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Thread: 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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    $\displaystyle x^2-1 = (x-1)(x+1)$

    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

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


    Edit: you can also plot $\displaystyle y=x^2-1$ (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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    e^(i*pi)'s Avatar
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    $\displaystyle 1 > ye^x$ which gives $\displaystyle y > e^{-x}$ is a special example due to the fact that $\displaystyle e^x > 0$ 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
    $\displaystyle x^2-1 = (x-1)(x+1)$

    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

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


    Edit: you can also plot $\displaystyle y=x^2-1$ (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 $\displaystyle x^2 > 1 \implies x > \pm \SQRT 1$ 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

    $\displaystyle 1-y e^x >0$
    $\displaystyle 1-y e^x <0$
    $\displaystyle 1-y e^x =0$

    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 $\displaystyle 1-ye^x > 0$.

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

    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: $\displaystyle y < \dfrac{1}{e^x}$


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

    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 $\displaystyle x^2 > 1 \implies x > \pm \SQRT 1$ 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 $\displaystyle \sqrt{x^2}=|x|$, so the result should be $\displaystyle |x|>1$.
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