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I am not sure I understood why the 0^0=1 is indicated as an oddity of the calculators you describe.

This is true if you define it to be so and I don't see it getting in the way of any normal exponent laws. We do this for non-zero integers so that

1=x^n x^(-n)=x^(n+ -n)=x^0.

So perhaps it appears you have a problem along the lines of

1=0^0=0^(n + -n)=0^n 0^(-n)=0.

But for 0 you don't have 0^(-n) so you cannot actually write this and so you don't have a real problem defining 0^0=1 just as we also define 0!=1 and so forth. After all, x^0 is just as much a matter of notation as is 0^0, there really isn't a 0 exponent until you define it. This is different from x^2 or x^9 where there is a prescribe solution already determined by multiplication.

As far as real numbers go, the problem with 0^0 is that the
power function is discontinuous at the point (0,0) --- depending
on how you take the limit as x and y go to zero, x^y could be
made to converge to anything or to diverge, so there is no natural
choice of value for 0^0 so it is a matter of convention how we
define it.

For fans of "new math", there is a sense in which we can show that
0^0 = 1 in set theory, although it is based on a loophole in the
definitions. Recall that, if A and B are sets, then A^B is the set
of all functions from B to A, so 0^0 should be the set of all
functions from the empty set to the empty set. What is that?
Technically, a function f from a set B to a set A is a subset of
the product B x A such that, for every element x of B, there
exists exactly one element y of A such that (x,y) belongs to f.
Now the product of the empty set with itself is just the empty set
again because there are no elements with which to form pairs.
Moreover, the only subset of the empty set, itself, technically
counts as a function --- since it has no elements, it is trivially
true that for every element of the the empty set, there exists
another element of the empty set such that the pair of these
elements belongs to the empty set! Hence, there exists exactly one
function from the empty set to itself, so we have 0^0 = 1.

rspuzio writes:

> there is a sense in which we can show that
> 0^0 = 1 in set theory, although it is based
> on a loophole in the definitions.

I wouldn't call this a loophole. In cardinal arithmetic x^0 = 1 for all x, and this is essentially the statement that the empty set is an initial object in the category of sets. More like a fundamental property than a loophole.

Nothig is wrong with ... You must calculate expression containing division by zero correctly!
Expression (x^n)*(x^-n) = (x^n)/(x^n) is for x equals zero expression 0 / 0 which is for division by zero defined. You must use lim(limes) for x->0 of ((x^n) / (x^n)). Using l'hospital rule(derive numerator and denominator by x repeatly) you get lim{x->0}(n! / n!) = 1. So your expression is valid!
Alternatively you can convert your first expression to 1 * (x^n) = (x^n) which for 0 equals to 0 = 0, which is valid too.
Stefan

x^0 (including zero) equals to 1. So 1 = (x^n)*(x*-n) for x equals to 0: 1 = (0^0)*(0^-0) = 1 * 1 = 1.

I know of a lot of non-continuous functions on the reals, .. that doesn't really seem to be problem...After all this discontinuity is at the end point which is a fairly easy to work around discontinuity.

I guess I am glad to see most people so far have agreed, there is nothing wrong with 0^0=1 as a definition, and it is infact just natural....so I guess my only unanswered question is why the surprise that on a calculator 0^0=1?.....that is probably for PrimeFan to answer.

Can be x^n an element of some group? If so, it is always certain that (x^n)(x^n)^{-1}=1, whenever 1 be the identity element?, or necessarily do we need de concept of limit?

I'm not sure what you mean by writing 0/0, or "division by zero correctly." That is something I try never to say when I teach calculus. Students hear only that you can divide by 0, even though you show them that doing so makes all numbers =0. So I really keep limits separate from 1/infty or 1/0, or other missleading notations. After all, even if one wishes to define 1/0 consistently through limits, you run into problems of sign, is it +infty or -infty? That is, 1/x or 1/x^2 that has x->0....in the end I find it too difficult to give made up rules to handle this so it seems best simply to describe it the way the theorems do: if the top and bottom go to 0 individually then you may apply the following tricks .....having never written 0/0.

[By the way, I'm not sure 1 * (x^n)= x^n shows 0^0=1 or that lim_{x->0} n!/n! = 1 matters, there is no x in that limit, and lim_{n->0} n!/n! =1 would not force 0!=1 --- beyond that, n! is discrete so I don't know what limit n->0 would mean there.]

