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Thread: Double Inequalities

  1. #1
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    Double Inequalities

    Need help to solve this:

    0 < √1-x^2 < 1

    answer is -1
    < x < 1

    How does that work?

    Thanks.


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  2. #2
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    Quote Originally Posted by anonymous_maths View Post
    Need help to solve this:

    0 < √1-x^2 < 1

    answer is -1
    < x < 1

    How does that work?

    Thanks.


    I take it that this is

    $\displaystyle 0 \leq \sqrt{1 - x^2} \leq 1$.

    Squaring everything maintains the inequality...


    $\displaystyle 0 \leq 1 - x^2 \leq 1$

    $\displaystyle -1 \leq -x^2 \leq 0$

    $\displaystyle 1 \geq x^2 \geq 0$

    $\displaystyle 0 \leq x^2 \leq 1$


    Now look at each inequality separately.

    $\displaystyle x^2 \leq 1$

    $\displaystyle |x| \leq 1$

    $\displaystyle -1 \leq x \leq 1$.


    But since $\displaystyle 0\leq x^2$

    $\displaystyle 0 \leq x$.


    Putting it together, that means

    $\displaystyle 0 \leq x \leq 1$.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    But since $\displaystyle 0\leq x^2$

    $\displaystyle 0 \leq x$.
    That is false. I.e.,$\displaystyle x = \frac{-1}{2}$. $\displaystyle x < 0$ yet $\displaystyle x^2 > 0$.

    $\displaystyle 0 \leq x^2 \leq 1$
    All you have to do from here is square root everything
    $\displaystyle 0 \leq x \leq 1 $
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  4. #4
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    yes, but that's different to the answer supplied of -1< x < 1
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  5. #5
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    Quote Originally Posted by Haven View Post
    That is false. I.e.,$\displaystyle x = \frac{-1}{2}$. $\displaystyle x < 0$ yet $\displaystyle x^2 > 0$.

    $\displaystyle 0 \leq x^2 \leq 1$
    All you have to do from here is square root everything
    $\displaystyle 0 \leq x \leq 1 $
    Correct in finding the mistake in my logic.

    However, since all square numbers are nonnegative, that means if

    $\displaystyle 0 \leq x^2$

    Then $\displaystyle -\infty \leq x \leq \infty$.


    So NOW putting the inequalities together you find that

    $\displaystyle -1 \leq x \leq 1$.
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  6. #6
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    Quote Originally Posted by anonymous_maths View Post
    Need help to solve this:

    0 < √1-x^2 < 1

    answer is -1
    < x < 1

    How does that work?

    Thanks.


    The crucial point with double inequalities is that you must have $\displaystyle 0\le \sqrt{1- x^2}$ and $\displaystyle \sqrt{1- x^2}\ge 1$.

    Personally, I think the simplest way to solve complicated inequalities is to solve the associated equation first.

    If [/tex]0= \sqrt{1- x^2}[/tex] then $\displaystyle 0= 1- x^2$, $\displaystyle x= \pm 1$.

    If $\displaystyle \sqrt{1- x^2}= 1$ then $\displaystyle 1- x^2= 1$, x= 0.

    The whole point of that is that the three points, -1, 0, and 1 separate intervals where $\displaystyle \sqrt{1- x^2}$ is less than 0, greater than 0, or does not exist. Try one value of x in each of the intervals x< -1, -1< x< 0, 0< x< 1, and x> 1.

    For example, if x= -2< -1, then $\displaystyle \sqrt{1- x^2}= \sqrt{-3}$ which is not a real number so does not satisfy the inequality. No value of x less than -1 satisfies the inequality.

    If x= -1/2, between -1 and 0, then [tex]\sqrt{1- x^2}= \sqrt{1- 1/4}= \sqrt{3}{2}. which is positive and less than 1. Every value of x between -1 and 0 satisfies the inequality.

    If x= 1/2, between 0 and 1, then [tex]\sqrt{1- x^2}= \sqrt{1- 1/4}= \sqrt{3}{2}. which is positive and less than 1. Every value of x between 0 and 1 satisfies the inequality.

    Finally, if x= 2, between 0 and 1, then $\displaystyle \sqrt{1- x^2}= \sqrt{-3}$ which is not a real number so does not satisfy the inequality. No value of x larger than 1 satisfies the inequality.

    Since the numbers -1, 0, and 1 make the two sides equal, the solution set is $\displaystyle -1\le x\le 1$.
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  7. #7
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    $\displaystyle \sqrt{1-x^2}\ge0$ it's true as long as $\displaystyle |x|\le1.$

    as for $\displaystyle \sqrt{1-x^2}\le1,$ (1) fix $\displaystyle |x|\le1$ and by squaring we get $\displaystyle x^2\ge0$ which holds for any $\displaystyle x,$ thus the solution set for (1) is again $\displaystyle |x|\le1,$ and the final solution set it's just $\displaystyle |x|\le1.$
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