1. ## Double Inequalities

Need help to solve this:

0 < √1-x^2 < 1

< x < 1

How does that work?

Thanks.

2. Originally Posted by anonymous_maths
Need help to solve this:

0 < √1-x^2 < 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

$\displaystyle 0 \leq x \leq 1$.

3. Originally Posted by Prove It
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
$\displaystyle 0 \leq x \leq 1$

4. yes, but that's different to the answer supplied of -1< x < 1

5. Originally Posted by Haven
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

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

6. Originally Posted by anonymous_maths
Need help to solve this:

0 < √1-x^2 < 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 $\displaystyle -1\le x\le 1$.

7. $\displaystyle \sqrt{1-x^2}\ge0$ 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 $\displaystyle |x|\le1.$