Depends what group we want here. If we write x^n we usually mean then that the operation is multiplication, and 0 is not in the multiplicative group of the reals so it would not be covered in the discussion of the multiplicative group. But of course you could use many other aglebra (no limit) arguments to verify 0^0=1 is perfectly natural, and I think many people here have.

Standard mathematical convention is that 0^0 = 1. The reasons for this have less to do with all the arguments provided so far than with convenience. The fact is, defining 0^0 = 1 means far fewer "exceptions" need to be made in formulas and theorems than other choices (either defining it to be 0, or leaving it undefined). The most well-known example is found in power series. For example,

e^x = \sum[n=0...] x^n/n!

would require an exception for x = 0 if we did not define 0^0 = 1. Mathematics is replete with situations like this where 0^0 = 1 is consistent with the notation, while 0^0 = 0 or any other convention would cause problems. Situations that go the other way are rare.

There is nothing wrong with the arguments given for why 0^0 = 1 (though that "new math" is about 150 years old). But they fail to be convincing when one can also come up with similar arguments that 0^0 = 0. The real reason we choose 0^0 = 1 is simply that it makes our lives easier.

"0^0 = 1" may be a convention, practical for expressing some formulae comfortably (in the case of the series of e^2 it has something to do with the fact that lim_{x\to 0+}x^x = 1), but there is no reason to set such definition. 0^0 is not defined, equally much as e.g. 0/0 is not defined. Or could we define 0/0 = 1 because 1 times 0 is equal to 0?
If 0^0 were a genuine and proper POWER, then we probably could say that it is exp(0 ln 0)!

"0^0 = 1 is standard mathematical convention" means that this IS the accepted definition of 0^0. All your arguments show is that x^y is not continuous at (0,0) (and ln x is not continuous at 0). "Not continuous" is not the same thing as "not defined". We could define values for any singular point on any function, if we wanted. It doesn't make the function continuous (unless the singularity was removable), but there are times when it is useful to do so.

The reason we don't define a value for 0/0 is not that it would be impossible. There is nothing impossible about a definition. We don't define a value for 0/0 because there is nothing to be gained by such a definition, and there is much to be lost. Because when 0/0 forms occur (any time we take a derivative, for instance), the values we get are distributed throughout C. Assigning a particular value to 0/0 would gain us nothing.

However, with 0^0, in 99.9% of situations where it occurs in mathematical work, we want it to have the value 1. There is much to be gained by this definition, but very little to be lost. On relatively rare occasions we might run into a curve approaching (0,0) in just the right fashion to take a limit of 0 (yes, x=0 provides just such a path), or even some other number in [0,1]. But when we do, we can just shrug, say "well, we already knew x^y is not continuous there", and move on.

> Because when 0/0 forms occur (any time we take a derivative, for
> instance), the values we get are distributed throughout C. Assigning a
> particular value to 0/0 would gain us nothing.

If 0^0 = 1 then assuming that the usual rules of algebra hold we
would get

0^0 = 1 => 0 * log(0) = 0 => 0 / (1/log(0)) = 0 => 0/0 = 0

so 0/0 would also get a value. Of course, you could define 0^0 to be
1 anyway even though the usual properties do not hold, but then you
would have to insert a special note every time you want to state or
use those rules. Hardly worth the effort.

In short, 0^0 is undefined for the same reason that 0/0 is.

> However, with 0^0, in 99.9% of situations where it occurs in
> mathematical work, we want it to have the value 1.

Just curious, how did you calculate that percentage?

Your calculation assumes that log(0) is somehow finite. It is not. The "rules of algebra" you have applied in fact do not hold for infinite values.

Whether you define a value for 0^0 or not, you have to make an exception for 0 when stating the rules for exponents. This happens simply because x^y is discontinuous there. The only difference in this that defining 0^0 = 1 makes is that you only have to make ONE exception instead of two:
(0^0 undefined):
0^x = 0 unless x = 0, where it is undefined
x^0 = 1 unless x = 0, where it is undefined
(0^0 = 1):
0^x = 0 unless x = 0, 0^0 = 1.
x^0 = 1 for all x.

You would only need to make special comment when using the properties in cases where a 0^0 form comes up and you don't want a value of 1. But again, you still need to comment in such cases if you leave 0^0 undefined.

