Wikipedia:Reference desk/Archives/Mathematics/2007 September 10

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September 10

Vector product

why is the product of two vectors , perpendicular to those two? if it's an axiom why isn't it self evident? —Preceding unsigned comment added by 203.145.156.9 (talk) 02:00, 10 September 2007 (UTC)

It is a definition, rather than an axiom. It happens to be a very convenient definition. Phils 02:51, 10 September 2007 (UTC)

It's definitely not an axiom, just part of the definition of cross product. See Cross product#Cross product as an exterior product. It's not fundamentally a vector (oriented line element), but an oriented area element (bivector). But a plane can be defined by a vector perpendicular to it, so there you go. —Keenan Pepper 04:18, 10 September 2007 (UTC)

Quintics and Up

I've been reading about Galois and the invention of group theory, and I keep running across the same thing that never gets developed - once they knew most large polynomials couldn't be solved in the radicals, what did they do next? It's implied that they quickly found other ways of getting answers, but nothing ever explores what those new ideas were. What did they come up with? Black Carrot 03:40, 10 September 2007 (UTC)

I think your history is a little mixed up. Numerical approximation methods such as fixed point iteration and Newton's method were known long before the Abel-Ruffini theorem was published. Solving high-order polynomials in radicals never had much practical application. Even for a cubic equation it's often much easier to form a sequence that converges to the solution than to go through the whole process to find an exact answer in radicals. Knowing that an exact answer is possible only for degree less than five is mostly of theoretical/philosophical interest.

If instead you're asking about ways to represent exact roots of polynomials that cannot be expressed in ordinary radicals, see Bring radical. —Keenan Pepper 04:44, 10 September 2007 (UTC)

Probabilities

No, this is not a homework question; I'm not in school.

Say you have a pizza shop that has 11 toppings: Pepperoni, mushrooms, onions, green peppers, pineapple, sausage, tuna, olives, ham, artichokes and anchovies. A pizza can have zero, 1, 2, 3, 4... up to all 11 toppings. For example, it can have just mushrooms; olives and onions; pepperoni, green peppers and tuna, etc.

How many different combinations of toppings are there? Please explain how you come up with the number. Thanks -- Mwalcoff 07:31, 10 September 2007 (UTC)

The order in which the toppings are applied does not matter (a pepperoni & mushroom pizza is the same as a mushroom & pepperoni pizza). So you want the count the number of subsets of the set of 11 toppings - not forgetting to include the possibility that there are no toppings at all, which is the empty subset. The set of all subsets of a set is called its power set. Read our articles on subset and power set and you will be able to work out the answer for yourself. Gandalf61 08:07, 10 September 2007 (UTC)

Easy! First assume that the pizza can only have two toppings, mushroom and/or olives.

The solution is 4 types. (no topping), (mushroom only), (olives only) and (mushroom and olives).

So the answer is mushroom(yes/no) * olives(yes/no) = 2*2 = 4

So for 11 toppings the answer is

2*2*2*2*2*2*2*2*2*2*2 = 2^11 = 2*2^10 = 2*1024 = 2048

Solved. 211.28.131.182 13:28, 10 September 2007 (UTC)

Do you assume that no-one wants a pizza with olives and, erm, more olives? Not a safe assumption with pizzas. --Dweller 13:47, 10 September 2007 (UTC)

So, let's allow double olives ... triple olives ... triple olives and hundred-fold mushrooms ... looks like we've discovered multisets. These pizzas sure are educational ! Gandalf61 14:31, 10 September 2007 (UTC)

Thanks. -- Mwalcoff 22:47, 10 September 2007 (UTC)

wording Exponential_function#Formal_definition

e is defined in the article e , but at the above link it states that e^x is defined as the power series given; shouldn't this be a proof that the two are equivalent (for reals), and not a definition.

Additionally is there notation that distinguishes between e^0.5 = |sqrt(e)| and +/- |sqrt (e)| ?87.102.77.35 13:04, 10 September 2007 (UTC)

The usual (because easiest) approach to exponentials is the following: define e^x by its power series. Define e to be e¹. Define log to be the inverse of e^x. Define a^x to be e^{x log a}. Check that this agrees with the previous definition for a=e, and that exponentiation has all the required properties, in particular a^{x + y}=a^x a^y. Thus the power series for e^x is central to the definition of exponentiation in general.

