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Let a and b be real numbers with a < b.

  1. Oct 7, 2013 #1
    1. The problem statement, all variables and given/known data

    Let a and b be real numbers with a < b.

    a. Derive a formula for the distance from a to b. Hint: Use 3 cases and a visual argument on the number line.

    b. Use your work in part (a) to derive a formula for the distance between (a,c) and (b,c) in a plane.

    c. Use the Pythagorean theorem to derive a formula for the distance between the points(a,c) and (b,d) in the adjacent sketch. The sketch is a line on x-y coordinate with two endpoints (a,c) and (b,d) where b > a and d > c.

    d. Generalize the distance formula to R^n.

    2. Relevant equations
    3. The attempt at a solution

    I tried doing (a):

    Derive a formula for the distance from a to b. Hint: Use 3 cases and a visual argument on the number line.

    Case 1: Assume that 0 ≤ a < b. Since a ≥ 0, the distance from a to 0 is a. Since b ≥ 0, the distance from b to 0 is b. The distance from a to b is the distance from b to 0 minus the distance from a to 0 which is b-a.

    Case2: Assume that 0 ≥ b > a. Since 0 ≥ a, the distance from 0 to a is -a. Since 0 ≥ b, the distance from 0 to b is -b. The distance from a to b is the distance from 0 to a minus the distance from 0 to b which is b-a.

    Case3: Assume that b ≥ 0 ≥ a. Since b ≥ 0, the distance from b to 0 is b. Since 0 ≥ a, the distance from 0 to a is -a. The distance from a to b is the distance from b to 0 plus the distance from 0 to a which is b-a.

    Does it look correct?

    Thanks.
     
  2. jcsd
  3. Oct 7, 2013 #2

    Office_Shredder

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    You're missing a ≥ 0 ≥ b for sure, and also what happens if a = b? I believe the three cases they are referring to is when a>b, when b>a, and when a=b.
     
  4. Oct 7, 2013 #3
    Thanks for the answer.

    When a ≥ 0 ≥ b. since a ≥ 0, the distance between a and 0 is a and the distance between 0 and b is -b because 0 ≥ b. The distance between a and b is the distance between a and 0 plus the distance between 0 and b: a - b.

    If a = b and 0 ≥ a, 0 ≥ b, the distance from 0 to a is -a. The same is true for b. The distance from a to b is -a -(-b)= b -a. Since a = b, b - a = a - a = 0.

    If a = b and a ≥ 0, b ≥ 0, the distance from 0 to a is a. The same is true for b. The distance from a to b is a - b. Since a = b, a - b = a - a = 0.

    Does it look correct?

    Thanks.
     
  5. Oct 7, 2013 #4

    Office_Shredder

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    I think that all of your formulas are correct, but you should notice that you never really needed to know whether a and b are positive or negative, just which one is larger.
     
  6. Oct 7, 2013 #5
    Thanks. (a) says I should derive a formula, but there are 3 cases so- 3 formulas. Honestly, i have no idea what to do with these formulas. Do they all magically converge into one :) Can you elaborate a little on this, please?
     
  7. Oct 7, 2013 #6

    Office_Shredder

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    I mentioned it in my previous post: consider the cases a<b, a=b and a>b. It doesn't matter whether a or b are positive or negative (you can check that all your cases where a<b agree, where a=b agree, and where a>b agree).
     
  8. Oct 7, 2013 #7
    It still sounds cryptic to me. Does it pertain to case 1, too? I just copied that from the back of my textbook. The rest of the formulas I modeled after that. Thanks.
     
  9. Oct 8, 2013 #8

    Ray Vickson

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    He was given the hypothesis that a < b.
     
  10. Oct 8, 2013 #9
    Can you give some clues and tips for (a)? Thanks.
     
  11. Oct 8, 2013 #10

    Ray Vickson

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    No need. You already did it in your original message.
     
  12. Oct 8, 2013 #11
    I am not sure how (b) is drastically different from (a). Would the distance from (a,c) to (b,c) be b - a? The only thing that changed is, while y was equal to 0 in (a), y = c in (b). Maybe there's a more formal way of writing it down, taking into account y = c? If there's, I am not seeing it :)

    Does (c) look like this: d = sqrt((b - a)^2 + (d - c)^2) ?

    Thanks.
     
    Last edited: Oct 8, 2013
  13. Oct 8, 2013 #12

    Ray Vickson

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    (b) is basically the same as (a); the only difference is that the line at elevation c instead of at elevation 0. And yes, you have done (c) correctly.
     
    Last edited: Oct 8, 2013
  14. Oct 8, 2013 #13
    Seeing as how much work went into defining the distance in (a), maybe, something to that effect needs to be done in (b) and (c). I mean it, probably, wouldnt be acceptable to just write down "the distance between (a,c) and (b,c) is a - b by analogy with the case 1 in (a) since y = c doesnt effect the distance between (a, c) and (b, c) because of bla bla" for (b). So how would you stylize (b) in math speak?

    Could I write "According to the case 1 of (a) the distance between (a,c) and (b,c) is a - b. Likewise, the distance between (b,d) and (b,c) is d - c. Then by Pythagorean, the distance between (a,c) and (b,d) is d = sqrt((b - a)^2 + (d - c)^2)" for (c)?

    Thanks.
     
  15. Oct 8, 2013 #14

    Ray Vickson

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    I hope you mean distance = |a-b| in question (b), since distance is always ≥ 0 and (b) does not tell us whether a < b or b < a. However, if we understand that a < b still applies, then you have written the negative of the distance: the distance is b-a, not a-b. And, yes: just saying that setting y = c does not affect the distance is good enough; that follows from some elementary geometry (but, unfortunately geometry is largely untaught now in the modern curriculum, or so I have been told).
     
  16. Oct 8, 2013 #15
    Thanks. Does putting it for (c) the way I did look acceptable?
     
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