As for the percentage, it is an estimation based on my own experience and knowledge of the field. In particular, it is based on the quite common practice of expressing functions as power series, both for specific functions and for generic expressions. Every power series formula with non-zero constant term expressed using the sigma sum notation - expressly assumes a value of 0^0 for 1. If we did not make this definition, every time we wrote down a power series, we have have to say that it does not hold for x=0, or else always express it as (constant term) + \sum[n=1]. That is what is not worth the effort!

I see such power series, and numerous other notational expressions dependent on 0^0 = 1 all the time. However, just about the only place where 0^0 comes up where 1 is not desired are limit expressions, almost all of which come up in homework assignments, not real problems.

But last, I suggest you actually start pulling out a few calculus and intro analysis books and see what they say. I promise you that every last single one of them uses 0^0 = 1 either explicitly, or implicitly.

I started thinking about this because of Graeme McRae, a respected contributor to Sloane's OEIS. After reading my page on the arguments for and against the primality of 1, he suggested I tackle the question of the value of 0^0 in the same mock trial format I used for pro/con 1 as prime. For some reason this topic doesn't interest me as much as other topics having to do with exponents. If anyone cares to do for 0^0 = 1 what I did for pro/con 1 as prime, I'll let Graeme know.

Wow, a debate about 1 not a prime.

Judge: "Please read the definition of a prime into the record."
"A positive integer with exactly two positive integer factors."

Side 1: "But 1 is divisable by 1 and itself, so it is prime."
Side 2: "These two factors are the same, they are both 1, so no."

Judge: "Yes, 1 is not prime, thank you, you are dismissed."

...I say that in jest, but yes, it is by definition, but as Icarus suggest, we define things to match the most useful way of using the definition. 1 is not prime and 0^0=1.

> Your calculation assumes that log(0) is somehow finite.

Not really.

My calculations (as the quoted text above them tried to indicate)
was in the context of limit calculations of indeterminate forms. I
assumed that your comment "Because when 0/0 forms occur (any time we
take a derivative, for instance), the values we get are distributed
throughout C" was alluding to the L'Hospital rule. In that context
the 0's stand for limits of functions and it is common practice in
such case to write things as if one applies operations to the limits
instead of the function. My comment could be rephrased as follows

Let f, g be positive functions with lim f(x) = 0 = lim g(x) as x->0.
Then the limit behaviour of f(x)^g(x) as x->0 is as ambiguous as
that of f(x)/g(x).

> You would only need to make special comment when using the
> properties in cases where a 0^0 form comes up and you don't want a
> value of 1. But again, you still need to comment in such cases if you leave
> 0^0 undefined.

My comment was in the context of limit calculations (see above).
In this context you don't want 0^0 to be 1 by default. You want
to calculate the value.

> As for the percentage, it is an estimation based on my own
> experience and knowledge of the field.

I see.

> Every power series formula with non-zero constant term expressed
> using the sigma sum notation - expressly assumes a value of 0^0 for
> 1.

This, together with the set theoretic example somebody else
mentioned earlier in this thread are the only examples that I know
where it make sense to define 0^0 to be 1.

> If we did not make this definition, every time we wrote down a power
> series, we have have to say that it does not hold for x=0, or else
> always express it as (constant term) + \sum[n=1].

Actually some books do exactly that. Others just state that in the
context of sigma notation for power series 0^0 = 1 somewhere in the
beginning of their treatment. Others yet, just ignore the issue
altogether.

> My comment was in the context of limit calculations (see above).
> In this context you don't want 0^0 to be 1 by default. You want
> to calculate the value.

Just because we define 0^0 = 1 does not in any way mean that we are demanding that 0^0 limits must be 1. This what the phrase "not continuous" means! The limits are not the function value.

> I see.

From your phrasing, I doubt it.

> This, together with the set theoretic example somebody else
> mentioned earlier in this thread are the only examples that I know
> where it make sense to define 0^0 to be 1.

There are more, though often related to these. In particular, the set-theoretic argument is only a more-abstract version of something that comes up in combinatorics. There is another field where formulas must be expressed with exceptions if 0^0 is not defined to be 1.

But even if these were the only ones, I argue that they are enough. Power series are ubiquitous. The only reason I have heard for not defining a value for 0^0 is that x^y is not continuous there. This hardly seems adequate cause, when we deal regularly with other discontinuous functions with defined values at the discontinuity.

