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**Hannibal lecter****Member**- Registered: 2016-02-11
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Hi,

I study the absolute value maybe more than 2 times in my life,, when I was in the school and now and on the internet for myself

now I read (Absolute Value how far a number is from zero) .. then it is mean abs for 6 is 6 and -6 is 6

what is the meaning of that? and if it was about how far the number 6 from zero is the meaning of abs

then why -6 is 6 ? why the abs -6 is -6?

I apologize for such questions again and again.. I don't have another way to understand! sorry

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**Mathegocart****Member**- Registered: 2012-04-29
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Absolute Value is its distance from 0, for instance...

<--- We count the steps from 0( 1,2 and then 3.) We ignore the sign and count the net distance from -3 to 0(aka 3).

The absolute value of negative 6 is 6, since it is 6 away from 0 on the number line..

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**bobbym****bumpkin**- From: Bumpkinland
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(Absolute Value how far a number is from zero)

Good Lord! Mathematicians and how they love to jargonize. Turning the simple, into the horribly complex. I have done several million calculations that have used the absolute value and never thought about it that way. Do I have a pea sized brain or are they incredibly deep or both? History will decide.

If you are using the absolute value to do a computation then all you have to remember is that the absolute value is always positive. Abs[5] = 5, Abs[-5] = 5. Always a positive value.

I see now that Mathegocart has provided a nice answer.

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More generally you can think of the absolute value as the Euclidean norm with That is, given a point then the Euclidean norm is defined as Sometimes it's denoted or just Then if the norm is just This is another way of defining "distance" from 0 in the -dimensional world. (In other words, becomes what is called a "metric space" when endowed with this metric.)

*Last edited by zetafunc (2016-12-16 21:06:47)*

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**Hannibal lecter****Member**- Registered: 2016-02-11
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#Mathegocart

yes thank you very much I got it now.

but can we say the abs -6 is -6 too? cause it is just a sign. what will happen if we say that

*Last edited by Hannibal lecter (2016-12-16 23:01:40)*

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**Hannibal lecter****Member**- Registered: 2016-02-11
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thank you bobbym,

and #zetafunc sorry but I don't understand these symbols you provided, I'm not like you I just do simple mathematics I don't even know what is

and I wish to know what is these beautifull symbols

sorry

*Last edited by Hannibal lecter (2016-12-16 23:19:01)*

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is just notation -- it usually means "norm of x". A norm is something which tells you how "big" something is. For instance, you already know how to measure the length of a line. The Euclidean norm -- which is basically an extension of Pythagoras' Theorem -- just tells you how to find the length of a line between two points in general. So in that sense, the absolute value is sometimes called the "absolute value norm". But there are plenty of other types of norms out there which are very interesting to study.

Let me know if there's anything you want me to explain, I'm happy to help.

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**Mathegocart****Member**- Registered: 2012-04-29
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Hannibal lecter wrote:

#Mathegocart

yes thank you very much I got it now.

but can we say the abs -6 is -6 too? cause it is just a sign. what will happen if we say that

Previously said, the absolute value is the distance from 0. Distance, consequently, cannot be negative, and so the absolute value of -6 is simply 6.

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**Hannibal lecter****Member**- Registered: 2016-02-11
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**Monox D. I-Fly****Member**- From: Indonesia
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Because u can either be a or -a.

Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away. May his adventurous soul rest in peace at heaven.

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Hannibal lecter wrote:

Remember that the absolute value is never negative. We can always say that Now, suppose that someone tells you they want to know where is on the number line. That's all the points which are a distance from . But there are two such points: (by counting along from ) and (by counting down from to ). Then more generally, corresponds to two points which are a distance from -- and .
thank you very much Mathegocart

can I ask another question?

what is that mean :

http://store2.up-00.com/2016-12/1482020053631.png

what i u ?

and why +-a

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**Hannibal lecter****Member**- Registered: 2016-02-11
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thank you for this great example,

can I ask you another question it was difficult to me to solve it :-

<=or like this

<= 1please may I get the full solution for this? with its interval notation

*Last edited by Hannibal lecter (2016-12-19 09:04:01)*

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**Mathegocart****Member**- Registered: 2012-04-29
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This was initially irritating to solve( some paradoxes sprung forth ), though once you convert an absolute value into its alternative form, it becomes much more tolerable.

(convert to tolerable forme)

(square both sides)*

(isolate..)

(expand both convoluted functions)

(add a negative)

(divide, while noting the particular fact to switch signs.)

We are finished, with our bombardment of mathematical operations, we were able to find the solution out of this gargantuan inequality.

*Last edited by Mathegocart (2016-12-19 14:08:30)*

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**Hannibal lecter****Member**- Registered: 2016-02-11
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Believe me if there is another way to help myself with, I'm sure I'll follow this way, but this is the only place that I can get help with a specific questions in math.

thank you very much.

so....is this the end of the solution? but how did you do that with an absolute value? ((( I mean we must take two possibilities once in >=0 and once in < 0 )))))???

please see this following picture, is the the solution of our lecturer in the university :- what is this? I really hop you to help me with full details about that

I always asking him about problems but our lecturer don't show respect to who ask him about that he learning us he just writing these problems without careful and caring ...

*Last edited by Hannibal lecter (2016-12-20 06:03:06)*

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The important part to take away from your lecturer's solution is that is equivalent to (Make sure you understand why this is true: you'll be seeing this inequality used again and again if you take any analysis courses.) All he's done from that point is consider the first and second inequalities separately, then combined them to obtain all the values that satisfy the original inequality.

