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What's wrong with this proof?

  • 22-06-2007 2:56pm
    #1
    Farouk.Bulsara
    Closed Accounts Posts: 50 ✭✭


    Hi,

    Can anyone see any flaws in this proof? I want to show that x + [ 9 / (x + 2 ) ] is greater than or equal to 4, given that x + 2 is strictly positive. Here goes:

    Given: x + 2 > 0
    Hence: [ (x + 2) + 3]^2 > 0 (the sum of two positive quantities squared is positive)
    Hence: (x + 2)^2 + 6(x + 2) + 9 > 0 (squaring the terms)
    Hence: x + 2 + 6 + [ 9 / (x + 2) ] > 0 (dividing across by x + 2)
    Hence: x + [ 9 / (x + 2) ] > -8 (bringing the constants to the RHS)

    I was expecting to end up with x + [ 9 / (x + 2)] >= 4 (and not > -8) as this was the expected result. Note: this was asked on this year's higher level Leaving Cert maths exam. Perhaps there was a misprint on the paper?

    Any thoughts or comments would be much appreciated.

    Fred


Comments

  • #2
    Crumbs
    Registered Users, Registered Users 2 Posts: 1,080 ✭✭✭Crumbs


    Farouk.Bulsara wrote:
    Hence: [ (x + 2) + 3]^2 > 0 (the sum of two positive quantities squared is positive)
    Make this [(x + 2) - 3]^2 > 0 and it will work out.


  • #3
    Farouk.Bulsara
    Closed Accounts Posts: 50 ✭✭Farouk.Bulsara


    Crumbs wrote:
    Make this [(x + 2) - 3]^2 > 0 and it will work out.

    Crumbs,

    Aha! Thanks a lot! It's sooooooooo obvious, now that you point it out! :)

    Fred


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