# Thread: inequality and an interval

1. ## inequality and an interval

I have the following inequalities:

-2 < h*2/x < 0

-2.78 <h*2/x < 0

where x is 1 < x < 2 ?

x > h > 0 and 1.39x > h > 0

or do i have problems with the way I have placed the signs?

and How does the interval affect the outcome?

2. Originally Posted by iwishiunderstood
I have the following inequalities:

-2 < h*2/x < 0

-2.78 <h*2/x < 0

where x is 1 < x < 2 ?

x > h > 0 and 1.39x > h > 0

or do i have problems with the way I have placed the signs?

and How does the interval affect the outcome?

There are only few rules which you have to observe when dealing with unequalities:

a) adding or subtracting a term doesn't change the relation-sign

b) dividing or multiplying by a positive term doesn't change the relation-sign

c) calculationg the reziprocal of fractions could change the relation-sign depending of the sign of the fractions in question. Examples:

$\displaystyle \dfrac13 > \dfrac14~\implies~3<4$

$\displaystyle -\dfrac13 < -\dfrac14~\implies~-3>-4$

BUT:

$\displaystyle -\dfrac13 < \dfrac14~\implies~-3<4$

$\displaystyle -2<h\cdot \frac2x <0~\wedge~x>0$ that means x is positive

$\displaystyle -2x<2h<0~\wedge~x>0$ divide through by -2 (the divisor is negative). The relation-signs have to be changed. Therefore your answer is correct.

$\displaystyle -2<h\cdot \frac2x <0~\wedge~x>0~\implies~-2x<2h<0$ . Dividing thriough by +2 will yield:

$\displaystyle -1.39x<h<0$

3. just notice in both my inequalities h * 2/x should be negative.

is this just a case of changing the signs round in my answers?

4. Originally Posted by iwishiunderstood
just notice in both my inequalities h * 2/x should be negative.

is this just a case of changing the signs round in my answers?
You probably have noticed that I've made a silly mistake. Here comes the corrected calculation:

$\displaystyle -2<h\cdot \frac2x <0~\wedge~x>0$ that means x is positive and h must be negative
$\displaystyle -2x<2h<0~\wedge~x>0$ divide through by -2 (the divisor is negative). The relation-signs have to be changed. Therefore the answer is:
$\displaystyle x>-h>0$