# Double Inequalities

• March 5th 2010, 12:46 AM
anonymous_maths
Double Inequalities
Need help to solve this:

0 < √1-x^2 < 1

< x < 1

How does that work?

Thanks.

• March 5th 2010, 01:27 AM
Prove It
Quote:

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

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

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

$0 \leq x \leq 1$.
• March 5th 2010, 01:32 AM
Haven
Quote:

Originally Posted by Prove It
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
$0 \leq x \leq 1$
• March 5th 2010, 02:01 AM
anonymous_maths
yes, but that's different to the answer supplied of -1< x < 1
• March 5th 2010, 02:18 AM
Prove It
Quote:

Originally Posted by Haven
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

$-1 \leq x \leq 1$.
• March 5th 2010, 03:34 AM
HallsofIvy
Quote:

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 $0\le \sqrt{1- x^2}$ 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 $-1\le x\le 1$.
• March 5th 2010, 07:22 AM
Krizalid
$\sqrt{1-x^2}\ge0$ 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 $|x|\le1.$