There is an alternative approach, which you might be happier with: define aⁿ as a multiplied by itself n times, then define a^n/m as the mth root of aⁿ, then extend to all real x by continuity. Then you can define e however you want, and define e^x in accordance with your previous definition of exponentials. However, I've never seen a book that actually does this, and I believe it to be substantially more effort to prove all the relevant facts. Algebraist 13:26, 10 September 2007 (UTC)

But by the definition given e^x is singly valued - which is(seems) wrong since e^x can be multiply valued - it shouldn't be quoted as a definition, but as being equivalent for the positive real values of e^x (when x is real) (when x is imaginary only a single value results so that doesn't matter here)87.102.77.35 13:49, 10 September 2007 (UTC)

e^x is single-valued. It has no branch points. Now x^0.5 might be multi-valued, but that is irrelevant. --Spoon! 15:17, 10 September 2007 (UTC)

What I said above concerned only real numbers. For complex numbers, one defines the function sometimes called exp by the power series for e^x, then defines the multi-valued inverse function Log, then the multi-valued functions a^x. One then has to distinguish between the functions exp (which is a single-valued function given by a power series) and e^x, which is multi-valued and is by definition exp^{x Log e}=exp^x(1+2niπ). Algebraist 17:44, 10 September 2007 (UTC)

I'm really confused here. If e^x is single value then what is the single value of e^0.5? Doesn't it have two values? ie. 1.64 and -1.64 202.168.50.40 00:45, 11 September 2007 (UTC)

There are two things commonly called e^x. One is also called exp, and is a well-defined function, given by a well-known power series. The other is a multi-valued function on the complex plane. exp(0.5) is 1.65 or thereabouts, while e^0.5 is 1.65 or -1.65, depending on which branch you choose of this multi-valued 'function'. Algebraist 01:13, 11 September 2007 (UTC)

I wonder, is $e^{x}$ really a "multi-valued function" in the common sense in complex analysis? Sure, you can take different n and end up with different single-valued functions, but for any particular choice there are no branch points, branch cuts or anything similar. Not like $x^{0.5}$, for which you can clearly define a Riemann surface and obtain different branches by taking different sheets of it, or by choosing a branch cut. Perhaps this term is more general than I am accustomed to?

As for the definition of the exponential function, defining it as a power series seems too artificial. My favorite is defining Log as an integral and then Exp as its inverse. My other, similar, choice, would be to define Exp by a differential equation. -- Meni Rosenfeld (talk) 08:14, 11 September 2007 (UTC)

How do I distinguish between e^x as a single valued function and 2.7...^x which can also be written e^x which can have more than one value83.100.251.220 08:42, 11 September 2007 (UTC)

$\exp(x)$ is always the single-valued function; $e^{x}$ might be used to allow for several values. Perhaps we can help more if you explain what do you need a multi-valued notation for, as this is one of those things that require extra care. -- Meni Rosenfeld (talk) 08:45, 11 September 2007 (UTC)

ok I admit I've never needed to use a multivalue e^x that I can think of, it was the semantics of the description that bothered me ie defining e^x as singly valued, when it looks exactly the same as 2.71..^x .. see Exponential_function#Formal_definition, more specifically the linked article seems to defined e^x as a specific function when it already has a meaning (talking e=2.71..) - nobody seems to be understanding what I'm saying - shouldn't the mathematical notation distinguish between the singly valued and multilply value functions...? (otherwise I need to use contextual information to decide...)83.100.251.220 09:55, 11 September 2007 (UTC)

In a perfect world, every piece of mathematical notation would have a clear, unambiguous meaning. However, mathematicians are still humans, and they use whatever notation is comfortable for them - hoping that others would be able to "translate" to a more rigorous formulation should a doubt arise. As it happens, $e^{x}$ should mean the "multi-valued" function - as a special case of the $x^{y}$ notation - but it really isn't all that common to need that, so it is used to refer to the single-valued function in a way which is slightly less clumsy than $\exp(x)$. In short, whenever you run into $e^{x}$, you can assume that it means $\exp(x)$, unless stated otherwise. -- Meni Rosenfeld (talk) 10:27, 11 September 2007 (UTC)

