I need help with Interval notation

Question

anonymous · Answer

Okay , what kind of help ?

theeric · Answer

Hi! Do you have some part of it in particular you need help with? Do you just want to understand it in general?

anonymous · Answer

$$3/2 \ge x < $$

anonymous · Answer

0 at the end

theeric · Answer

So, you want to say that $\frac{3}{2}$ is greater than or equal to $x$, and $x$ is less than $0$?

So, $\frac{3}{2}\ge x$ and $x\lt0$?

anonymous · Answer

Yes

anonymous · Answer

wait

anonymous · Answer

$$3/2 \ge x > 0$$

theeric · Answer

I would try to visualize it on a number line quick. That usually helps.
Okay, I have to draw the picture quick.

theeric · Answer

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amistre64 · Answer

the hardest part tends to be in remembering which brackets to use

theeric · Answer

Oh, here's a link! http://www.mathsisfun.com/sets/intervals.html

amistre64 · Answer

think of the square bracket as an equal sign connected by a bar

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theeric · Answer

Now, in my picture, I drew the two inequalities. Where do they over lap? They overlap to the right of $0$ up to where $\frac{3}{2}$ is!

theeric · Answer

Haha, nice @amistre64 !

amistre64 · Answer

$$a\le x<b$$

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amistre64 · Answer

its the only way i know to keeo track of all these notations :)

theeric · Answer

For that interval, $\frac{3}{2}\ge x\gt0$, it starts just to the right of $0$, so you want to NOT INCLUDE $0$. You know? So you have to use a $($.

anonymous · Answer

So it's (3/2,0] ?

theeric · Answer

And your interval include everything between $0$ and $\frac{3}{2}$, and $\frac{3}{2}$. Since in includes $\frac{3}{2}$, you want to use $]$.

theeric · Answer

@Juliaxo9 I think we always want the number on the left to be the lesser number, and the one on the right to be greater!

theeric · Answer

You go from $0$ to $\frac{3}{2}$, then!

anonymous · Answer

[0, 3/2) ?

theeric · Answer

Close! So they're in the correct order!
$\frac{3}{2}\ge x\gt0$
Now look at the signs. $x\gt0$, but not equal to $0$. So, should you use $($ to show "not equal," or $[$ to show "equal to."

theeric · Answer

?*

anonymous · Answer

Sorry I went to eat

theeric · Answer

Haha, no problem!

anonymous · Answer

(0, 3/2)

theeric · Answer

Okay, so, I'll just read what that says, and you tell me if it is correct.

You are going from $0$, but not including $0$, to $\frac{3}{2}$, but not including $\frac{3}{2}$. So that's where $x$ must be. Is that right?

theeric · Answer

Based on $\frac{3}{2}\ge x\gt0$?

anonymous · Answer

Yes