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Homework Help: Why some repeating decimals have nonrepeating part? I think its blasphemy! Help!

  1. Nov 30, 2011 #1
    Why some repeating decimals have nonrepeating part? I think its blasphemy! :D Help!

    Example: Fraction 1/6 has a decimal notation of 0.1(6)

    Why do other primes in denominator, aside from 2 and 5, cause a
    nonrepeating part (numeral 1 in 0.1(6)) in decimal notation?

    Ok, so my book says that if the denominator of a fraction contains,
    aside from primes 2 and 5, some other primes like 3, 7 or others, than
    only after first few nonrepeating numerals comes the repeating part,
    called period.

    I have this in my mind: 1/6=(1/2)x(1/3),

    so, because 1/3 is 0.3333....,

    we have 1/6=(1/2)x 0.3333....,

    by distribution 1/6=(1/2)x(3/10)+(1/2)x(3/100)+(1/2)x(3/1000)+...

    and now i cant see why it gives that nonrepeating part...
  2. jcsd
  3. Nov 30, 2011 #2


    Staff: Mentor

    Re: Why some repeating decimals have nonrepeating part? I think its blasphemy! :D Hel

    Another way to look at it is
    1/6=(1/2)x 0.3333....,
    = (1/2) * 3 * .1111...
    = 1.5 * .1111...
    = .1111... + (.5)*.1111...

    That term on the right is .055555..., so the sum of both terms is

    I'm not sure what all the fuss is about. The repeating part doesn't have to come at the beginning of the decimal representation. The important thing is that if there is an endlessly repeating pattern, whether the pattern starts at the decimal point or not, then the number can be expressed as the quotient to two integers.
  4. Dec 1, 2011 #3
    Re: Why some repeating decimals have nonrepeating part? I think its blasphemy! :D Hel

    Thanks for your reply. See, i know that whenever we have some prime factor in the denominator aside from 2 and 5 then we will have a group of numerals repeating forever. Its the nature of primes (that is, consequence of their non-divisibility by anything aside from them self and 1) and it is also due to our number system, based on product of primes 2 and 5, which implies that only those fractions having denominators composed solely of those (2 and 5) primes will yield a terminating decimal. So if we have 3, 7, 23 or any other prime (or product of those) in our denominator we will have a repetend repeating forever. I understand that. But when we have denominator containing both 2´s, 5´s and other primes, then we will also have that non repeating part followed by repetend. I can see that in this case decimal notation of that fraction will also have repetend (because denominator contains primes other than 2 or 5) but i cant see why that extra 2´s and 5´s are responsible for the non-repeating part. When i do the long division for 1:6 i know that 6 goes into 10 one time, remainder 4, now 6 goes into 40 six times, remainder 4, and now we came to that loop. What i dont know (cant relate) why is that factor of 2 (in denominator) causing this "postponement" of the repeating part. I know it sounds trivial to you, but that lack of insight is really bugging me, and i cant get a hold of it.
  5. Dec 1, 2011 #4


    User Avatar
    Science Advisor

    Re: Why some repeating decimals have nonrepeating part? I think its blasphemy! :D Hel

    1/6 = 1/10 + 1/15 = 1/10 + 2/30 = 1/10 + (1/10)(2/3).

    let's see how the "2 factor" makes this happen:

    1/6 = (1/2)(1/3) = (5/10)(1/3) = (1/10)(5/3) = (1/10)(1 + (2/3)).

    when we add a 5 to the numerator to get a 10 in the denominator (because powers of 10 are "transparent" to our decimal system, they just shift the place of digits, not the digits themselves), we see that 1/6 is really 5/3 in disguise (shifted by a decimal point).

    the integer part of 5/3 isn't going to be a repeating anything, it's just an integer. the 2/3 left-over from taking out the integer part, is what controls the repeating part.

    try this same logic with 1/12, which has 2 factors of 2 in the denominator.

    1/12 = (1/4)(1/3) = (25/100)(1/3) = (1/100)(25/3) = (1/100)(8 + 1/3).

    so, without even doing the division, you can see that:

    1/12 = 0.08 + (something)

    and that "something" is 0.333333..... shifted over 2 decimal places (2 for 100),

    so 1/12 = 0.083333.....

    anything with a factor of 2 or 5 in the denominator will display this behavior, because we need to "turn the 2's and 5's to 10's" before we can see the "repeating part".
  6. Dec 2, 2011 #5
    Re: Why some repeating decimals have nonrepeating part? I think its blasphemy! :D Hel

    Thank you Deveno, i can see it more clearly now. I was becoming desperate, laughing hysterically, wanting to pull out my eyeballs (later on i have calmed myself... and pulled out someone elses :D).

    Anyway, i can see now, just as you explained, that i can decompose any denominator, for example 7/15 = (1/5) x (7/3), now i can write 1/5 as fraction whose denominator is a power of ten, that is, 2/10. Now because of commutative property i can write (1/10) x (14/3) giving me
    0.1 x 4.6666... = 0.46666... and i can see why i have that nonrepeating numeral (4). Is my reasoning valid here?

    As you called it "transparency", i have noticed too that whenever there is equal number of factors of 2 and 5 in the denominator it causes only a shift of decimal point. So when we have, say 3 factors of 2 and 3 of 5, its just 10 x 10 x 10, therefore shifting the decimal point 3 places to the left. Am i right?

    And finally, i have one more thing that isnt so clear: i cant relate, when doing the same thing (like the example above) but with a long division algorithm (without decomposing denominator like you shown), how that 2, or 5 in the denominator is causing right away to give me that non-repeating part. For instance, here it would be: 7 x 10 divided by 15 gives us 4 (actually 4/10), remainder 10, 10 x 10 divided by 15 gives us 6 (actually 6/100), remainder 10 (and so on). How is this different?
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