Talk:Absolute value

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1 Needs disambiguation
2 Fromat issue (old discussion prior to Feb 25, 2002)
3 bar notation
4 Link suggestions
5 "informal" definition in the lead
6 Is the seventh property ok?
7 Rewrite of article
8 terminology should be precise
9 tvexiall?
10 Multiplicative Absolute Values
11 Absolute value a degenerate hyperbola?
12 Defferentiability
13 Division
14 Another Definition of Absolute Value

[edit] Needs disambiguation

It seems to me that this page needs disambiguation as absolute values is a term that is also used in philosophy.

[edit] Fromat issue (old discussion prior to Feb 25, 2002)

I call that ugly and further, the symbols are no longer symmetric and it is harder to read for someone not accustomed to the greater than and equal sign. I can't imagine why you changed it. RoseParks

I made the major changes to this page, mainly to wikify what appeared to be HTML-based text and to make it more readable from my own esthetic perspective, so if you think there is something that doesn't work, I'd be happy to fix it, but you'll have to be more specific. I don't know what you mean by "the symbols are no longer symmetric", and I don't know what you mean by someone being "unaccustomed" to the ≥ sign. We have to express that idea somehow, and the only two reasonable ways are "≥" and ">=". But I think the former is more readable for students of mathematics, while the latter is more readable for computer geeks. I can't imagine the article being of any use at all to someone who doesn't know the basic symbols of elementary mathematics; if you think it might be useful for readers that elementary, perhaps the article could contain links to other basic articles explaining the comparison operators? Are you perhaps talking about the alignment of the blockquote? How do you think it should appear? --LDC---- I had used ≤ and ≥ to match "<" and ">." Actually, someone else took the font size out before you. And, I just put it back...:-(..RoseParks

OK. That's a very font-specific thing (on my machine the ≥s look a bit too small), and it makes the text a real pain to edit, but if you think it makes a real readability difference on your machine, and the text doesn't need to be edited much, go for it. I won't remove any further ones I see. I would hesitate to make that a standard practice for math pages here in general unless we do end up using something like TtH to do the conversions automatically so we won't have to edit all those font commands. I do think agree that esthetic details can make a big difference in the readability of math formulas (I really wish there were an easy way to vertically center the internal ||s within the larger enclosing ||s in rule 3 above), but the limitations of HTML are pretty severe, so you can't have everything you might want, and what works on one machine might not work on others. --LDC

[edit] bar notation

what's the difference between $|x|\,$ and $\|x\|\,$ ? It should be specified in the article. - Omegatron 18:04, Sep 26, 2004 (UTC)

The latter notation is usually reserved to represent some kind of norm, such as in a normed linear space. Revolver 03:50, 27 Sep 2004 (UTC)

[edit] Link suggestions

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[edit] "informal" definition in the lead

I've changed the following:

In mathematics, the absolute value (or modulus) of a number is the difference between that number and 0. Simply speaking, it is the number without a negative sign. So, for example, 3 is the absolute value of both 3 and −3.

to:

In mathematics, the absolute value (or modulus) of a number is its numerical value without regard to its sign. So, for example, 3 is the absolute value of both 3 and −3.

My reasons are that I find the former both confusing and misleading. For example, for the number three, "the difference between that number and 0", can be construed as either "0 - 3" or "3 - 0". And saying that "it is the number without a negative sign" leads to the misimpression that |-a| = a.

Paul August ☎ 13:46, Apr 17, 2005 (UTC)

I think the change is good. We should mention the misimpression |-a| = a. I suspect it may not be obvious to those with little math knowledge. -- Taku 16:42, Apr 17, 2005 (UTC)

[edit] Is the seventh property ok?

Is is ok that |x|=sqrt(x^2) as it is written in the article? As far as I understand, the square root has two solutions: one that is positive and the other one that is negative (except for sqrt(0)). That's why I would like to fix the article but I'm not completely sure about it because the one who wrote it must have had some idea in mind. (Unsigned comment by User:Erast)

Yes, the seventh property is correct. While it is true that for every non-negative real number x, there are two (provided x ≠ 0) numbers whose square equals x, the symbol " $\sqrt x$ " denotes the principal square root of x, that is the non-negative real number whose square is x see: Square root. Paul August ☎ 15:46, Apr 19, 2005 (UTC)

[edit] Rewrite of article

I've just completed a major revision of this article in which I have:

Expanded slightly the lead, better placing (I hope) the absolute value in context.
Combined the "Definitions" and "Properties" sections into one section called "Real numbers".
Named, organized, reformatted, and added to the properties in the "Real numbers" section.
Moved all the content about the complex absolute value to one section called "Complex numbers", and expanded.
Moved content pertaining to absolute value functions to its own section called "absolute value functions", and expanded.
Created new section about absolute value in ordered rings.
Created new section called "Distance", describing the relationship between the absolute value and distance.
Created new section about absolute value in fields.
Moved content about norms and vector spaces to new section called "Vector spaces", and expanded.
Added a "References" section, and some notes.

