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Help with a discrete math homework problem

  1. Jun 4, 2012 #1
    1. The problem statement, all variables and given/known data
    show that for positive r and s, with r<s, we always have:

    r<(r+s)/2<s and 2/[(1/r)+(1/s)]^2< 2rs< (r+s)^2


    2. Relevant equations



    3. The attempt at a solution
    i have shown that r<s because r+r<s+r, 2r<s+r, r<(s+r)/2 and 2r< (2s+2r)/4
    (r+s)^2= r^2+2rs+s^2 where are<r^2+2rs+s^2

    not sure if I'm even on the right track or if i have even finished the problem. I'm kinda confused.
     
  2. jcsd
  3. Jun 4, 2012 #2

    Mark44

    Staff: Mentor

    You've wandered off the right track. You don't need to (and shouldn't) show that r < s. You are given that r < s.

    Some of what you wrote makes sense, but a lot doesn't (e.g., "(r+s)^2= r^2+2rs+s^2 where are<r^2+2rs+s^2"), and it's not well organized.


    For each part, you should start with r < s. For the first part, the last line should be r < (r + s)/2 < s. For the second part, the last line should be 2/[(1/r)+(1/s)]^2< 2rs< (r+s)^2. Each line between the start and end should make sense logically and mathematically.
     
  4. Jun 5, 2012 #3
    okay so if i start with r<s, i have r<s.

    then, how do i show r<s for r+s/2 ? Would I start with r+r<r+s, if so, how do i incorporate the 2?
     
    Last edited: Jun 5, 2012
  5. Jun 5, 2012 #4
    Just break up the inequality into 2 parts, that is it.
     
  6. Jun 5, 2012 #5

    Mark44

    Staff: Mentor

    You don't show r < s for anything. You are to assume this. You need to show that r < (r + s)/2 < s.

    What you started with was OK, but you need to take it a bit further.

    r < s
    => r + r < s + r and s + r < s + s -- Do you understand why this follows from r < s?
    => ? What can you do with the two inequalities above?
     
  7. Jun 5, 2012 #6
    Re: help with a discrete math homework problem

    okay so,

    r + r < s + r < s + s, this is true because if r < s then, r < s +r and because r < s then r+r<s, so we have what follows,

    r + r < s + r < s + s , now i believe if we add 2 ,

    2r < 2(s+r) < 2s then,

    r< (s+r)/2 < s
     
  8. Jun 5, 2012 #7

    Mark44

    Staff: Mentor

    Re: help with a discrete math homework problem

    Here's a better way to say it.
    r < s (given)
    r + r < s + r (by adding r to both sides of the given inequality)

    r + s < s + s (by adding s to both sides of the given inequality)

    Combining these, we get
    r + r < r +s < s + s
    ??? You're not adding 2.
    No. The expressions at either end are right, since r + r = 2r and s + s = 2s, but you can't just stick a factor of 2 in the middle.
    This doesn't follow from your previous inequality.
     
  9. Jun 5, 2012 #8
    how to i prove the middle part then? I understand the other part now.

    would i just divide out by two to get,

    r<(r+s)/2<s
     
  10. Jun 5, 2012 #9

    Mark44

    Staff: Mentor

    Yes.
    2r < r + s < 2s ==> r < (r + s)/2 < s

    What this is saying is that if r < s, the average of r and s will be between r and s.
     
  11. Jun 5, 2012 #10
    okay now where do i start with the second part?

    2/[(1/r)+(1/s)]^2< 2rs< (r+s)^2

    Would I start by supposing r<s again? and then,

    would i start by expanding out (r+s)^2?
     
    Last edited: Jun 5, 2012
  12. Jun 5, 2012 #11
    so i've tried to figure it out and this is what i have come up with,

    2rs< r2+2rs+s2, (by expansion of (r+s)2)

    2/(2/rs) < 2rs, (by expansion of (1/r + 1/s)2) and finding a common denominator
     
  13. Jun 5, 2012 #12

    Mark44

    Staff: Mentor

    Starting from r < s, how do you get to this?

    When you expand (r + s)2, you get r2 +2rs + s2. IOW, (r + s)2 = r2 +2rs + s2.

    How did you get an inequality out of that?


    This doesn't make any sense to me.
     
  14. Jun 5, 2012 #13
    well I was just comparing it to the other part in the problem
     
  15. Jun 5, 2012 #14

    Mark44

    Staff: Mentor

    You have to be able to justify each step. Each inequality you write has to be a direct consequence of the one before it.
     
  16. Jun 5, 2012 #15
    so how do i go about doing that then? I have to start with:

    2/[(1/r)+(1/s)]2
     
  17. Jun 5, 2012 #16

    Mark44

    Staff: Mentor

    The second part is quite a bit easier; namely, showing that 2rs < (r + s)2

    As before, start with the given statement: r < s

    Notice that r2 + s2 > 0 -- can you explain why?
    What can you add to the left side (and the right side) to make the left side (r + s)2?
     
  18. Jun 5, 2012 #17
    its greater than zero because there are fractions
     
  19. Jun 5, 2012 #18
    would i add r2+s2 to both sides?
     
  20. Jun 6, 2012 #19

    Mark44

    Staff: Mentor

    No, that's not why. If r < s, it does not necessarily follow that 1/r + 1/s > 0. Consider r = -1 and s = 2. We have r < s, but 1/r + 1/s = 1/(-1) + 1/2 = -1/2 < 0.
    Yes.
     
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