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Proof to find fraction inbetween to fraction

  1. Aug 26, 2012 #1
    I keep getting slightly confused with the algebraic method of finding a fraction between, two other fraction. Here is an example question, I have been doing

    Find the fraction between 13/15 and 14/15? I personally convert both to percentages and find the average between the two, the convert back to a fraction. So in this case I did:

    13/15 = 0.8666666 = 87%
    14/15 = 0.933333 = 93%
    87+93 = 180/2= 90
    90/100 = 9/10: which I believe is correct

    The method I do not understand is the proof behind this method: 13/15+14/15 = 27/30 / 3 top and bottom you get 9/10. But its the proof that confuse me. I will show where I get confused:

    I understand this part: a/b < c/d cross multiply ad < bc

    This is the part I don't understand: Add ab to both sides ab+ad < ab+bc

    why add ab to both sides, where dose this come from?

    then factor : a(b+d)<b(a+c)⇒ a/b < a+c/a+b

    Then you add cd to both sides, once again why? then you factor out again.
     
  2. jcsd
  3. Aug 26, 2012 #2
    Think of it this way:

    You start with 13/15 and 14/15

    13/15 is equal to 26/30
    and
    14/15 is equal to 28/30

    This is done by a simple multiplication of two on both the numerator and the denominator.

    Looking at your new fractions, it is obvious that 27/30 is in between the two of them, and 27/30 simplifies to 9/10
     
  4. Aug 26, 2012 #3

    Mentallic

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    Homework Helper

    Because it works. Notice in the next line the factoring worked out such that we can get a/b on its own on the left side.

    Just a typo but it should be a/b < (a+c)/(b+d)

    I guess when you say you add cd to both sides you're talking about

    [tex]ad<bc[/tex]

    [tex]cd+ad<cd+bc[/tex]

    [tex]d(a+c)<c(b+d)[/tex]

    [tex]\frac{a+c}{b+d}<\frac{c}{d}[/tex]

    Which again work exactly the way we want it to. We've now just shown that [tex]\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}[/tex]

    By using algebraic manipulations that were cleverly used to give us the answer we were looking for.
    But keep in mind that this value [tex]x=\frac{a+c}{b+d}[/tex] is not always exactly in the middle of a/b and c/d. When b and d are different, it doesn't turn out to be the average of the two fractions.

    If you wanted the average of a/b and c/d as your in-between fraction, then you'd have

    [tex]x=\frac{\frac{a}{b}+\frac{c}{d}}{2}[/tex]
    [tex]=\frac{ad+bc}{2bd}[/tex]

    Which is a lot more calculations than the value of x obtained from the proof above.
     
  5. Aug 27, 2012 #4

    pwsnafu

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    Science Advisor

    Trivia: the fraction ##\frac{a+c}{b+d}## is called the mediant of ##\frac{a}{b}## and ##\frac{c}{d}##.

    Further the mediant does not preserve order. Suppose that ##\frac{a}{b} < \frac{A}{B}## and ##\frac{c}{d} < \frac{C}{D}##, but it is possible to have ##\frac{a+c}{b+d} > \frac{A+C}{B+D}##. This is known as Simpson's paradox.
     
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