Wikipedia:Reference desk/Archives/Mathematics/2010 November 17

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November 17

Function

Is it correct (or rigorous, I should say) to say that functions are not in fact rules (defined by elementary functions such as log, power, multiplication, addition, etc., for a random example f(x)=3x^2+ln(x)) but infinite collections of points, which on certain intervals can sometimes be generated by a rules? I'm supposed to be doing a presentation project and this would help clear a certian point up. 24.92.78.167 (talk) 01:04, 17 November 2010 (UTC)

No, I would say that neither statement is rigorous. A function need not have a rule that we can easily write down; when there is a such a rule, we would say it is an elementary function. (The idea of elementary functions is not actually that useful for most mathematicians; and when it is used, may give different lists for the building blocks used to construct elementary functions. The article gives a reasonable definition, but not the only possible one.) Your second idea would more often be called piecewise elementary functions. (The idea is that if we take pieces of elementary functions and glue them together, we get a new function.)

The rigorous definition of a function is fairly simple, but abstract (which means it won't seem simple to anyone not comfortable with abstract language). You seem to be thinking about functions from (some subset of) the set of real numbers to the set of real numbers, which is the kind of function most people think about. (But you can define functions from any set to any other set.) This Math.Stackexchange discussion gives the rigorous definition, and several nice ways to explain it. I really like the "function monkey" idea. 140.114.81.55 (talk) 02:18, 17 November 2010 (UTC)

The reason I say neither statement is rigorous is that there are many functions that don't fit either definition. There are functions so messy that no finite description with the language we have can give the "rule" for the function. Suppose you find a black box lying on the street some day. You can give it any real number, and it gives you back a real number. It doesn't matter if anyone ever understands the pattern of how the black box works: as long as it is consistent with itself, it is a function. (I.e., it always gives you the same result any time you give it 1, or pi, or -24.32325.)

One other comment: when you say "an infinite sequence of points," it's not quite clear to me what you mean. "Sequence" usually means a countable list, like {1,2,3,4,5,...}. If you are thinking of something like all real numbers from 0 to 1, I wouldn't call that a sequence because it's not a countable set.140.114.81.55 (talk) 02:32, 17 November 2010 (UTC)

oops, sorry I didn't mean all functions. I fixed it to say certain intervals.—Preceding unsigned comment added by 24.92.78.167 (talk) 02:41, 17 November 2010 (UTC)

The rigorous definition of function that I learned is, IIRC, this: A function $f$ from set $A$ to set $B$ is defined as a subset of $A\times B$ satisfying that for each $a\in A$ there's precisely one element of $f$ whose first term is $a$. Not a rule, not a monkey, not a black box, just a subset of a product. (Then you need to know the definition of $A\times B$. IIRC that can most readily be defined as the set of $\{a,\{a,b\}\}$ where $a$ ranges over $A$ and $b$ over $B$; the "first term" referred to above is then the $a$ in $\{a,\{a,b\}\}$.)—msh210℠ 06:29, 17 November 2010 (UTC)

I think that should be $\{\{a\},\{a,b\}\}$, so that it contains exactly one singleton set and one doubleton (unless a = b) See Ordered pair. AndrewWTaylor (talk) 14:34, 17 November 2010 (UTC)

Not necessarily. See Ordered pair#Variants.—Emil J. 14:41, 17 November 2010 (UTC)

Right, AndrewWTaylor, sorry. The "first term" is then $a$ anyway.—msh210℠ 16:46, 17 November 2010 (UTC)

ring problem

Hey guys. Just a quick question with a problem on rings. Here's the problem I need to solve/prove:

For any arbitrary element x of a given ring <R, +, •>, it satisfies x•x=x. Prove that for any arbitrary element x, x+x=0

A simple enough problem, yet I'm stuck and can't go any further. The problem is that all I know about rings is from the little I could find on wikipedia. Sure, I've learnt about operators and identities and inverses at school (I'm currently in 11th grade), but that's about all I know in this field. Anyways, despite my laughable amount of knowledge on rings, I managed to reach this conclusion: if we denote e as the additive identity of this ring, then x+x=e for any arbitrary element x.

(Here's how I reached this conclusion: Since x•x=x, we can say that (x+x)•(x+x)=(x+x). But since the distributive law holds, (x+x)•(x+x)= x•x+x•x+x•x+x•x = x+x+x+x. Therefore, x+x=x+x+x+x. Hence, x+x=e.)

After that, I tried to prove that e=0, but to no avail. So my question is this: in this particular problem (or generally speaking), is 0 denoted as the additive identity, or the actual real number 0?

And if 0 isn't the additive identity, how can I go about finishing off this proof?Johnnyboi7 (talk) 14:33, 17 November 2010 (UTC)

0 denotes the additive identity in this context.—Emil J. 14:36, 17 November 2010 (UTC)

Use your common sense, how can 0 be the real number 0? It's not necessarily in the set R. This is called abuse of notation, and as you progress in math you'll find many authors abuse notations to the limit. Money is tight (talk) 19:36, 17 November 2010 (UTC)

Ah, see how unfamiliar I am with the concept of rings here? I actually thought that the set R denoted the set of real numbers R, since the problem didn't specify what R was. And so naturally, my initial guess was that 0 denotes the real number 0. My bad. Now I can see that the set R is called R in reference to the "Ring" it is related to. Thanks EmilJ and Money is tight.Johnnyboi7 (talk) 03:19, 18 November 2010 (UTC)

