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Math Help - Understanding logic of proof

  1. #1
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    Understanding logic of proof

    Prove: If x^2 + x - 6 \geq 0, then  x \leq  -3 or  x \geq 2.


    Im not understanding the logic behind this questions answer. My professor went ahead and:

    Suppose that x^2 + x - 6 \geq 0 and  x > -3. It follows that (x-2)(x+3) \geq 0 and, since x + 3 > 0, it must be that x -2 \geq 0. That is, x \geq 2.

    im not understanding why they suppose x > -3, and how the bolded makes sense. I think im just not getting something easy here, please help.

    Thanks!
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello,
    Quote Originally Posted by p00ndawg View Post
    im not understanding why they suppose x > -3,
    For any real number x we have either x\leq -3 or x>-3, in particular this is also true if x satisfies x^2 + x - 6 \geq 0 . Thus instead of showing "If x^2 + x - 6 \geq 0, then  x \leq  -3 or  x \geq 2." one can show these two statements:
    • If x^2 + x - 6 \geq 0 and x\leq -3, then  x \leq  -3 or  x \geq 2.
    • If x^2 + x - 6 \geq 0 and x>-3, then  x \leq  -3 or  x \geq 2.

    As the first one is obviously true they do not mention it and only prove the last one.
    Quote Originally Posted by p00ndawg View Post
    It follows that (x-2)(x+3) \geq 0
    This is true because (x-2)(x+3) = x^2-2x+3x-6 = x^2+x-6\geq0.
    Quote Originally Posted by p00ndawg View Post
    and, since x + 3 > 0
    This is a consequence of this inequality: x>-3.
    Quote Originally Posted by p00ndawg View Post
    it must be that x -2 \geq 0.
    We have  (x-2)(x+3)\geq 0 hence x-2 and x+3 must have the same sign. Since x+3 is non-negative, x-2 has to be non-negative too.
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  3. #3
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    Quote Originally Posted by p00ndawg View Post
    Prove: If x^2 + x - 6 \geq 0, then  x \leq  -3 or  x \geq 2.


    Im not understanding the logic behind this questions answer. My professor went ahead and:

    Suppose that x^2 + x - 6 \geq 0 and  x > -3. It follows that (x-2)(x+3) \geq 0 and, since x + 3 > 0, it must be that x -2 \geq 0. That is, x \geq 2.

    im not understanding why they suppose x > -3, and how the bolded makes sense. I think im just not getting something easy here, please help.

    Thanks!

    Put...( x^2 + x -6\geq 0) = p, ( x\leq -3)= q ,( x\geq 2) =r.

    and we asked to prove :

    if p ,then q or r which in symbols is :  p\Longrightarrow (q\vee r).................................................. ......................................1

    But q or r is logically equivalent to \neg q\Longrightarrow r then (1) becomes :

    if p then ( if not q then r) which in symbols is:  p\Longrightarrow (\neg q\Longrightarrow r).


    But.

    p\Longrightarrow (\neg q\Longrightarrow) is logically equivalent to (p\wedge\neg q)\Longrightarrow r.

    That in logic means that whether we prove the 1st or the 2nd formula it is the same thing.

    And in using the 2nd formula we must assume p and not q ( p\wedge \neg q) and try to prove r.

    OR .

    assume : x^2 + x - 6\geq 0 and not x\leq -3....... \neg (x\leq -3).... and try to prove x\geq 2.

    But since  \neg(x\leq -3)\Longrightarrow x>- 3 ,because  x\leq -3 or x>-3,instead assuming :

     x^2 + x -6\geq 0 and \neg(x\leq -3) we can assume:

     x^2 + x -6\geq 0 and x>-3 or x+3>0.

    Now since  x\geq 2 or x<2 or x-2<0,if x-2<0 and since x+3>0 we have (x-2)(x+3) >0 or x^2 + x -6< 0 which contradicts the fact that x^2 + x -6\geq 0.

    THUS  x\geq 2
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