> Actually some books do exactly that. Others just state that in the
> context of sigma notation for power series 0^0 = 1 somewhere in the
> beginning of their treatment. Others yet, just ignore the issue
> altogether.

Those who do "exactly that" do not exploit the notation to it's full advantage, and as such are very restricted in the level of material they represent. Those who "ignore the issue altogether" do no such thing. They use the definition 0^0 = 1. They don't bother to mention it likely because they never saw it as an issue. I've never come across a book where the author said 0^0 = 1 is for power series only. If I did, I would be aghast at an author who wants expressions to have different values in different contexts, even though all the elements of the expressions mean the same thing in each of the contexts.

> Just because we define 0^0 = 1 does not in any way mean that we are
> demanding that 0^0 limits must be 1. This what the phrase "not
> continuous" means! The limits are not the function value.

I know the meaning of the phrase "non-continuous". The point is
that the usual algebraic operations when defined are continuous, so
giving them arbitrary values is introducing more problems than it
solves by creating artificial exceptions to well known rules. If
you don't see that as a problem, then go ahead! define 0^0 =1 and
0/0 = 0 or 24 or 143.7 or whatever else you fancy.

I don't find it particularly satisfying to keep repeating the same
points to people who don't seem to listen. AFAIAC this conversation
is over. I have far more creative ways of procrastinating than
arguing over the net with semi-ignorant arrogant fools.

If we assume that 0^0 = 1 is true, we can simply get to the statement, that 0/0 = 1. This is one way to get that:
1 = 0^0 = (n - n)^{1 - 1} = (n - n) / (n - n) = 0/0
Where n is any number you like.

Nice, Mephisto!

> If we assume that 0^0 = 1 is true, we can simply get to the
> statement, that 0/0 = 1. This is one way to get that:
> 1 = 0^0 = (n - n)^{1 - 1} = (n - n) / (n - n) = 0/0
> Where n is any number you like.

Hmm! I think that this only shows that negative powers of 0 are
undefined. You can carry a similar argument starting from any power
of 0. For example

0 = 0^7 = (n - n)^{8 - 1} = (n - n)^8/(n - n) = 0/0

That sounds like a Texas judge. A more enlightened judge would give both sides a chance to read their definitions and then the two sides would both get a chance to argue their definitions.

It's not an open and shut case, and from the looks of all the discussion here lately, neither is the value of 0^0.

> 1 = 0^0 = (n - n)^{1 - 1} = (n - n) / (n - n) = 0/0

Perhaps we, after all, can see from this, that if one gives to 0^0 some properties of power (i.e. if one thinks that 0^0 is a POWER), it leads to absurdities.

Are we asserting that the equality (n-n)^{1-1}=(n-n)/(n-n) makes sense?

Well the judge part is a joke of course, but I think one would be hard pressed to find a reference in print where 1 is prime. All the number theory books I find have that, in fact, the algebra books in my office which generalize primes to ideals also don't allow the whole ring to be prime, so that again 1 is not prime. (Includes Eynden, Hungerford, Rotman, the MIT Encyclopedic Dictionary of Mathematics, Vinberg] That is why I think there is no real debate on 1 as a prime. [Also 0^0, if books even bother with it, probably uniformly treat this as 0^0=1, but that is conjecture on my part.]

My office is not that big, but do little library digging and I wager the same result will be had.

Yes, indeed.
So we can assume that 0^0 is an undefined number, as you can get various results.

time not space

The most ``seasoned'' PlanetMath users were taught as children that 1 is prime.

Me, I was taught the 2-distinct divisors definition, and I was also taught that a woman who knows too much about the sciences scares away some potential husbands. Both of these are true, the latter I questioned until I realized that some men are better scared off and thus narrowing the dating pool.

The former I didn't question until I saw PrimeFan's page on the subject. All this time I had simply accepted the definition on the strength of what amounts to little more than a "Because I said so." It doesn't matter if it's a thousand people each saying "Because I said so." If PrimeFan said:

"2^8 - 1 is prime"

I would immediately counter "It's divisible by 3 AND 5." And after checking with my calculator, I'd add 17 to the list. What about when Mersenne said

"2^61 - 1 is prime"

? It wasn't until the 20th Century that anyone could contradict him, when the factors were finally found. But when someone says

"1 is prime"

was there some previously unknown prime factor that was discovered at the turn of the 20th Century? No.