In general (though certainly not always) the real numbers are $\mathbb {R}$ while an arbitrary ring is just R. 67.158.43.41 (talk) 04:10, 18 November 2010 (UTC)

Integer

Is 0 an integer? —Preceding unsigned comment added by 24.92.78.167 (talk) 22:32, 17 November 2010 (UTC)

Yes. Algebraist 22:33, 17 November 2010 (UTC)

Zero is always an integer and an even number. It is neither positive nor negative. The only debatable category is whether or not it is a natural number. Dbfirs 08:46, 18 November 2010 (UTC)

Well, it depends on which zero you mean. Arguably the zero of the real numbers is a distinct object from the zero of the integers or natural numbers, even though in most cases it's expedient to elide the distinction. --Trovatore (talk) 09:58, 18 November 2010 (UTC)

I've also seen the convention that zero is both positive and negative, rather than neither. I don't think that's very common, though. Algebraist 12:09, 18 November 2010 (UTC)

I think that's the usual rule in French. I suppose there's nothing inherently wrong with it. I hope it doesn't diffuse into English, though; I wouldn't like to have to start doubly disambiguating, saying strictly positive on the one hand or positive or zero on the other. --Trovatore (talk) 19:45, 18 November 2010 (UTC)

There are already common words for "positive or zero" and "negative or zero": nonnegative and nonpositive. So I don't think there need be much fear of such diffusion.—msh210℠ 20:53, 18 November 2010 (UTC)

It can happen. Mathematicians taught in French publish in English. People translate from Bourbaki. --Trovatore (talk) 22:06, 18 November 2010 (UTC)

Not the only debatable category, Dbfirs. Also debatable is whether it's a whole number.—msh210℠ 20:53, 18 November 2010 (UTC)

Well, whole number is not really a mathematical term. In the United States (and for all I know maybe elsewhere) it's used in mathematics education, but not in mathematics itself. --Trovatore (talk) 22:05, 18 November 2010 (UTC)

Right, not a term of professional math (AFAIK), but Dbfirs, to whom I was responding, didn't specify he'd been restricting himself to that domain. I guess whether it's "used in math" or not depends on whether you're a member of the AMS or the MAA.  :-) —msh210℠ 06:46, 19 November 2010 (UTC)

Points taken, though I would have thought that there could be little debate about whether zero is a whole number. If your definition of "whole number" is "integer" or "non-negative integer" then zero is included, and if your definition is "positive integer" then it isn't. I've always taken the term to be a synonym of "integer", but perhaps some American educationalists disagree? The debate would be about what you mean by the term, not about whether zero is included. (I suppose the same could be said about "natural number", but there is much more debate about where is the natural place to start the counting numbers.) Dbfirs 09:29, 19 November 2010 (UTC)

In the Houghton–Mifflin books used in my junior high and high schools, a whole number was what I now call a natural number (that is, including zero) whereas the natural numbers excluded zero. They also had some weird notations that are hardly ever seen in research mathematics. I think the whole numbers were W, which is natural enough, and maybe the naturals (without zero) were N, but the integers were J, I think? Hard to say how that came about. Maybe they wanted to use I, but worried that it would be confused with the numeral 1? --Trovatore (talk) 09:58, 19 November 2010 (UTC)

The textbook I use for high school now adopts J for the integers. This practice is quite mystifying, given the prevalence of Z and the fact that J is sometimes used to represent the irrational numbers. —Anonymous Dissident^Talk 12:12, 19 November 2010 (UTC)

Table of mathematical symbols uses Z for integers.—Wavelength (talk) 04:44, 20 November 2010 (UTC)

My high school algebra class (and book?) used whole numbers to mean either positive integers or nonnegative integers, I can't remember which. I think they went with "whole" for "counting" and so started them at 1, with naturals starting at 0. For my money, I go with $\mathbb {Z} ^{+}$ much of the time anymore to avoid possible confusion, though I haven't found a simple-to-write, unambiguous symbol for the nonnegative integers ($\mathbb {Z} ^{+}\cup \{0\}$ is gross). 67.158.43.41 (talk) 03:14, 20 November 2010 (UTC)

I use ω. Unambiguously includes zero. Unfortunately it might not be understood outside of a set-theory context. --Trovatore (talk) 00:56, 21 November 2010 (UTC)

[1] uses $\mathbb {N}$ for the natural numbers without zero and $\mathbb {N} _{0}$ for the natural numbers with zero. I personally think this is a very clean convention. JamesMazur22 (talk) 14:30, 20 November 2010 (UTC)

Except that there isn't really all that much use for the natural numbers without zero. The modern trend is to assume that the natural numbers, represented by N (or the bbb equivalent) do include zero. Of course both conventions still exist, so in the (fairly rare) case that it actually matters, you do need to clarify. --Trovatore (talk) 00:19, 21 November 2010 (UTC)

Except that automated proof checker programs I've looked at seem to have the natural numbers starting with 1 rather than 0. For some reason that seems to be more useful for them. Dmcq (talk) 10:29, 21 November 2010 (UTC)

Flavor

Is chocolate a flavor? —Preceding unsigned comment added by 128.62.81.128 (talk) 23:08, 17 November 2010 (UTC)

This is the mathematics reference desk, and the question is not a mathematical one. Bo Jacoby (talk) 23:42, 17 November 2010 (UTC).

No.—msh210℠ 01:59, 18 November 2010 (UTC)