Suppose you were to look at your "multi-valued" e^x or whatever. What would it look like? e^1/2 would have 2 values? e^1/3 would have 3 values? e^π would have infinite values? That's not a very "function" looking thing to me; and is not a very useful definition. --Spoon! 17:58, 11 September 2007 (UTC)

What makes you think that e^x "can be multiply valued"? And what do you mean here by "can"? I have the ability to proclaim that henceforth 2+2 shall be defined to mean 5, and since I'm able to do so, I might claim that 2+2 "can" be 5. Is that the meaning of "can" you have in mind? Or do you think some existing definition of e^x says or implies it is a multivalued function?  --Lambiam 19:56, 11 September 2007 (UTC)

Doesn't the discussion above clarify the legitimacy of the OP's questions? -- Meni Rosenfeld (talk) 20:35, 11 September 2007 (UTC)

Not really. The meaning of e^x where x stands for a real number is obviously the same as exp x. After all, I hope you're not going to tell me that it is "legitimate" to maintain that (for example) 2² is multivalued, ambiguous, and needs some notation to indicate which possible choice we are taking. If we leave that domain, and move to the complex domain, what is the meaning of e^z for complex z? Simply exp z, and we can use the identity e^x+iy = e^x(cos y + i·sin y). I'd say that in general, for a > 0, the meaning of a^b is exp(b ln a), which is still single-valued. Thus, e^z = exp(z·ln e)= exp z. Multivaluedness only rears its unseemly head when we're attempting to extend the meaning of exponentiation to forms a^b in which a is not a positive real number. Then, because of properties of the logarithm function, ln a is multivalued, which then may make exp(b ln a) multivalued. For e^x that does not play a role.  --Lambiam 14:33, 12 September 2007 (UTC)

What about x^m/n where n and m are integers, there are n values that can result, what I was saying was that if x=eulers constant - the formula reads e^m/n, which is indistinguishable from the exponential function as written "e^m/n" - yet the function returns only one value yet there are n roots. Maybe italics is in order here!87.102.16.32 15:55, 12 September 2007 (UTC)

I'm not convinced. Surely the complex function $x^{0.5}$ is multivalued? And, while both $x=2e^{0i}$ and $x=2e^{2\pi i}$ correspond to the complex number 2, we have $(2e^{0i})^{0.5}={\sqrt {2}}$ and $(2e^{2\pi i})^{0.5}=-{\sqrt {2}}$? Likewise for e. I don't see how being positive exempts a complex number from the multivaluedness of the power. -- Meni Rosenfeld (talk) 14:51, 12 September 2007 (UTC)

This then would imply that 2^1/2 means a multivalued ±√2 (why exempt complex numbers that happen to be positive integers?), which is not, however, how this is commonly defined. In the commonly used definition of a^b you treat a positive real a differently from a negative or non-real a, with still special treatment for negative a and integral and some rational values for b. The generalization a^b = exp(b ln a) with a multivalued log function is so mathematically totally useless, that it would be better not to attempt to wring a last drop of meaning out of exponentiation and then get into trouble because of inconsistencies between the useless "generalization" and the original, restricted but useful definition.  --Lambiam 22:35, 12 September 2007 (UTC)

I guess I'll have to think this over. -- Meni Rosenfeld (talk) 08:13, 13 September 2007 (UTC)

I forgot to say thanks (especially those who read what I was saying!!!)Thanks.87.102.16.32 08:52, 12 September 2007 (UTC)

I think the easier way to the exponential function is this. [[1]]. It was once part of exponentiation, but some editors did not like it. Bo Jacoby 10:02, 12 September 2007 (UTC).

(Taking the equation out of context)- I think that starts off as a multiply valued function eg (lim ... )^x and ends up as a singly valued function lim ( ...ⁿ ) ,so it goes from 2.718...^x (more than one value) to a function (the exponential function) that only gives one value.87.102.16.32 13:04, 12 September 2007 (UTC)

There was some confusion in my mind between the usage of e^x since it contains the constant e .. and so could be misinterpreted to mean that there is more than one value when x is a fraction, In common use I can see that taking multiple values would mean that I was "jumping" from path to path on a set of continuous functions...87.102.16.32 13:17, 12 September 2007 (UTC)

coordinates of a point

Hello. I'm not sure that this question has the complete information but anyway. Let X and Y be two points in R^n and Z be a point on the line segment joining X and Y such that |X-Z| = a. What will be the value of t if Z=tX + (1-t)Y and 0<t<1? Thanks.--Shahab 16:10, 10 September 2007 (UTC)

