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?

Re: Logic behind interval notation signs in inequalities

Quote:

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.

Re: Logic behind interval notation signs in inequalities

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

Thank you.

Re: Logic behind interval notation signs in inequalities

Quote:

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.