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Prove every rational number has a never ending decimal view

  1. Oct 17, 2011 #1
    Its all in the title: How can we prove every rational number has a never ending decimal view. Since I translated the question from another language, I didn't know how to put it.


    I know it's true because:
    1=0.9999....
    and i.e.:
    1/3=0.333...
    or
    24.67=24.669999...

    The problem is proving it. Any ideas?
     
  2. jcsd
  3. Oct 17, 2011 #2

    SammyS

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    Welcome to PF .

    Look at a long division algorithm .

    (What is this doing in the Intro. Physics section ??)
     
  4. Oct 17, 2011 #3
    Sorry. My bad about posting it here.
    Can you expand on the "long division"?
     
  5. Oct 18, 2011 #4

    HallsofIvy

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    By "long division" algorithm, he just means the standard way you divide one number by another: to divide 3 by 7, you would write out a 3.00000000000000000, with as many 0s as you felt you needed and start dividing:

    7 divides into 30 4 times with remainder 2 so you write "4" above and "bring down" a 0 beside the 2.

    7 divides into 20 twice with remainder 6 so you write "2" above and "bring down" a 0 beside the 6.

    7 divides into 60 8 times with remainder 4 so you write "8" above and "bring down" a 0 beside the 4.

    7 divides into 40 5 times with remainder 5 so you write "5" above and "bring down" a 0 beside the 5.

    7 divides into 50 7 times with remainder 1 so you write "7" above and "bring down" a 0 beside the 1.

    7 divides into 10 once with remainder 3 so you write "1" above and "bring down" a 0 beside the 3.

    7 divides into 30 4 times with - but wait! That is exactly the division we did to start all this! We will get exactly the same quotient and remainder, of course, and because we will also "bring down" a 0, exactly the same sequence of divisions will repeat.

    At this point we know the 3/7 is just a decimal point and "428571" repeated. Further, we knew that had to happen because the remainder at each step, when we divide by 7, must be a non-negative number less than 7. There are only 7 possible remainders so one must repeat within no more than 7 divisions.

    And this is true for any rational number. To reduce the rational number m/n to decimal form, we divide m by n. But any remainder must be a non-negative number less than n. which means that within n divisions, a remainder must repeat and we will just repeat everying again, or will be 0, a "terminating" decimal, which we can then think of as "repeating 0s" or, as you have it, "repeating 9s".
     
  6. Oct 18, 2011 #5
    Thanks HallsofIvy...
    I totally completely understand the whole terminating and repeating decimal concept.
    The thing is I just wanted to know if there is any written mathematical form of proof for it. By the answers I've gotten so far it means it's just a matter of understand the question.
    I guess the proof is just verbal and a simple explanation.
     
  7. Oct 18, 2011 #6
    From wiki (http://en.wikipedia.org/wiki/Rational_number):
    "The decimal expansion of a rational number always either terminates after finitely many digits or begins to repeat the same finite sequence of digits over and over. Moreover, any repeating or terminating decimal represents a rational number. These statements hold true not just for base 10, but also for binary, hexadecimal, or any other integer base."

    This means that some rational numbers actually do terminate at some decimal, thus you can't prove it for any rational number [itex]\mathbb{Q}[/itex]. You must first find out which rational numbers has an infinite decimal sequence [itex]\left(\text{say } Q_{\infty}\right)[/itex], and why it is so, and then rephrase your original statement to prove it for them.
     
  8. Oct 18, 2011 #7
    Thanks for taking your time Hioj;
    But even terminating decimals have an infinite decimal sequence:
    Take 0.3, the infinite decimal sequence would be 0.2999...
    These two are exactly equal (if you transform 0.2999... to a fraction, it would be 3/10)
    So all rational numbers do have an infinite decimal sequence.
     
  9. Oct 18, 2011 #8

    SammyS

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    You are correct here.

    In fact, every rational number has a unique non-terminating decimal representation. Moreover, the sequence of digits ends in a finite length repeating pattern.

    (I think that made sense.)
     
  10. Oct 19, 2011 #9
    SammyS thanks for agreeing...
    Though I'm still looking for a mathematical way of proving it.
     
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