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What guarantees the 'greater than' sign between r/(r+b)>(r-1)/(r-1+b)?

  1. Jun 8, 2013 #1
    Hi all,
    I've been trying to figure out why this statement is true:

    [tex] \frac{r}{r+b} > \frac{r-1}{r+b-1} \quad \text{for all } b>0 [/tex]

    I can't seem to reason it out. I've plugged in a few values greater than 0 and yes, it works out. But I don't understand how I can look at this and find it true.

    I'm looking at it as such:
    frac{r}{r} where we minus b from the denominator.
    frac{r-1}{r-1} where we minus b from the denominator as well.

    The only difference is that perhaps frac{r-1}{r-1} is a smaller 1 than frac{r}{r}?

    If anyone could perhaps show me a more reasonable way of looking at this, I would greatly appreciate it! :)

    Thanks in advance!
     
    Last edited by a moderator: Jun 8, 2013
  2. jcsd
  3. Jun 8, 2013 #2

    D H

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    It's not necessarily true. Try r=1/4, b=1/4.
     
  4. Jun 8, 2013 #3

    DrClaude

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    Have you tried multiplying both sides by ##\frac{r+b}{r}##?
     
  5. Jun 8, 2013 #4
    What happens when r = 1/2 and b = 1/3?
     
  6. Jun 8, 2013 #5
    Hm. I see your point. The thing is - this is part of a solution to a probability problem. Here's the problem:

    A drawer contains red socks and black socks. When two socks are drawn at random, the probability that both are red is 1/2. (a) How small can the number of socks in the drawer be? (b) How small if the number of black socks is even?

    I got as far to this part:

    Let r be the number of red socks.
    Let b be the number of black socks.

    So,

    [tex] \frac{r}{r+b} * \frac{r-1}{r-1+b} = \frac{1}{2} [/tex]

    And looking at the solutions, they continued with this:

    Notice that,

    [tex] \frac{r}{r+b} > \frac{r-1}{r+b-1} , for b > 0 [/tex]

    Therefore we can create the inequalities:

    [tex] (\frac{r}{r+b})^2 > \frac{1}{2} > (\frac{r-1}{r+b-1})^2 [/tex]
    [tex] \frac{r}{r+b} > \frac{1}{\sqrt{2}} > \frac{r-1}{r+b-1} [/tex]

    From the first inequality we get

    [tex] r > \frac{1}{sqrt{2}} * (r+b) [/tex]

    or

    [tex] r > \frac{1}{sqrt{2} - 1}*b = (sqrt{2} + 1)* b [/tex]

    From the second we get

    [tex] (sqrt{2}+1)*b > r - 1 [/tex]

    or all told

    [tex] (sqrt{2} + 1)*b + 1 > r > (sqrt{2} + 1)*b[/tex]

    For b =1, r must be greater than 2.414 and less than 3.414, and so the candidate is r = 3. for r = 3, b=1, we get

    [tex]P(2 red socks) = \frac{3}{4} * \frac{2}{3} = \frac{1}{2} [/tex]

    And so the smallest number of socks is 4.

    Questions I have about the above solution:
    1. Why by taking the square of the two components that I could have 1/2 being between the two components? I don't understand how they got there at all. And why did they take the square of the components in the first place? What are they trying to get to?

    2. How did they get this: [tex] r > \frac{1}{sqrt{2} - 1}*b = (sqrt{2} + 1)* b [/tex] from [tex] r > \frac{1}{sqrt{2}} * (r+b) [/tex]?

    3. I get how they got [tex] (sqrt{2} + 1)*b + 1 > r [/tex] from [tex] (sqrt{2}+1)*b > r - 1 [/tex], but I don't understand why they had to add the third component [tex] (sqrt{2} + 1)*b + 1 > r > (sqrt{2} + 1)*b[/tex] with [tex] r > (sqrt{2} + 1)*b[/tex]


    Thanks in advance! :D
     
  7. Jun 8, 2013 #6

    DrClaude

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    So, have you tried multiplying both sides by ##\frac{r+b}{r}##?
     
  8. Jun 8, 2013 #7
    @DrClaude - Yea I tried multiplying both sides by [tex] \frac{r+b}{r} [/tex], and got 0 > -b, which is false. So are you saying that the solution in the book is actually incorrect?
     
  9. Jun 8, 2013 #8

    DrClaude

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    How did you get that? On the LHS, for instance:
    $$
    \frac{r}{r+b} \frac{r+b}{r} = 1
    $$
     
  10. Jun 8, 2013 #9
    @DrClaude - Yes I got 1 for the LHS, but the RHS became quite complicated. For the RHS, I got

    [tex] \frac{r^2+br-r-b}{r^2+br-r} [/tex]

    So what do you do with that?
     
  11. Jun 8, 2013 #10

    DrClaude

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    You have
    $$
    1 > \frac{r^2+br-r-b}{r^2+br-r}
    $$
    which should be obvious why it is true for ##r \ge 1## and ##b>0##.
     
  12. Jun 8, 2013 #11
    @DrClaude - I don't see how you got to the conclusion. How did you get there?
     
  13. Jun 8, 2013 #12

    DrClaude

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    Set ##x \equiv r^2 +br - r##. For ##r \ge 1## and ##b>0##, you have ##x > 0##. Consider
    $$
    1 > \frac{x - b}{x} = 1 - \frac{b}{x}
    $$
    for ##b>0## and ##x>0##.
     
  14. Jun 8, 2013 #13
    @DrClaude - Ok, that makes sense. So you're saying that, in the solution the statement with b > 0 is only true when r >= 1 is guaranteed?

    Also - would you happen to know the answer to the 3 questions I posted as quoted below:

     
  15. Jun 8, 2013 #14

    DrClaude

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    You have
    $$
    x \times y = \frac{1}{2}
    $$
    Knowing that ##x > y##, you can replace ##y## by ##x## and write
    $$
    x^2 > \frac{1}{2}
    $$
    or replace ##x## by ##y## and get
    $$
    y^2 < \frac{1}{2}
    $$
    You then combine both inequalities into one.

    This is just basic algebra. Start by collecting all the terms in ##r## on the LHS.

    You have two inequalities concerning ##r## which you can combine to put both an upper and a lower bound on its value.
     
  16. Jun 8, 2013 #15

    DrClaude

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    As others have shown, the inequality doesn't hold at least for some values where ##r < 1##. I haven't spent the time to check what is the minimum value of ##r## for which it holds, but ##r \ge 1## is guaranteed to work. As you are dealing with a number of socks, i.e., ##r \in \mathbb{N} ##, this is good enough!
     
  17. Jun 8, 2013 #16
    @DrClaude - Thank you for such an elaborate explanation and to have done it so promptly as well! This forum is amazing!
     
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