Assuming X and Y are at least distance a apart, there is indeed a unique solution for t. To find it, we have |(t-1)X + (1-t)Y|=a, i.e. (1-t)|X-Y|=a, so t=1-a/|X-Y| Algebraist 17:48, 10 September 2007 (UTC)

Modular arithmetic

Whilst attempting a question involving mod 5 arithmetic, I noticed that for a select few numbers

if

$a+b=5\!$

then

$a^{2}\equiv b^{2}{\text{ (mod }}5)$

Is this always true and, if so, can it be generalised to other mods and powers of 'a' and 'b'? asyndeton 19:14, 10 September 2007 (UTC)

As difference of two squares says: $a^{2}-b^{2}=(a+b)\cdot (a-b)$. PrimeHunter 20:02, 10 September 2007 (UTC)

Well think about this, we have $a+b\equiv 0{\pmod {n}}$ so $a\equiv -b{\pmod {n}}$ Can you get the rest of the way to proving the general case for your hypothesis? Donald Hosek 21:30, 10 September 2007 (UTC)

Quadratic formula

• How do i show that this: $x={\frac {-3\pm {\sqrt {(5^{2})-4x-3x2}}}{2x3}}$ = two thirds or minus five thirds
• How do i use the quadratic formula,$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$ and these numbers a=1, b= -3, c= -10 to make the answers x=5 x=-2
• How do i use the quadratic formula to solve the equation X²+4x+2=0, whilst leaving the answer in surd form,

I dont understand my homework questions, was wondering if any of you could help me, thank you

For the second question, just plug in your values for a, b, and c. --LarryMac | Talk 20:02, 10 September 2007 (UTC)

• The plus and minus thing confuses me, can somebody help? —Preceding unsigned comment added by 82.36.182.217 (talk) 20:11, 10 September 2007 (UTC)

You want two answers, right? You know that. The answer is either one thing or the other... so what do you think the plus or minus means? Gscshoyru 20:16, 10 September 2007 (UTC)

• Does it mean i have to both add it and subtract it? —Preceding unsigned comment added by 82.36.182.217 (talk) 20:28, 10 September 2007 (UTC)

Does that sound helpful to you? Not sure what you mean. If by both adding and subtracting you end up with two different numbers, then yes. If you mean adding and subtracting from the same number, than no -- adding and then subtracting something from the same number will get you the same thing back, of course. There's a reason it says or, and not and. Basically, in one answer you'll add, the other you'll subtract. Gscshoyru 20:38, 10 September 2007 (UTC)

The symbol "plus or minus" (±) means that you perform both operations (the plus and the minus) (separately and independently).

So, for example, consider this mathematical equation: x = 100 ± 25

In English words, you would say: "x equals one hundred plus or minus twenty-five"

And, that is just a combination of the following two English sentences:

• "x equals one hundred plus twenty-five" OR "x equals one hundred minus twenty-five"

So, back to mathematical symbols: x = 100 ± 25

That statement is simply a short-hand way to combine the following two separate, independent mathematical statements:

• x = 100 + 25 (in other words, x = 125 as ONE answer/result)

OR

• x = 100 - 25 (in other words, x = 75 as ANOTHER answer/result)

So, in other words: if x = 100 ± 25 ... then that means that x is equal to BOTH the value of 125 as well as the value of 75.

Thus, in correct mathematical wording, we would say: "x = 100 ± 25" means that x = 125 OR that x = 75.

Does that make sense to you? (Joseph A. Spadaro 20:41, 10 September 2007 (UTC))

• I think but when i finished my GCSE's i took a gap year, so i forgot lotsa stuff :-( —Preceding unsigned comment added by 82.36.182.217 (talk) 22:45, 10 September 2007 (UTC)

If it's easier to understand, you can eliminate the ± by considering two equations: $x={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}$ and $x={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}$. This is all the ± means anyway. Tesseran 02:15, 11 September 2007 (UTC)

