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Math Help - 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

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

    Squaring everything maintains the inequality...


    0 \leq 1 - x^2 \leq 1

    -1 \leq -x^2 \leq 0

    1 \geq x^2 \geq 0

    0 \leq x^2 \leq 1


    Now look at each inequality separately.

    x^2 \leq 1

    |x| \leq 1

    -1 \leq x \leq 1.


    But since 0\leq x^2

    0 \leq x.


    Putting it together, that means

    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 0\leq x^2

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

    0 \leq x^2 \leq 1
    All you have to do from here is square root everything
     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.,  x = \frac{-1}{2}. x < 0 yet x^2 > 0.

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

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

    0 \leq x^2

    Then -\infty \leq x \leq \infty.


    So NOW putting the inequalities together you find that

    -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 0\le \sqrt{1- x^2} and \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 0= 1- x^2, x= \pm 1.

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

    The whole point of that is that the three points, -1, 0, and 1 separate intervals where \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 \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 \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 -1\le x\le 1.
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  7. #7
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    \sqrt{1-x^2}\ge0 it's true as long as |x|\le1.

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