# Thread: Logic behind interval notation signs in inequalities

1. ## Logic behind interval notation signs in inequalities

I'm comparing two sums here and can't see the logic why I need to use less than/more than signs or less than-equal to/more than-equal to.

Sum 1 -
(8-x)/(9-x) <= 0

Sum 2 -
(8-x)(x-2) >= 0

For both of these I need to find the roots of the inequality, illustrate them on a number line, then pick a number within the interval and check whether it fits with being less than or equal to zero - or more than or equal to zero.

Now, where I'm picking my intervals is where my problem begins.
Take Sum 1, my intervals are:
x<=8 , 8<x<=9 and x>9
The signs here are correct, my answer is [8;9)

What is the logic for putting <= instead of < in such a sum during my intervals? Because I'm not seeing it.
In my Sum 2, my intervals are:
x<=2, 2<x<=8 and x>8 -- And the answer tells me that this is wrong! The answer should be [2;8], which means that the intervals should be: x<2, 2<=x<=8 and x>8

Is there some set of rules I'm not aware of?

2. ## Re: Logic behind interval notation signs in inequalities

Originally Posted by Aluminium
I'm comparing two sums here and can't see the logic why I need to use less than/more than signs or less than-equal to/more than-equal to.

Sum 1 -
(8-x)/(9-x) <= 0

Sum 2 -
(8-x)(x-2) >= 0

For both of these I need to find the roots of the inequality, illustrate them on a number line, then pick a number within the interval and check whether it fits with being less than or equal to zero - or more than or equal to zero.

Now, where I'm picking my intervals is where my problem begins.
Take Sum 1, my intervals are:
x<=8 , 8<x<=9 and x>9
The signs here are correct, my answer is [8;9)

What is the logic for putting <= instead of < in such a sum during my intervals? Because I'm not seeing it.
In my Sum 2, my intervals are:
x<=2, 2<x<=8 and x>8 -- And the answer tells me that this is wrong! The answer should be [2;8], which means that the intervals should be: x<2, 2<=x<=8 and x>8

Is there some set of rules I'm not aware of?
Check the boundaries of the intervals separately. For example in the fist problem, consider take the intervals as x<8, 8<x<9 and x>9. So you will see that the inequality is satisfied in the region, 8<x<9. Now check what happens at the boundaries. For this problem the boundary points are, x=8 and x=9. When x=8 the inequality is satisfied but not at x=9. Therefore your answer is [8,9). Hope you could do the second problem similarly.

3. ## Re: Logic behind interval notation signs in inequalities

Wow. Goodness, this was so simple! And none of the books explained it.

Thank you.

4. ## Re: Logic behind interval notation signs in inequalities

Originally Posted by Aluminium
I'm comparing two sums here and can't see the logic why I need to use less than/more than signs or less than-equal to/more than-equal to.

Sum 1 -
(8-x)/(9-x) <= 0

Sum 2 -
(8-x)(x-2) >= 0

For both of these I need to find the roots of the inequality, illustrate them on a number line, then pick a number within the interval and check whether it fits with being less than or equal to zero - or more than or equal to zero.

Now, where I'm picking my intervals is where my problem begins.
Take Sum 1, my intervals are:
x<=8 , 8<x<=9 and x>9
The signs here are correct, my answer is [8;9)

What is the logic for putting <= instead of < in such a sum during my intervals? Because I'm not seeing it.
In my Sum 2, my intervals are:
x<=2, 2<x<=8 and x>8 -- And the answer tells me that this is wrong! The answer should be [2;8], which means that the intervals should be: x<2, 2<=x<=8 and x>8

Is there some set of rules I'm not aware of?
You should note that neither of the inequations you posted are "sums", as a sum is what you get after you add something.

For your second inequation, I nearly always complete the square to solve quadratic inequalities, as I find it the easiest method.

\displaystyle \begin{align*} (8-x)(x-2) &\geq 0 \\ 8x - 16 - x^2 + 2x &\geq 0 \\ -x^2 + 10x - 16 &\geq 0 \\ x^2 - 10x + 16 &\leq 0 \\ x^2 - 10x + (-5)^2 - (-5)^2 + 16 &\leq 0 \\ (x - 5)^2 - 25 + 16 &\leq 0 \\ (x - 5)^2 &\leq 9 \\ |x - 5| &\leq 3 \\ -3 \leq x - 5 &\leq 3 \\ 2 \leq x &\leq 8 \end{align*}