Generally YES, BUT... If there are two or more $\pm$ or $\mp$ symbols in the same equation, they do not work independently, so you do not get 2^{no. of symbols} cases. Instead take two solutions, one for all upper signs and the other for all lower signs. For example $100\pm 20\pm 5$ means $100+20+5$ OR $100-20-5$ but NOT $100+20-5$ or $100-20+5$. Another example is a trigonometric identity:
$\tan(\theta \pm \phi )={\frac {\tan \theta \pm \tan \phi }{1\mp \tan \theta \tan \phi }}$
which means either
$\tan(\theta +\phi )={\frac {\tan \theta +\tan \phi }{1-\tan \theta \tan \phi }}$
or
$\tan(\theta -\phi )={\frac {\tan \theta -\tan \phi }{1+\tan \theta \tan \phi }}$
CiaPan 06:58, 11 September 2007 (UTC)

See also Plus-minus sign and Quadratic equation.  --Lambiam 14:12, 11 September 2007 (UTC)

Multiplication of Negative Integers

Let's say that I want to offer an "intuitive" example to illustrate integer multiplication. Example 1: For (+3) x (+5) = +15 ... I can intuit adding 3 dollars (a gain of 3 dollars, thus +3) into my bank account every day for 5 days. Thus, I will have gained 3 dollars 5 times and thus I will have gained 15 dollars (that is, +15). Example 2: For (-2) x (+4) = -8 ... I can intuit subtracting 2 dollars (a loss of 2 dollars, thus -2) from my bank account every day for 4 days. Thus, I will have lost 2 dollars 4 times and thus I will have lost 8 dollars (that is, -8). What is a good, clear, real-life, concrete, intuitive example to illustrate why a negative number multiplied by a negative number yields a positive product? How can I visualize or intuit why, say, (-7) x (-6) = +42 ...? Yes, mathematically, it makes sense and everything falls into place (with regard to negation and opposites and integer rules, etc.). But, can anyone offer an intuitive example (perhaps analogous to my two bank account examples) that clarify and concretely illustrate why a negative times a negative should yield a positive? Thanks. (Joseph A. Spadaro 22:33, 10 September 2007 (UTC))

Perhaps you could say that you know how much money you have in your bank account today, and you know that for the past six days you've been subtracting 7 dollars per day. Then from (–7)×(–6) you know you had 42 dollars more in your bank account then than you do now. Strad 23:03, 10 September 2007 (UTC)

I didn't follow. I got confused. In your example, what exactly does the -6 represent? ... what exactly does the -7 represent? ... and what exactly does the +42 represent? Thanks. (Joseph A. Spadaro 23:15, 10 September 2007 (UTC))

6 days ago. As in, if we plug in -6 for time. Gscshoyru 23:18, 10 September 2007 (UTC)

So, to translate the integer operation of (-6) x (-7) = +42 ... you are saying (in plain English): "Withdrawing seven dollars (-7) for the past/previous six days (-6) yields a balance that initially was 42 dollars greater (+42) than today's balance." Is that it? OK. It would still take an elementary student a bit of doing to wrap their mind around that. (I think.) But, it's OK. Does any one have anything else perhaps more "obvious" / easier to "see"? Thanks. (Joseph A. Spadaro 00:40, 11 September 2007 (UTC))

For each tree that you bring into the Biodome research station, the daily production of oxygen in the sealed facility is increased by 3 litres. Bring in 5 trees and the resultant increase in daily O₂ creation is (+5 trees) × (+3 litres/tree) = +15 litres. Each mule in the Biodome consumes 7 litres of O₂ per day. So if you bring in 4 mules, the resultant increase in daily production of oxygen is (+4 mules) × (-7 litres/mule) = -28 litres, or a net loss of 28 litres/day. But if you remove 6 mules, then the effect on the rate of O₂ creation is (-6 mules) × (-7 litres/mule) = +42 litres, or a net gain of 42 litres every day. PaulTanenbaum 02:37, 11 September 2007 (UTC)

And I might add that removing 3 mules has the exact same effect as introducing 7 trees. Saying that algebraically, (-3 mules) × (- 7 litres/mule) = (+7 trees) × (+3 litres/tree) = +21 litres. Either one of these changes to what organisms are in the Biodome will increase the oxygen level within at the rate of 21 litres/day. PaulTanenbaum 02:51, 11 September 2007 (UTC)