I'd appreciate any comments/criticisms anyone might have. Thanks, Paul August ☎ 19:49, July 16, 2005 (UTC)

Wikipedia logo appears on top of article text in the following browsers: Mozilla 1.7.8 (Debian Sarge), Konqueror (Debian Sarge), Netscape 7.1 (Windows XP). --Trovatore 21:20, 16 July 2005 (UTC)

Travatore: Can you isolate what the problem is? Paul August ☎ 21:24, July 16, 2005 (UTC)

For the record, it looks ok to me in Safari, Firefox and IE (MacOS X). Paul August ☎ 21:31, July 16, 2005 (UTC)

Experiments in my sandbox suggest that it's the Ent and Rf templates. Removing just the Rf's doesn't solve it, but removing both Ent and Rf does. I haven't tried removing just Ent. --Trovatore 21:42, 16 July 2005 (UTC)

I've replaced each template with it's substitution, did that fix it? Do other articles use thes templates like Demetrius of Pharos have this problem? Paul August ☎ 21:54, July 16, 2005 (UTC)

Didn't fix it. It must be a very strange bug--removing any 2 of the 3 Ent templates fixes the problem, but removing 1 of 3 (no matter which one) does not. --Trovatore 21:58, 16 July 2005 (UTC)

Demetrius of Pharos looks fine. --Trovatore 21:59, 16 July 2005 (UTC)

I've made another change, added ending div tags, does this help? I'm afraid I'm flying blind, grasping at straws and mixing my metaphors (when i'd much rather be mixing my drinks). Paul August ☎ 22:17, July 16, 2005 (UTC)

Yes, it worked --Trovatore 22:21, 16 July 2005 (UTC)

The last change (inserting the <div>s) worked --Trovatore 22:15, 16 July 2005 (UTC)

The older version Trovatore mentioned at Wikipedia talk:WikiProject Mathematics still works fine for me in Mozilla 1.6 and in Konqueror 3.2.2-4 on Fedora Linux. Oleg Alexandrov 22:51, 16 July 2005 (UTC)

Hm. I don't know why that would be. It certainly doesn't seem likely that it's a peculiarity of my machine, given that I observed the problem under both Linux and Windows, and in both a Gecko and a non-Gecko browser (I don't think Konqueror uses Gecko, does it?). --Trovatore 22:54, 16 July 2005 (UTC)

Thanks Oleg, and Trovatore, for helping out with this problem. I would very much like to get to the bottom of it, particularly since I'm the author of the "rf" and "ent" templates, and I use them a lot, so any further help would be greatly appreciated.

Just tried it on Netscape 4.79 (WinXP), but none of the Wikipedia really works very well there, so it's hard to tell.--Trovatore 23:36, 16 July 2005 (UTC)

One thing that strikes me, though: If Wikipedia is not putting on the closing </div> tags, then it's not producing compliant HTML, and that's something that should be fixed, irrespective of problems in any particular extant browsers. Sounds like a developer issue. --Trovatore 23:41, 16 July 2005 (UTC)

Regarding the content of the article, do either of you have in comments?

Paul August ☎ 23:13, July 16, 2005 (UTC)

Articles should ideally not be in proposition-definition form (proofs are fine, it's just the layout of content that's the issue), but that's just my POV. Dysprosia 01:37, 17 July 2005 (UTC)

Yes, I don't usually use this "proposition" style, but I chose to in this case, because I wanted to group and name the properties of the real absolute value, so I could refer to them further down in the article. Proposition 1, is the identity, which can be seen to be related to the definition of the complex absolute value, Proposition 2, is the group of properties used to motivate the definitions a metric, and of the absolute value in the case of fields and vector spaces, and Proposition 3, is a group of properties which "come for free" as a consequence of adopting the properties in Proposition 2 as axioms. Also I think listing them in this layout helps to show the various structural relationships. Having said all that, feel free, of course to make or suggest any changes you think appropriate. I'm very open to better ways of presenting this material. Paul August ☎ 02:46, July 17, 2005 (UTC)

