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Prove that the rationals are dense

  1. Jun 15, 2015 #1
    • Member warned about posting homework problems in non-homework sections
    Show that there are infinitely many rational numbers ## \frac{a +m*c}{b + m*d} ## between the two rational numbers ## a/b## ##c/d ##. m is any positive integer.

    My attempt:

    first make common denominator

    ##
    \frac{a*d}{b*d}##
    ##\frac{c*b}{d*b}
    ##
    all numbers going from ##a*d## to ##c*b## divided by ##b*d## is in this interval. but these are finite so we have to increase the resolution. say multiply numerator and denominator by 1000 so now ## a*d*1000 ## to ## c*b*1000 ## divided by ## b*d*1000 ## .

    It can be 100000 or anything very large say multiplied to infinity. how do I write this mathematically?
     
  2. jcsd
  3. Jun 16, 2015 #2

    RUber

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    I am not sure I understand what you are trying to argue. Is ad<cb? How are you going to show the form stated in the problem?
    Try looking at values for m from zero to infinity. You should be able to conclude that they are all in the interval [a/b,c/d).
     
  4. Jun 16, 2015 #3

    Mark44

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    I would do this in steps.
    1. Show that ##\frac{a + mc}{b + md}## is less than c/d, for any choice of a positive integer m.
    2. Show that ##\frac{a + mc}{b + md}## is greater than a/b, for any choice of a positive integer.
    I am assuming that a/b < c/d.
    If you can prove the above propositions, you're done, since m can be any integer in the set {1, 2, 3, ...}
     
  5. Jun 16, 2015 #4

    I want to know how to derive the form ## \frac{a +m*c}{b+m*d} ##
     
  6. Jun 16, 2015 #5

    Mark44

    Staff: Mentor

    I don't believe that you need to derive it, just show that there are an infinite number of rationals of this form between a/b and c/d.

    I think that how it works is that given a/b and c/d, where a/b < c/d, the rational number with m = 1, ## \frac{a +1*c}{b+1*d} ##, is halfway between a/b and c/d. The same fraction with m = 2 gives a number that is 2/3 of the way between a/b and c/d. When m = 3, the fraction gives a number that is 3/4 of the way between a/b and c/d, and so on. These are the results I get when I choose a/b = 1/4 and c/d = 1/2, for a few selected values of m.
     
  7. Jun 16, 2015 #6
    Is not halfway between ## a/b ## and ## c/d ## equal to ## \frac{a*d+c*b}{2*b*d} ##?
     
  8. Jun 16, 2015 #7

    SammyS

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    Yes it is.

    Furthermore, the result for ##\displaystyle \ \frac{a+m\cdot c}{b+m\cdot d} \ ## depends not only on the two rational numbers, a/b and c/d but depends on the particular values used for each of a, b, c, and d .

    For Mark's example with a/b = 1/4 and c/d = 1/2:
    Using a = 1, b = 4, c = 2, d = 4 the result with m = 1 is: ##\displaystyle \ \frac{1+1\cdot 2}{4+1\cdot 4}=\frac38 \ .##

    Using a = 1, b = 4, c = 1, d = 2 the result with m = 1 is: ##\displaystyle \ \frac{1+1\cdot 1}{4+1\cdot 2}=\frac13 \ .##​

    Each of these results is between 1/4 and 1/2. Only the first is halfway between .
     
  9. Jun 17, 2015 #8
    Can I say that my way to find rational numbers inbetween ## a/b ## and ## c/d ## is
    ## \frac{a*d+c*b}{m*b*d} ## where m is an integer from 1 to infinity. I realize that this will miss certain rational numbers but there is no way to encompass all the rational numbers using any formula involving integers like m? because between m = 1 and m = 2 there are infinitely many fractions? so the formula ##
    \frac{a+m\cdot c}{b+m\cdot d}
    ## is also similar to mine missing some rational numbers so whose formula has higher resolution?

    Thank you!
     
  10. Jun 17, 2015 #9

    haruspex

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    that only works for m=2. You need some more references to m in the expression.
    Anyway, the way I read the question, you are required to show specifically that there are infinitely many of the form ##\frac{a +m*c}{b + m*d}##
     
  11. Jun 17, 2015 #10

    RUber

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    It seems pretty straightforward that if m = 0, you have a/b and the limit as m goes to infinity will surely be c/d. From my vantage point, the only challenge would be to show explicitly that ##\frac ab \leq \frac{a+mc}{b+md} < \frac cd ## for all natural m.
    This might be easier if you look at this as some sort of average.
    Perhaps a is a total number of something done by b people, and c is the total number done by d people, the ##\frac{a+c}{b+d}## is the average done by all the people.
    If every additional group of d people can do c things, than you can continue to average these events out by taking the total number done (a+mc) and dividing by the total number of people (b+md).
    From this perspective, it should be clear that your mean can never exceed your largest value (c/d) and it can never be less than your smallest value (a/b).
     
  12. Jun 17, 2015 #11

    SammyS

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    I agree with haruspex here. (I usually do.)

    If you can show that ##\displaystyle \ \displaystyle \ \frac{a+c}{b+d} \ ## is between ##\displaystyle \ \frac ab \ ## and ##\displaystyle \ \frac cd \,,\ ## then that will show that ##\displaystyle \ \frac{a+m\cdot c}{b+m\cdot d} \ ## is between ##\displaystyle \ \frac ab \ ## and ##\displaystyle \ \frac cd \ ## in general. Can you see why?
     
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