Mathegocart's solution is fine, but as a point of caution, squaring an inequality (or an equation) can introduce solutions which don't satisfy the original inequality. If you do that, it's important in general to go back and check whether or not the solutions you've found are actually valid.

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**Hannibal lecter****Member**- Registered: 2016-02-11
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zetafunc wrote:

The important part to take away from your lecturer's solution is that is equivalent to

but mathegocart didn't do like the lecturer did, what does it changed? how the solution is

,and the sol of the teacher is :

, U {-1}.zetafunc wrote:

it's important in general to go back and check whether or not the solutions you've found are actually valid.

but how to do that? I mean how to check wether or not the solution is valid or not????

how did you know the solution was fine please help with that...

*Last edited by Hannibal lecter (2016-12-20 07:20:32)*

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**Mathegocart****Member**- Registered: 2012-04-29
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Hannibal lecter wrote:

Believe me if there is another way to help myself with, I'm sure I'll follow this way, but this is the only place that I can get help with a specific questions in math.

thank you very much.so....is this the end of the solution? but how did you do that with an absolute value? ((( I mean we must take two possibilities once in >=0 and once in < 0 )))))???

please see this following picture, is the the solution of our lecturer in the university :- what is this? I really hop you to help me with full details about that

http://store2.up-00.com/2016-12/1482256157981.jpg

I always asking him about problems but our lecturer don't show respect to who ask him about that he learning us he just writing these problems without careful and caring ...

Sometimes it is more intuitive to explain it in the alternate form, which is also applicable for absolute values.

And yes, heed zetafunc's warning, sometimes it can introduce extraneous solutions.

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**Hannibal lecter****Member**- Registered: 2016-02-11
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Mathegocart wrote:

Sometimes it is more intuitive to explain it in the alternate form, which is also applicable for absolute values.

And yes, heed zetafunc's warning, sometimes it can introduce extraneous solutions.

may you please solve it like my lecturer did but in simple way without these strange number lines and +++ or ---

if you have a free time please, I need you sol with your elegant words and short steps. with greater and smaller and with interval notations.

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**Mathegocart****Member**- Registered: 2012-04-29
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Hannibal lecter wrote:

Mathegocart wrote:Sometimes it is more intuitive to explain it in the alternate form, which is also applicable for absolute values.

And yes, heed zetafunc's warning, sometimes it can introduce extraneous solutions.may you please solve it like my lecturer did but in simple way without these strange number lines and +++ or ---

if you have a free time please, I need you sol with your elegant words and short steps. with greater and smaller and with interval notations.

Hmmm, those number lines are invaluable in visualizing the sets the inequalities include.

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can't be a solution, because while (So your lecturer was wrong about that.) I agree that what your lecturer wrote looks a little confusing. They write:The expression on the left is well-defined, since we know that Since multiplying both sides of an inequality by something negative reverses the sign, you need to consider two cases separately: and

(Since we know , then we don't need to worry about this possibility.Thus, if then

The intersection of these sets is . That is, the result of combining the above two inequalities is

(Since we know , then we don't need to worry about this possibility.Thus, if then

So now you know that the values for which your inequality hold are:

andThe intersection of these sets is . That is, the result of combining the above two inequalities is

To summarise, the left-hand side of what your lecturer wrote is fine. For the rest, see the above.

*Last edited by zetafunc (2016-12-20 12:10:27)*

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**Mathegocart****Member**- Registered: 2012-04-29
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Hannibal lecter wrote:

zetafunc wrote:The important part to take away from your lecturer's solution is that is equivalent tobut mathegocart didn't do like the lecturer did, what does it changed? how the solution is

,and the sol of the teacher is :

, U {-1}.zetafunc wrote:it's important in general to go back and check whether or not the solutions you've found are actually valid.

but how to do that? I mean how to check wether or not the solution is valid or not????

how did you know the solution was fine please help with that...

Put one value that supposedly satisfies the inequality(make sure it is convenient though), and see if you can verify it to be true. You could also graph it, but that is tedious.

*Last edited by Mathegocart (2016-12-20 07:37:56)*

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**Hannibal lecter****Member**- Registered: 2016-02-11
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zetafunc wrote:

can't be a solution, because while

and what is that mean if I mean from where did you get |-1+3| there is just |x-3| and |x+1| in the expression I don't see |x+3|?

and do you mean it can't be equal to -1 because when we replace -1 by |x-3| ≤ |x+1| it will be |-1-3| ≤ |-1+1| then |-4| ≤ |0| and that is false it can't be -1, ... is that right? is I'm doing a good step or not....

zetafunc wrote:

So now you know that the values for which your inequality hold are:

and

The intersection of these sets is . That is, the result of combining the above two inequalities isTo summarise, the left-hand side of what your lecturer wrote is fine. For the rest, see the above.

you said inequality are hold x > -1

and x ≥ 1.

so where is number zero? why the inequality sol is not [0,∞) ?

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Hannibal lecter wrote:

do you mean it can't be equal to -1 because when we replace -1 by |x-3| ≤ |x+1| it will be |-1-3| ≤ |-1+1| then |-4| ≤ |0| and that is false it can't be -1, ... is that right? is I'm doing a good step or not....

Yes, that's what I meant.

Hannibal lecter wrote:

you said inequality are hold x > -1

and x ≥ 1.so where is number zero? why the inequality sol is not [0,∞) ?

Which values of x satisfy both x > -1 and x ≥ 1?

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**Hannibal lecter****Member**- Registered: 2016-02-11
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I think I got is thanks for your help

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