Perhaps giving the results names instead of merely calling them propositions would help. It would also be advantageous to possibly tie the idea of norms and inner products in slightly, in some of the cases (esp. the real case). Dysprosia 06:38, 17 July 2005 (UTC)

Dysprosia, could you be more specific? Paul August ☎ 16:45, July 17, 2005 (UTC)

In the real case, we can define an inner product over R (which is a vector space) to be regular multiplication of reals. The absolute value of a number is then the same as the vector norm with respect to the inner product just defined. Likewise for the complex case. It would be nice to mention this in relation to the absolute value. In a sense the article does so, but it doesn't seem too unified. This may also take care of the first problem of mentioning things as "propositions" instead of giving them more illustrative names. Dysprosia 04:49, 18 July 2005 (UTC)

Well we could make the relationship between inner products and absolute values more clear, but I would be inclined to do that in the "Vector space" section. I would prefer to keep the "Real numbers" section as basic as possible, so that it could, for example, be read by a high school student. As you say the article does mention this relationship in the "vector space" section, but I agree, it is a bit disjointed. I will/may try to rewrite that section. Any suggestions on how best to do that (or doing it yourself) would be appreciated. However, this doesn't address your "propositions" concern. Could you explain your suggestion about giving "them more illustrative names"? I'm not sure I understand. Could you give examples? Paul August ☎ 15:04, July 18, 2005 (UTC)

I'm not saying rewrite the real numbers section completely in terms of vector space terminology but merely mentioning it somewhere to give a view of the big picture and to give the page more of a unified feel. Instead of saying "Proposition 1" one could say "Alternate definition", for "Proposition 2" one could say "Axiomatic properties" (thus linking with the properties of the norm), for "Proposition 3" one could say "Consequent properties", perhaps. Dysprosia 12:44, 19 July 2005 (UTC)

Shouldn't the fact that absolute value generalizes into norm be mentioned (more prominently) in the heading? I have also heard that absolute value on reals is related to valuation (I don't know any details, I may be completely wrong). Samohyl Jan 09:29, 17 July 2005 (UTC)

I've added a mention of the norm in the lead. Paul August ☎ 16:45, July 17, 2005 (UTC)

The article suggests that absolute values and valuations are synonyms, but this is wrong. The usual absolute value on the rationals does not correspond to any valuation.--Gwaihir 10:45, 17 July 2005 (UTC)

Well as defined in the article, a real-valued function on a field which is non-negative, positive-definite, multiplicative, and subbadditive is called both an "absolute value" and a "valuation" according to Eric Schechter in Handbook of Analysis and Its Foundation pp 259-260. see also PlanetMath: "Valuation", and the absolute value for the rationals is a "valuation, under this definition. I'm unfamiliar with the definition given in valuation (mathematics). Paul August ☎ 16:45, July 17, 2005 (UTC)

MathWorld has both, and IIRC Bourbaki (Algèbre commutative) uses the logarithmic definition.--Gwaihir 09:17, 20 July 2005 (UTC)

One thing that doesn't seem to make sense: In the section on distance, I believe there was a mistake. |a+b| doesn't equal $s q r t a 2 + b 2$ It's $s q r t (a + b) 2 = s q r t a 2 + a b + b 2$ Exponentation doesn't distribute additively.He Who Is 22:09, 27 April 2006 (UTC)

[edit] terminology should be precise

an "absolute value in vector spaces" seems completely out of scope to me. A norm is not an absolute value, it does not verify the multiplicative identity (except in some algebras). Although there should be a reference to norm (insisting on the necessity of having an abs value on the scalars), there should not be so much details about norms here, since it is out the subject.
the computer functions calculating the absolute value has many other names than abs(), e.g. fabs(), dabs(), mmabs(), ieeeabs()...
there IS a general definition for an abs value, which should be found here
a valuation is NOT an absolute value: it has another definition (see e.g. Lang's "Algebra").
there should be a note about important results that cam be deduced from the existence of an absolute value on a ring or on a field (e.g. that the latter contains the rationals).

