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I'm veryconfused with an a level maths q

  1. Apr 11, 2005 #1


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    the question:
    (im presuming n^2 means n squared)

    prove the following result: "a traingle with sides that can be written in the form n^2+1, n^2-1, and 2n (where n>1) is right angled.
    show, by means of a counter example, that the converse is false.

    this q was taken from the back of the book "the curious incedent of the dog in the night-time" and there was a full proof but i dont understand some of it.

    it start by exmplaining we need to prove which side is the longest by doing:

    n^2+1 - 2n = (n-1)^2

    if n>1 then (n-1)^2 >0

    therefore n^2+1 > 2n

    similarily (n^2+1) - (n^2-1) = 2

    therefore n^2+1 > n^2-1

    so n^2+1 > n^2-1.

    the rest is worked out using pythagoras but then the converse bit has completly lost me.

    if anyone would like to explain, then i would be very grateful. thanks!
  2. jcsd
  3. Apr 11, 2005 #2


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    He is simply showing which side is longer,

    [tex](n^2+1) - (2n) = (n-1)^2[/tex]

    The left hand side is the difference between 2 of the sides.

    However, [itex](n-1)^2 \geq 0[/itex]

    So you can replace [itex](n-1)^2 [/itex] with: [itex]\geq 0[/itex] and you get:

    [tex](n^2+1) - (2n) \geq 0[/tex]

    Adding 2n to both sides:

    [tex]n^2+1 \geq 2n [/tex]

    So you know that the n2 + 1 side is greater or equal to the 2n side.
  4. Apr 11, 2005 #3


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    Please,do not double post.It's not really fair.

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