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Math Help - Inequality x^2-1 > 0

  1. #1
    Senior Member bugatti79's Avatar
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    Inequality x^2-1 > 0

    Folks,

    For x^2-1 > 0 solution x>1 and x<-1

    For x^2-1< 0 solution -1< x < 1

    I can see what the solution is easily but how do we show this mathematically?

    For the first one we have x^2 > 1 implies x > + - SQRT 1 but this doesnt make sense...?

    Thanks
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  2. #2
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    x^2 - 1 = (x - 1)(x + 1). A product of two numbers is positive iff the numbers are both negative or both positive, i.e., iff x < 1, x < -1 or x > 1, x > -1. Obviously, x < -1 implies x < 1, and x > 1 implies x > -1, so the product is positive iff x < -1 or x > 1.
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  3. #3
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    e^(i*pi)'s Avatar
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    x^2-1 = (x-1)(x+1)

    As emakarov says, the product of two expressions are positive if they have the same sign. That is you need to solve over the following intervals

    x > 1\ ,\ -1 \leq x \leq 1 \ ,\ x < -1


    Edit: you can also plot y=x^2-1 (it's a simple graph - y=x^2 shifted down 1) and see where it's greater than 0 (the x axis)
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  4. #4
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by emakarov View Post
    x^2 - 1 = (x - 1)(x + 1). A product of two numbers is positive iff the numbers are both negative or both positive, i.e., iff x < 1, x < -1 or x > 1, x > -1. Obviously, x < -1 implies x < 1, and x > 1 implies x > -1, so the product is positive iff x < -1 or x > 1.
    Hi Emakarov

    Ok, so my approach is not correct then?

    How about 1-ye^x > 0. This implies y e^x <1.

    Is this the kind of exercise where you have to infer what x and y should be?

    I take it that x and y cannot be both negative real numbers here otherwise we get a positive.
    I know y cannot be 0
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  5. #5
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    e^(i*pi)'s Avatar
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    1 > ye^x which gives y > e^{-x} is a special example due to the fact that e^x > 0 for all real x

    In your original question we need to consider that a negative squared is a positive. For example, if you tried x = -3 in your OP you'd find the inequality is satisfied
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  6. #6
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    x^2-1 = (x-1)(x+1)

    As emakarov says, the product of two expressions are positive if they have the same sign. That is you need to solve over the following intervals

    x > 1\ ,\ -1 \leq x \leq 1 \ ,\ x < -1


    Edit: you can also plot y=x^2-1 (it's a simple graph - y=x^2 shifted down 1) and see where it's greater than 0 (the x axis)
    Despite its simplicity, I dont understand this. (x-1)(x+1) do not have the same sign, if they were the same sign ie ((x-1)(x-1) it gives x^2 -2x +1..
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  7. #7
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    When we say that x - 1 and x + 1 have the same sign, we mean that both numbers are positive or both are negative. The sign of x - 1 is not necessarily - .

    Ok, so my approach is not correct then?
    Which approach exactly?

    How about 1-ye^x > 0.
    This depends on what you want to do with this expression. This is a property that becomes true or false for each given x and y. What to do with it is a different issue.
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  8. #8
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by emakarov View Post
    When we say that x - 1 and x + 1 have the same sign, we mean that both numbers are positive or both are negative. The sign of x - 1 is not necessarily - .
    Ok, I never heard this before. Thanks


    Quote Originally Posted by emakarov View Post
    Which approach exactly?
    If x^2 > 1 \implies x > \pm \SQRT 1 ie x can be either > +1 or > -1. That was my initial attempt, how is it wrong?

    Quote Originally Posted by emakarov View Post
    This depends on what you want to do with this expression. This is a property that becomes true or false for each given x and y. What to do with it is a different issue.
    Need to determine the range of x and y values for when

    1-y e^x >0
    1-y e^x <0
    1-y e^x =0

    SO for the first one, e^(i*pi) qouted y>e^-x. Therefore if x can be any real number then y < 0 for the inequality to hold?

    Thats my interpretation...pardon my ignorance!

    Thanks
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  9. #9
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    e^(i*pi)'s Avatar
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    Just so I'm not misinterpreted the question is 1-ye^x > 0.

    I added ye^x to both sides: 1 > ye^x \implies ye^x < 1

    The inequality will change sign if we divide by a negative number so normally it's necessary to take care. This is a special example though, because the range of e^x is greater than 0 we can divide without worrying about the direction of the inequality changing.

    Dividing by e^x: y < \dfrac{1}{e^x}


    Finally, due to the laws of exponents y < e^{-x}

    My apologies I got my inequality wrong before
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  10. #10
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    Quote Originally Posted by bugatti79 View Post
    If x^2 > 1 \implies x > \pm \SQRT 1 ie x can be either > +1 or > -1. That was my initial attempt, how is it wrong?
    What you're attempting is valid. You can take the square root of both sides and keep the inequality, but remember that \sqrt{x^2}=|x|, so the result should be |x|>1.
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