— MFH: Talk 13:35, 28 September 2005 (UTC)

MFH: I've numered your points above. I will rely by the numbers ;-) — Paul August ☎ 15:56, 28 September 2005 (UTC)

1. I believe several authors refer to a norm as an "absolute value", but I don't have any references ready to hand. Perhaps we should de-emphasize that somewhat. However the norm is certainly a generalization of the idea of absolute value. You are correct, a norm in general does not satisfy the multiplicativeness property of the real and complex absolute values. It's "corresponding" property is positive homogeneity (also called positive scalability): $\|a \mathbf{v}\| = |a| \|\mathbf{v}\|$ . I think there is a good reason to have a thorough discussion and explication of the generalizations of the idea of absolute value here. — Paul August ☎ 15:56, 28 September 2005 (UTC)

2. Feel free to add those. — Paul August ☎ 15:56, 28 September 2005 (UTC)

3. What is that general definition? — Paul August ☎ 15:56, 28 September 2005 (UTC)

4. Eric Schechter's Handbook of Analysis and Its Foundation pp 259-260 and PlanetMath: "Valuation", define valuation in the same way that it is defined here, as a real-valued function on a field which is non-negative, positive-definite, multiplicative, and subbadditive. I don't have a copy of Lang's book, what is his definition? — Paul August ☎ 15:56, 28 September 2005 (UTC)

5. Feel free to add those too ;-) — Paul August ☎ 15:59, 28 September 2005 (UTC)

[edit] tvexiall?

I've removed from the article twice the following:

The V-shaped graph of an absolute value function is called a tvexiall.

This was first added by 68.162.147.182, and later by Emorgasm. I've never heard of this term and neither has Google. Anyone know anything about this? Paul August ☎ 17:01, 18 November 2005 (UTC)

Never. Sounds like some kind of bizarre neologism to me. It seems they intended it to be a term along the same lines as "parabola", describing the shape of the curve. It might be a misspelling of something, but I have no clue what. Deco 01:05, 12 December 2005 (UTC)

[edit] Multiplicative Absolute Values

Since for all x and y the actual 'difference of x and y can be represented as |x-y|, and that equals x-y if x is the greater, it only stands to reason thata similaroperator should be defined multiplicitively, such that:

$MAV\left( \frac{x}{y} \right) = \left\{ \begin{matrix} \frac{x}{y} & \mbox{if } x > y \\ \frac{y}{x} & \mbox{if } y > x \end{matrix} \right.$

This would give no results that lie on the interval [-1,1], 1 being the multiplicitive identity. This is analogous to the absolute value giving no results less than 0, whitch is the additive identity. Graphed, this would hyperbolic over [-x,x] and linear outside that. Anyone care to comment on the subject?

He Who Is 00:07, 27 April 2006 (UTC)

Here are my thoughts on your idea. A formula which suggests itself:

$MAV(x,y)=\exp \left | \log \left( \frac{x}{y}\right) \right |.$

Note that I have denoted it as being a function of two variables. If you want to define it as a function of one variable, as your notation suggests, then its definition is slightly problematic for irrational numbers and negative numbers. However, I don't think this formula is really what you want. You want a function like this:

$f(x) = \begin{cases} x & |x| >1\\ \frac{1}{x} & |x|<1 \end{cases}$

and then you can define MAV(x,y) = f(x/y). This seems to extend naturally to all complexes without difficulty:

$f(z) = \begin{cases} z & |z| >1\\ \frac{1}{\overline{z}} & |z|<1 \end{cases}$

and I take the complex conjugate to make the number's argument stay the same, though you don't have to do that if you don't want to. This is actually a bit reminiscent of the concept of radial ordering, which is used in quantizing fields on the worldsheet in string theory. -lethe ^{talk +} 00:40, 27 April 2006 (UTC)

The only problem I have with making MAV a function of two variables is that it loses its analogy to ABS in that it is no longer graphable on a requalar two-dimensional carteisian plane. Because of this, while it can be said that -5, for instance has an ABS of 5, -5 doesn't have a Multiplicitive Absolute Value unless it is paired with another number. And as for problems concerning negatives and irrationals, as far as I can tell MAV(x) = MAV(-x), and if you could elaborate on the problems with irrationals, that would be helpful, as they're not apparent to me. He Who Is 01:57, 27 April 2006 (UTC)

