Ferdinando

Typically, 0 is considered to be neither positive or negative. When you want to include 0, you typically use the phrase "nonnegative" or "nonpositive" instead. However, it is important to keep in mind that many people do use the word "positive" to include 0, even though it isn't precisely correct.

According to some Calculus textbooks, 0^0 is an "indeterminate form." What mathematicians mean by "indeterminate form" is that in some cases we think about it as having one value, and in other cases we think about it as having another. When evaluating a limit of the form 0^0, you need to know that limits of that form are "indeterminate forms," and that you need to use a special technique such as L'Hopital's rule to evaluate them. For instance, when evaluating the limit Sin[x]^x (which is 1 as x goes to 0), we say it is equal to x^x (since Sin[x] and x go to 0 at the same rate, i.e. limit as x->0 of Sin[x]/x is 1). Then we can see from the graph of x^x that its limit is 1. Other than the times when we want it to be indeterminate, 0^0 = 1 seems to be the most useful choice for 0^0 . This convention allows us to extend definitions in different areas of mathematics that would otherwise require treating 0 as a special case. Notice that 0^0 is a discontinuity of the function f(x,y) = x^y, because no matter what number you assign to 0^0, you can't make x^y continuous at (0,0), since the limit along the line x=0 is 0, and the limit along the line y=0 is 1. This means that depending on the context where 0^0 occurs, you might wish to substitute it with 1, indeterminate or undefined/nonexistent. Some people feel that giving a value to a function with an essential discontinuity at a point, such as x^y at (0,0), is an inelegant patch and should not be done. Others point out correctly that in mathematics, usefulness and consistency are very important, and that under these parameters 0^0 = 1 is the natural choice.

Yes (2,5) is what i found too. I am trying to resolve this equation formally. Thanks for your input.

Sorry, I was thinking of something else. I got distracted. 0^0 is undefined (Of course !)

Prove that for all positive x, y and z (x+y)^z + (y+z)^x + (z+x)^y > 2

Hi D H, 0^0 = 1, if we substitute, then the eq. would be valid And, if x = 1 and y = 0, it will work.

Could you find all positive integer solutions to ... x^y - y^x = x + y

Hi Ajb, Oh, cool ! you're going to a lecture. I look forward to reading your next posting.

"The Elegant Universe" made them reputable in the lay-audience world. Have you read the papers by Alan Sokal ? and, its counterpart in the Bogdanov affair ? The public may have difficulties taking theories seriously. String theorists don't know whether or not nature/reality can be described by this theory. But, they believe it has a very good chance of being true. The Large Hadron Collider (LHC), which is located in France and Switzerland, is scheduled to go online in 2007. It is expected to provide evidence to support aspects of string theory, such as supersymmetry, that is every known elementary particle and force carrier particle has an as-yet-undiscovered partner particle, known as a superpartner. It is a big deal. Because such findings would support string theory. And, even if string theory is not correct, supersymmetry could be a feature of the universe. The LHC may well discover that the principle of supersymmetry is one that is needed to describe our universe. Particle physicists are highly skeptical of string theory. String theory which brings the radical notion that there are more dimensions than we have traditionally accepted with essentially no experimental support, the built-in conservatism kicked in. And, there is the misconception of what is supersymmetry. It is not a thing, it is rather a relationship between what things are made of and the cause that makes those things clump together. Any thoughts ?

Has anyone worked on the P vs NP problem ?