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Every straight line has no length

  1. Oct 23, 2006 #1
    ok i stuck guys here is the question.

    Consideer n infintely long straight lines. none are parrelel and no three are lines have a common intersection. show that for N>=1, the lines divide the plane into (n^2+n+2)/2.

    kay I am stuck kay i now base case works.

    I know that k=(k^2+k+2)/2

    but how do relate that to k+1 lines... I am stuck.
    I can't get them to relate for some reason.. what the trick i am missing.
  2. jcsd
  3. Oct 23, 2006 #2


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    Homework Helper

    First of all, every straight line has no length. But if you pick two different points A and B on the line, then [tex]\overline{AB}[/tex] has a length.

    Now, for n = 1 it is obvious that the line 'divides' the plane into two parts. Assume it's true for some k. You have to show it's true for some k+1, too. If it is, then it is true for every n. So, you have to find a way to show that for k+1, you have [tex]\frac{(k+1)^2+(k+1)+2}{2} = \frac{k^2+3k+4}{2}[/tex]. Here's a way. First, let's assume [tex]P(k) = \frac{k^2+k+2}{2}[/tex] is true for some k. Then look at the difference P(k+1) - P(k) = k+1. That means that the plane is divided into k+1 more parts than it is divided for some k. For example, if you take k = 2, and k = 3, you get that for k = 2 the plane is divided into 4 parts, and for k' = 3 it is divided into 7 parts. So, 7 - 4 = 3 = k + 1. Now, use that fact to prove the statement is true for some k+1. You have [tex]\frac{k^2+k+2}{2} + k+1 = \cdots[/tex] The rest is obvious. :smile:
  4. Oct 23, 2006 #3
    ok.. but how.. not making sense. I how would I jsutiify p(k+1) - P(K). that is not as easy..
  5. Oct 23, 2006 #4


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    You don't have to justify it. It is not a formal part of induction. You showed that for some k+1 the statement P(k+1) holds, and that's all you need to do.
    Last edited: Oct 23, 2006
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