If you want a function of a single variable, see my second suggestion above. The problem with your definition is, –1/2 = (–1)/2 = 1/(–2). According to the first representation, –1 < 2, so MAV(–1/2) = –2. According to the second representation, 1 > –2, so MAV(–1/2) = –1/2. For positive rationals, each positive rational can be reduced to a fraction of the form p/q, where p and q are relatively prime. Irrational numbers generally can be written as fractions many different ways, and there's no way to choose one. The best thing to do is forget about fractions, and just check whether the real number is bigger than 1 or less than 1, which gives you the definition I gave. -lethe ^{talk +} 02:21, 27 April 2006 (UTC)

I think you mean check it it's absolutely greater than one. Also, in your example, you were comparing two numbers within a fraction, but the need not be represented as a fraction at all. Also, you said that "Irrational numbers generally can be written as fractions many different ways, and there's no way to choose one." This is inherently untrue, as by definition an rational number is: $\mathbb{Q} = \left\{\frac{a}{b}|\left\{ a,b\right\} \sub \mathbb{Z}\right\}$ and an irrational is it's complement on $\mathbb{R}$ . Therefore, Irrationals cannot be represented as any fractions. —The preceding unsigned comment was added by He Who Is (talk • contribs) .

I think you've finally understood my point, which is this: forget about fractions. I'm slightly bothered however by the fact that now you're representing this point as your own, and trying to teach it to me. It is not me who was comparing two numbers in a fraction, that was you. I was the one telling you that you should not be doing that, because some numbers can be represented more than one way as a fraction. In my definition, I compare a single number to the number 1. No fractions in sight. As for your suggestion that irrational numbers cannot be represented as fractions, let me show you a counterexample: π=π/1=3π/3 = –π/(–1). -lethe ^{talk +} 19:50, 27 April 2006 (UTC)

I apologize, as I must have misspoke. I meant that irrationals cannot be represented as fractions of integers. Also in what case was I comparing components of a fraction? If I was I hadn't meant to, unless you mean representing a reciprocal as 1/x, which is the only way I know of in which they can be represented. I fear this entire conversation may have just come from a few misunderstandings.He Who Is 20:06, 27 April 2006 (UTC)

You defined MAV(x) as x/y if x>y, in your first post in this thread (scroll up to see for yourself). Here, you take a real number, and compare the numerator with the denominator. For various reasons I mention, this is problematic. One way to get around it is to view the MAV function as a function of two variables; then you're simply comparing the first variable to the second. Another is to forget about fractions, and simply compare the real number to the number 1. I suggested to you that the second method was the better way: forget about fractions. You indicated that you didn't understand the difficulties with fractions, so I explained further. A few posts later, you tried making the same point to me, which struck me as odd, since it was the same point I made to you in my first post to this thread. Finally, your point about irrationals admitting no representations as fractions of integers is true, but is no longer a correction to anything I wrote. -lethe ^{talk +} 20:14, 27 April 2006 (UTC)

Oh... You're right. And I now see what you mean. Inherently, this is the same as your notation posted thereafter, but as irrationals cannot be represented as fractions of rationals, it presents a problem. I suppose I had forgotten how I represented it in my first post, and for that reason was confused by your subsequent replies.
This aside, I was recently considering this and something strange struck me. Where x plus its additive inverse equals zero, the additive identity, and x times its multiplicitive inverse equals 1, the multiplicitive identity, a theoretical exponential inverse seems slightly more complicated. It follows that 1 would also be an "exponential identity", but for all complex numbers z, z^0 = 1. This implies an exponential inverse of 0 for all numbers. But this does not follow in that x minus -x = 2x, x divided by 1/x = x^2, but the 0th root of x does not equal x^^2, as would implied.which seems contridictory... He Who Is 20:32, 27 April 2006 (UTC)

Yes, I see what you mean. Well, let's look more closely: z^1=z for all z, so 1 is a right-identity, meaning that when we put 1 on the right-hand side of the exponentiation, we get back what we started with. Unfortunately, it is not also a left-identity, since 1^z≠z, for all z. I guess there is no left-identity. When there are both a left-identity and a right-identity, then they have to be equal (at least when the operation is associative), which allows us a flexible definition of inverse. We don't have that, and exponentiation is not associative (since 1^(2^3) = 1 ≠ 8 = (1^2)^3), so our inverse is also in trouble. Nevertheless, we could try to imagine 0 as some sort of "exponential right-inverse". It's not an "exponential left-inverse", since 0^z≠1 for all z. But check this out, there are two inverse operations of exponentiation: extracting roots, and taking logarithms with a given base. Under addition, we get 0–x = -x, and under multiplication we get 1÷x = x^–1, and under exponentiation, we get log_x 1 = 0 (which is the exponential inverse of x). This doesn't work with root extraction: 1^1/x≠0. Thus, what we really want, instead of taking roots, is taking logarithms. Trying your equation again, I get log_x 0, which is probably some infinity. Alright, you're right. It just doesn't work.

