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Homework Help: Using only arithmetic prove

  1. Aug 27, 2010 #1
    1. The problem statement, all variables and given/known data
    (!) Assuming only arithmetic (not the quadratic formula or calculus), prove that
    [tex]
    \left\{x \in\Re: x^2-2x-3 < 0\right\} = \left\{x \in\Re: -1 < x < 3\right\}
    [/tex]


    2. Relevant equations

    [tex]
    \left\{x \in\Re: x^2-2x-3 < 0\right\} = \left\{x \in\Re: -1 < x < 3\right\}
    [/tex]

    3. The attempt at a solution
    Again, new and trying to learn how exactly to do this to proper mathematical standards.

    I think this will just be a simple plug'n'play:

    Test case, 0 and 2 since using boundaries results in 0 < 0:
    (0)^2-2(0)-3 < 0
    -3 < 0
    True

    (2)^2-2(2)-3 < 0
    4-4-3 < 0
    -3 < 0
    True

    Thoughts? This seemed too easy but when I manipulated with adding and subtracting/multiplying I started to lose focus from the core inequality ie:

    x^2-2x-3 < 0
    x^2-2x < 3

    -1 < x < 3
    0 < x+1 < 3+1
    0 < x^2+x < 4x

    Et cetera...

    Sincerely,

    NastyAccident
     
  2. jcsd
  3. Aug 27, 2010 #2

    ehild

    User Avatar
    Homework Helper

    It is a good start to factorise the expression.

    Hint: x^2-2x-3 =(x-1)^2-4.

    ehild
     
  4. Aug 27, 2010 #3
    note: the quadratic formula is achieved through algebra alone. ;)
     
  5. Aug 27, 2010 #4
    Going your way:
    x^2-2x+1-4 < 0
    (x-1)^2-4<0
    (x-1)^2<4

    -1<x<3
    0<x+1<4
    x+1<(x-1)^2
    x+1<x^2-2x+1
    0<x^2-3x
    0<x(x-3)
    x=0 or x=3?

    Somewhere near that?
     
  6. Aug 27, 2010 #5

    ehild

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    Homework Helper

    Was not that you wanted to arrive at?

    It is not that what I meant by factorize, but it is correct up to here.


    (x-1)^2-4 = (x-1)^2-2^2=((x-1)-2)((x-1)+2)=(x-3)(x+1)

    You have to find the range of x where (x-3)(x+1)<0

    The product of two numbers is negative if one of the factors is negative, the other is positive. As x+1>x-3, x-3<0 and x+1>0 that is equivalent to -1<x<3.


    ehild
     
  7. Aug 27, 2010 #6
    Okay, the way of factoring originally brought up confused me for a second since it appeared to be completing a square to a problem that could of already been factored.

    After reading your original suggestion, I had the thought of breaking it down to (x-3)(x+1) >0. Though, after that I was unsure of where to go. And now, with the range comment I am a bit lost as to why you were able to move from (x-3)(x+1)>0 to x+1>x-3. Logically, it does hold since (0)+1>0-3 = 1>-3.

    I guess, I'm just getting lost on the transition from (x-3)(x+1)>0 to x+1>x-3 to x-3<0 and x+1>0. In my mind, I see it as going with (x-3)>0 and (x+1) > 0 instead of the inequalities that you presented. If you could clarify that I would appreciate it.

    Sincerely,

    NastyAccident
     
  8. Aug 27, 2010 #7

    ehild

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    x+1 is always bigger than x-3, isn't it?

    I did not say that (x-3)(x+1)>0. You have to find the values of x which make (x-3)(x+1)<0 true, that is the product of (x+1) and (x-3) is negative. That happens if one of the factors is negative, the other positive. As x+1>x-3, if x+1<0, x-3 is also negative and their product is positive. So x-3 has to be negative and x+1 positive:

    x-3<0 --->x<3

    and x+1>1 --->x>1


    ehild
     
  9. Aug 27, 2010 #8
    First, I apologize for the (x-3)(x+1)>0 statement. It should have been (x-3)(x+1) < 0. I hit the wrong key.

    I now see and understand why the split is occurring like that since you are trying to find terms that create a negative which would yield something that is less than zero.

    Now, this proves that set A (x^2-2x-3 < 0) is included in Set B (-1<x<3). In order to ensure that both are equal sets, I now have to repeat this process so that set B is included in Set A.

    So, would it be fair to do the following:

    -1<x<3
    -1<x
    x<3
    0<x+1
    x-3<0
    (x+1)(x-3)<0
    x^2-2x-3<0

    Hence,
    Set A = Set B

    Sincerely,

    NastyAccident
     
  10. Aug 27, 2010 #9

    ehild

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    Well done!


    ehild
     
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