I think 1 is not a prime. The reason I think that today isn't because someone told me that what I should think. The reason I think that now is because I've carefully considered for myself every argument presented in PrimeFan's page (I even contributed one of my own) and I came to the conclusion that the arguments against it being prime are more convincing than those for. If other arguments are presented, I will consider those also.

I do wish PrimeFan would've worded the verdict against more strongly, but I recognize that this is not an open and shut case, and that it goes beyond semantics (in a way that say, the arguments for 1 being a repunit do not).

I'm not sure I have a problem studying the properties of 1, but who cares if things are given names "because I said so" That doesn't take away the properties of the numbers that don't have the name. Use of the name prime allows for simple hypotheses in theorems. So convention is what it is all about. So it has to be a "because I said so." Prime is not a concept all together intrinsic. In Lie algebra, abalian Lie algebras of dimension 1 are not considered simple, even though in group theory they are. Why the difference? Because people say so. Being simple has some basic properties but yes, we do get to label things however we want.

I used to think 1 was prime, now I know it is a unit. I don't call it prime because prime has a different definition. But that doesn't take away any use or value of 1. Once a definition is so strongly set in print I don't think one gets to freely debate it, that is all. But of course, one should continue to use the "left out" numbers.

Well, I suppose that compared to other "because I said so" situations (like a Saudi man saying "stone that woman") the non-primality of 1 doesn't seem that important. But here in America (where I am, and where you probably are too -- the digital divide, don't get me started on that) we can freely debate anything, regardless of how unimportant it might seem (though whether or not we want to debate a particular issue is a different matter).

But this issue of 1 as not prime could be important where the education of children is concerned. I can only imagine that those of you here with children, when helping them with their math homework, have probably come across things that make you wonder if the kid's teacher is qualified, or at least marvel at how differently math is taught these days.

Good point. Teaching this stuff as a series of rules makes it bad math, so debating it as a means to confer understanding has its place. But ultimately, I want my kids to know how to add more than to know some concept of addition which they cannot actually execute, so in the end I would insist they learn "the rules."

Anyway, I have my answer, some people don't like 0^0=1, but I'm not convinced by those who argued (mostly through division by 0 of all things) that it was impossible to do. I'll continue to use this assumption as it makes hypotheses easier and has no realy downsides.

Takeuti & Zaring, _Introduction to Axiomatic Set Theory_, p.57, Theorem 8.31 (1): 0^0 = 1. This is in the context of ordinal arithmetic, where the arithmetical operations are defined recursively. Since the natural numbers are the finite ordinals, it seems safe to say of the natural numbers 0 and 1 that 0^0 = 1, as ordinals. However, the natural numbers are also cardinal numbers. In general, ordinal arithmetic and cardinal arithmetic are quite different; however, for the natural numbers, the two arithmetics coincide. It seems safe, therefore, to say of the natural numbers 0 and 1, that 0^0 = 1.

It is common when developing the theory of natural numbers axiomatically using 0 and S (successor) as primitives, to define the arithmetical operations recursively for all naturals n and m:

n + 0 = n,
n + S(m) = S(n + m ).

n * 0 = 0,
n * S(m) = n * m + n.

n^0 = 1,
n^S(m) = (n^m) * n.

I have my own vision of this problem. :)

Everyone will agree with me, that 0^(-1) makes no sense, right? But what about trivial ring (ie. where 1=0)? 0^(-1) exists there. So now, will everyone agree that some things exist (or do not) only in "special" cases?

So if we need 0^0 to evaluate some limits it makes no sense (Lim x->0 (0^x) = 0, but Lim x->0 (x^0) = 1 ). If we need it in set theory, then 0^0 means cardinality of set of all functions from empty set to empty set - this is 1. Then we can think about group G acting on sets, where if H is a subgroup of G and X is a G-set, then X^H is set of points from X fixed by elements of H. Sometimes it is useful to assume that empty set is also a subgroup, but then 0^0 is again 0 (there are no fixed points in empty set, because it's empty :-) ).