It's a phenomenon that I've thought about in the past: addition is a nice operation. Iterate it, and you get multiplication, another nice operation. Iterate again, and you get exponentiation, which is not so nice. What happened in the second iteration that didn't happen in the first iteration? I never came up with a satisfactory answer to this question. -lethe ^{talk +} 21:19, 27 April 2006 (UTC)

Yes. It's a strange phenomenon. Why does exponentiation lack the commutativity of addition and mutiplication? This subject is closely related to the hyper operator, which I've been doing some work with lately. Namely, to generalize for things like $2 (1.5) 3$ ,which essentially would ask to perform the operation halfway between addition and multiplication. Perhaps understanding this could give a better reason for the differences from exponentation on up. And another related question: where does commutativity disappear? Does it exist only for addition and multiplication, or for everything in between them also? If so, this would mean that the functions hyper(a,x,b) and hyper(b,x,a) coincide from 1 to 2, but seperate outside that which doesn't seem to make sense.

And for the record, I believe log_x0 is in fact negative infinity, which makes sense looking at an exponential graph, as it diverges from 0.He Who Is 21:59, 27 April 2006 (UTC)

To try to understand this question, I turned to abstract algebra. Starting with an abelian group operation (addition), by iterating you get something that, while similar to multiplication, is not precisely a multiplication. That is to say, you get a left Z-module, mutliplication by "numbers" on the left by integers. For general modules, the left R-module is not isomorphic to the right R-module. Over abelian rings, they are isomorphic. So viewed in this way, the commutativity of multiplication is actually a restatement of the commutativity of addition. This framework provides a little bit of insight as to why iterated addition is still commutative, but it doesn't really support any definition of exponentiation. Perhaps this is the reaon, though I don't find this explanation completely satisfying.

As for log_x 0, over the reals, that quantity is undefined. Over the extended reals, negative infinity is indeed the natural definition for values of x greater than 1, while positive infinity is a natural definition for values of x between 1 and 0, while I'm not sure what to do with 0, 1, or negative numbers. I thought perhaps some insight might be provided by the projective real line, where there is only an unsigned infinity. In the end, I just gave up and left the vague phrase "some infinity". -lethe ^{talk +} 22:49, 27 April 2006 (UTC)

For negative numbers, the answers are simply complex. For 1, it in undefined even over complex numbers, because 1 does not change with exponentation, and as for zero, z^x is zero for all complex numbers, so that could be anything.He Who Is 20:30, 28 April 2006 (UTC)

The logarithm is single-valued on its Riemann surface, infinite-valued on the complex plane (because log –1 = iπ + 2iπn for any integer n), and undefined on the reals. There is, as far as I know, no way to add signed infinities to the complexes, so if you want log 0 to be negative infinity, you can't also want log –1 to be complex. (Maybe you could add an infinity along every line through the origin in the complex plane, getting the extended complex numbers, homemomorphic to a closed disk?) May I make a suggestion? You should be more careful with the domains of your functions. I think that's the third time in this thread that your disregard for the domains of your functions has led you to make statements which might get you into trouble. -lethe ^{talk +} 21:26, 28 April 2006 (UTC)

I apologize for that. I should be more careful. But either way I think a more formal definition for MAV(x) could be found (anologous to the sqrt(x²) for |x|) if another function for which the value at x and 1/x are equal could be found. Then that nested within its own inverse would equal to MAV(x). Do you know of any such function? He Who Is 00:17, 13 May 2006 (UTC)

[edit] Absolute value a degenerate hyperbola?

I have a question... Correct me if I'm wrong, but isn't an absolute value graph just a hyperbola of a circle radius 0? Because if

$y^2 = \sqrt x^2$

Then this would lead to the conclusion that

$y 2 - x 2 = 0$

Thereby being a hyperbola, albeit a degenerate one.

