a_1a_2...a_n$ is the non-repeating decimal part, and $b_1b_2...b_m$ is the part that now repeats endlessly.

Let $x= N.a_1a_2...a_nb_1b_2...b_mb_1b_2...b_m...$. Then $10^nx= 10^nN+ a_1a_2...a_n.b_1b_2...b_mb_1b_2...b_m...$ and $10^{n+m}x= 10^{n+ m}N+$$a_1a_2...a_nb_1b_2...b_m.b_1b_2...b_mb_1b_2...b_m...$ so that the difference is
$(10^{n+m}- 10^n)x= 10^{n+ m}N$$+ a_1a_2...a_nb_1b_2...b_m- 10^nN+ a_1a_2...an$

That is, an integer times x is equal to an integer so x is a fraction.