I think that expression "0^0=1 makes no sense" is referred to limits, where it makes no sense. By the way... if we assume that 0^0=0 then can someone prove that it is wrong? Algebraically of course. I doubt it. Then again you can assume that a^b=0 for all real a,b. Will this be wrong? No, it won't. But in that case you loose something... good properties. Everything you can define as you wish, but the real problem is: will this definition be useful to anything? In this case I think that's why 0^0 is not defined - mostly it occurs in evaluating limits. Other special cases are just useless. :-)

joking

About the "other special cases being useless", I sure hope you're joking. :)

In analysis, we're already used to many functions being discontinuous. Defining 0^0 = 1 does little more than add a discontinuity in the domain of the function f(x,y) = x^y. That's okay, we know how to deal with those, and the limits involved, we still had to be careful about them anyway. If it's any consolation, there's a sense in which the limit as (x,y) -> (0,0) is almost always 1 -- if you approach along any curve not tangential to the y-axis, you'll get a limit of 1. "0^0" is still an "indeterminate form", it just happens to be one that has a numeric value when interpreted as an expression. Defining it doesn't really hurt anything.

The desire to satisfy those algebraic and combinatorial properties however is pretty important. One ends up writing things in awkward and unnatural ways without it. For instance, consider exp(x) = sum over k >= 0 of x^k/k!. Oops, you can't write that, you have to write exp(x) = 1 + sum over k >= 1 of x^k/k!, because otherwise exp(0) wouldn't be defined. This will happen all over the place, with almost any function that has any regularity at all in its power series. You already know about the other combinatorial reasons, for sure, but to reiterate, n^m is the number of functions from a set of size m to a set of size n. Failing to define 0^0 breaks that slightly.

You raise a good point that people can define things however they like, but for sanity's sake, it's nice to have a few standards which can be relied upon. Telling people that 0^0 is undefined will just add confusion when people who define it to be 1 use it that way, whereas telling people that it's defined to be 1 will not do any harm in the other direction, because you're also going to notify them that x^y is discontinuous at the origin.

- Cale

> Telling people that 0^0 is undefined will just add confusion

> because you're also going to notify them that x^y is discontinuous at the origin

Well... in my opinion this brings confusion. So you will say "0^0=1 almost always, except (...) where it is 0 and except (...) when it is infinity" (and who knows what other values it can have).

You say "x^y - it's just one function". Have you ever heard about Riemann hypothesis? To find only one non-trivial root (of one special function) with real part not equal to 1/2 would be disastrous for theory of numbers. My point is that sometimes those "small, little things" can be very important. That's why we shouldn't ignore them. On the other hand some "big things" (like cardinality of o^o where o is the empty set) aren't so important.

Correct me if I'm wrong. :-)
joking

Yet another argument that operates by pretending that a discontinuous function ought to be continuous if it is defined at all.

A function does NOT NEED to be continuous to be defined at a point. A great many functions that are used all the time (step functions, for example) are defined everywhere but discontinuous at many points (in the case of the Dirichlet function and similar ones, at every point).

0^0 = 1 is the standard convention because it is far more convenient notationally than leaving it undefined, not because because formulas that work for other values somehow extend continuously to (0,0). The function is discontinuous there, and formulas break down at the discontinuity. This is true whether you define a value for 0^0 or not, so the fact that this holds is no reason to not define a value for 0^0.

Conversely, 0/0 is not defined in standard convention simply because unlike 0^0, there is little advantage to be gained by giving it any particular value.

As a side note (this really has nothing to do with why 0^0 = 1 is preferred, but I think it is interesting): if you approach (0,0) linearly from any direction in the right half-plane except along x=0, lim x^y = 1 (the right half-plane restriction is because x^y is not well-defined for x < 0).

The exception is not artificial. It exists whether you define a value for 0^0 or not. So defining one does not introduce any exceptions. It just removes large numbers of exceptions you have to make when 0^0 = 1 is exactly the thing you want (and as I have indicated, and no one has been able to deny, this is by the far the more common situation - just from polynomials and power series expressed with sigma notation alone).

I will certainly grant you the bit about "people who don't seem to listen", though! And I have my own opinion about people whose argument devolves into "insult them and leave".

What is the sign of (-x)^(a/b) when (a/b) is not an integer? It tried (-2)^1.5 in the Windows Calculator and it said "Invalid input for function."

(-2)^1.5 is not real:

$(-2)^1.5=(-2)^{3/2}=\sqrt{(-2)^3}=\sqrt{-8}=2i\sqrt{2}$

If $x$ is negative, $a$ is an integer, and $b$ is a positive integer with $\gcd(a,b)=1$, then $x^(a/b)$ is only real when $b$ is odd.