Fisrt of all, you really should sign your messages. But either way, I see three problems in your above statement. First of all, $y^2 = \sqrt x^2$ becomes $y=\sqrt x$ , not $y 2 - x 2 = 0$ . But even simpler that you seem to have misread the formula. Absolute value is graphes as $y=\sqrt x^2$ not $y^2 = \sqrt x^2$ . Finally, y^2-x^2=0 is not a hyperbola, and a hyperbola does not have a radius. In fact if you were to simplfy that, it bould be equivalent to $\pm y= \pm x$ , which would be a large x, made up of x=y, and -x=y.He Who Is 02:21, 18 May 2006 (UTC)

I give my apologies for both not signing my post and also for making a syntactical error. I meant to say $y=\sqrt x^2$ , the squaring of y was an error. This explains both the first and second fallacies in my logic; as for the third, isn't a geometric point considered a circle radius zero? If this is true, then absolute value would, I believe, be the hyperbola of said circle. --ThatOneGuy 07:19, 19 May 2006

Ah. I'm sorry. I read hyperbola and thought parabola. As for your error, You ar in fact correct, come to think of it. The reason this doesn't show in the graph at first glance may lie in the difference between sqrt(x^2-r^2) and sqrt(r^2-x^2). One is your hyperbola, and the other is the circle of the same radius. If the absolute value is taken before extracting the squared root, you get both at once. In the case of the absolute value, the radius is zero, and the hyperbola's slope starts at $\pm 1$ , it does not converge, and simple creates a right angle at the nonexistently small circle that normally seperates both parts. In the case of absolute value, the squared rootis taken to mean the principal root, so only the top half is shown, and because with a redius of zero, neither the circle nor the hyperbola are made up of complex points, they both show up. (even though since it is degenerate, as you said, the circle cannot be seen.) I've never thought about absolute value that way.He Who Is 00:55, 20 May 2006 (UTC)

[edit] Defferentiability

I think a section, or at least a mention, of differential and antiderivitive of |x| should be added. It's differentibility is mentioned, but not it's actual derivitive. (That is, $\frac{d}{dx}|x| = sgn{x} = \frac{x}{|x|}$ . Also, $\int |x| dx = x|x|$ . Which is x² for all x greater than or equal to 0, and -x^2 otherwise.)He Who Is 22:08, 23 May 2006 (UTC)</math>

I think those would be good additions to the article. The derivative of the absolute value might actually be a useful formula (mention the Heaviside step function), but I don't think I've ever seen the integral used. Of course you have to make sure to mention that it's not differentiable at the origin. -lethe ^{talk +} 00:20, 24 May 2006 (UTC)

Added it. It could probobaly use some work, but I think its good for now.He Who Is 01:27, 24 May 2006 (UTC)

I copy-edited the new addition. It's mostly OK, but I also decided to get rid of the stuff about multiple integrals, which seems a bit weird to me (not to mention badly formed). -lethe ^{talk +} 01:53, 24 May 2006 (UTC)

[edit] Division

I'm tempted to remove a certain statement, but decided to see what other wikipedians have to say about it first. It is the statement that claims preversation of division is equivalent to multiplicativeness. This is because of the following: (Remember that this assuming the person performing the calculations doesn't know about preversation of division in absolute value.)

$\frac{|a|}{|b|} = |a| \cdot \frac{1}{|b|}$

You see, preversation of division is a result of multiplicativeness as well as the fact that 1/|b| = |1/b|, which is not implied by multiplicativness. Otherwise you don't have the product of two absolute value, you have an absolute value times the reciprocal of another. -- He Who Is^{[ Talk ]} 22:30, 20 July 2006 (UTC)

But $|b||1/b| = |b \cdot 1/b| = |1| = 1$ by multiplicativity too. Dmharvey 22:54, 20 July 2006 (UTC)

Just so. Paul August ☎ 04:22, 21 July 2006 (UTC)

[edit] Another Definition of Absolute Value

I am not at all advanced in Math, or absolute values, so please correct me if I am wrong. I would like to know if it makes more sense to define Absolute Value as "An integer's distance from zero". Yeah, it's basic, but absolute value isn't just disregarding the integer's sign, is it? --Chrishy Man 02:02, 29 September 2006 (UTC)

First note that absolute value is defined for all real (and complex) numbers not just integers. Second the two definitions amount to the same thing, but I think the article's previous definition is better, since it doesn't involve a geometric interpretation of the real numbers. I've reverted to the original definition. Paul August ☎ 21:15, 16 October 2006 (UTC)

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