1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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


    User Avatar
    Homework Helper

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Every straight line has no length