Paul Tanenbaum -- Thanks. That is an example that I think will work well to illustrate the concept that negative times negative yields positive. Can you offer any type of similar analogy to the adding/subtracting money from the bank account examples? Thanks. (Joseph A. Spadaro 06:17, 11 September 2007 (UTC))

OK... we're going to be reasoning about how the balance of our bank account on various dates compares to today's balance, under the assumption of some fixed daily change in balance. We'll be quantifying two things: the number of days (into the future) since today and the number of dollars deposited every day. In quantifying the date, if we're interested in a year from today, that's +365; if it's a week ago, then it's -7. In quantifying the daily change in balance, if it's a dozen-dollar deposit, that's +12; if it's a hundred-buck withdrawal, then it's -100. So, how does our balance on John's birthday compare to our balance today? Well, if John's birthday is 5 days from now and we withdraw $7/day, then the birthday balance is (-7 $/day) × (+5 days) = -35 dollars higher than today's. Similarly, if John's birthday was 8 days ago and we've been withdrawing $7/day, then the birthday balance is (-7 $/day) × (-8 days) = 56 dollars higher than today's.—PaulTanenbaum 01:46, 12 September 2007 (UTC)

Paul -- thanks. Just so I am clear ... isn't your example exactly the same as Strad's example (above), while just using different numbers? (Joseph A. Spadaro 05:53, 12 September 2007 (UTC))

Yes, the application is essentially the same, but that was essentially imposed by (the way I interpreted) your request for similarity. I'd hoped my way of mofe fully describing how the model works would help you get students past the "clarifying why" problem. If I've misunderstood, then what do you wish? I mean what would constitute "any type of similar analogy to adding/subtracting money" if the trees-mules-and-oxygen model doesn't fit?—PaulTanenbaum 10:15, 12 September 2007 (UTC)

It might help to speak a little more generally first. Let's begin with the operation itself. There are two pieces - the first bit, and the second bit, with a symbol between them. Let's take the second bit to indicate an amount, and the first bit to indicate repeated addition. So, 3x5 indicates three additions of five - either to zero, or to whatever came previously in the expression. In other words, something+(3x5) = something+5+5+5 = something+15. It follows immediately that something+(3x-5) = something+(-5)+(-5)+(-5) = something-5-5-5 = something-15. Now, what do we do with 0 in the first bit? By exension, figure that that corresponds to performing the action 0 times. something+(0x5) = something, which also equals something+0. What about a negative in the first bit? That would correspond to undoing the action, or performing its opposite, that many times. That can get a bit hairy for many actions, but not for addition, since we have subtraction. something+(-1x5) = something-5. something+(-3x5) = something-5-5-5 = something-15. Notice that I subtracted a positive, I didn't add a negative, though it amounts to the same thing. Finally, negative times negative is straightforward. something+(-3x-5) = something-(-5)-(-5)-(-5) = something+5+5+5 = something+15. It is possible from there to prove that switching the two numbers produces the same answer, something that I noticed you assumed from the beginning. Naughty. Now, you can take as an example absolutely anything at all that fits that description. One would be the tree-and-donkey example, where removing things is taken as opposite to adding them, and donkeys are opposite to trees. You could take it as moving along the number line, where one number indicates jumping (or "unjumping") a certain number of times, and the other indicates distance to the right (or left). You could take a bank example, where perhaps withdrawing is the opposite of depositing and fines are the opposite of interest. It only works when you have two distinct pairs of opposites. If all you have is two things that can be described by the positive numbers, like a positive amount and repetition, this doesn't come up. If all you have is one thing that's described by positive numbers and another that's described by negative numbers, like adding positive amounts of orange juice and removing positive amounts of orange juice, or like adding positive and negative amounts of juice, this still doesn't come up. It's only when both things actually naturally have opposites of some kind that it truly fits. Usually, one will be repetition vs. repeated undoing, and the other will be amount vs. negative amount. What the negative-times-negative-equals-positive idea indicates is that when you do this, you introduce redundancy. Undoing the opposite of something has the same effect as doing the thing itself. Not only that, it's so similar an idea that it's hard to accept that the two are different, which is part of why it's tough to get a good example. I like the number line, because it's easy to think of both walking forwards or backwards, and moving to the east or west, as distinct ideas that give the same result. Black Carrot 02:33, 14 September 2007 (UTC)