So to answer your question, it should not have a sign since, to the best of my knowledge, imaginary numbers do not have signs.

> (-2)^{3/2}=\sqrt{(-2)^3} = ...

This equation is wrong; see http://planetmath.org/encyclopedia/FractionPower.html, the paragraph 4.

> If $x$ is negative, $a$ is an integer, and $b$ is a positive integer with $\gcd(a,b)=1$, then $x^(a/b)$ is only real when $b$ is odd.

This is not true; e.g. (-1)^{1/3} is not well-defined (1/3 = 2/6).

Ah, the thing is so difficult to learn =o(

Cf. also "generalpower".

Jussi

Jussi,

Apparently, we are playing by different rules. In my graphing calculator, when I type in $(-1)^{(1/3)}$, it gives the answer -1. I highly doubt that Texas Instruments would have programmed the calculator to do that if it were not true according to our set of rules.

In any case, I will look up more sources on this when I get a chance.

Warren

Sadly, only one book that I own discusses negative numbers to fractional exponents:

Precalculus Mathematics: A Graphing Approach
by Franklin Demana and Bert K. Waits
Reading, Massachusetts: Addison-Wesley, 1989.

This book agrees with me that $(-1)^{(1/3)}=-1$, as they indicate that the graph of $y=\sqrt[3]{x}$ and the graph of $y=x^{(1/3)}$ are identical (pp. 370 and 517). There is no discussion of expressions such as $(-2)^{(3/2)}$.

Warren

Dear Warren, I guess that your calculator don't have a cube root button. The TI thus is forced to identify cube roots and the powers with exponent 1/3. What the calculator says of 2/6 and 1/2 as exponents?
Jussi

> This book agrees with me that $(-1)^{(1/3)}=-1$, as they indicate that the graph of $y=\sqrt[3]{x}$ and the graph of $y=x^{(1/3)}$ are identical (pp. 370 and 517). There is no discussion of expressions such as $(-2)^{(3/2)}$.

I guess that the authors have not thought the matter all the way to the bottom. I see that such powers simply are not well-defined as real numbers. Another thing is that e.g. (-1)^{1/3} as a general complex power has three distinct values, one of them is real (= -1).

Jussi

No, this calculator does have a cube root button as well. This calculator yields the following:

$\sqrt[3]{-1}=-1$
$(-1)^{(1/3)}=-1$
$(-1)^{(2/6)}=-1$

The last one is a little surprising. I suspect that its order of operations is to first reduce the fraction, then compute the fractional power.

> In my graphing calculator, when I type in $(-1)^{(1/3)}$, it gives the answer -1. I highly doubt that Texas Instruments would have programmed the calculator to do that if it were not true according to our set of rules.

My CVS-brand scientific calculator just gives an error message. I don't know who they license to make their electronics, but it seems safe to say it's not Texas Instruments.

Until just now did I think of trying this out on Mathematica. I input $(-1)^{(1/3)}$ and it spat back the same thing formatted slightly nicer. Now, $(-2)^{(3/2)}$ gives a more interesting answer: $-2i\sqrt{2}$. I find all this a little mind-bending, but then so is almost everything else having to do with negative numbers.

Interesting! The calculator thus seems to take the given exponent as a _number_ and not to fasten its attention on the _form_ of the exponent. Very smart! It lacks only that it should also give the imaginary values of e.g. (-1)^{1/3}.
Jussi

Dear PrimeFan, I think you have a better calculator than the one of Wkbj79: as a real power (-1)^{1/3} is not well-defined, and the machine seems to grasp this.

As for (-2)^{3/2}, which has nothing to do with real numbers, your calculator gives the one of the imaginary values in a smart form. As a failing, it however doesn't give the other value.

What do we learn of the behaviour of the machines? That we cannot rely on them! It is better that we ourselves think and perform such quite simple calculations by using mathematics!

Jussi

> Now, $(−2)^{(3∕2)}$ gives a more interesting answer: $−2�i\sqrt{2}$.

My calculator is weird: $(-2)^{(3/2)}$ yields an error message, but $(2i^2)^{(3/2)}$ yields $−2�i\sqrt{2}$. It does not seem to like to give a nonreal answer when the argument does not have an $i$ in it.

As for $(−2)^{(3∕2)}=−2�i\sqrt{2}$, this means that the calculator is taking the square root first, then cubing. In general, this means that $x^{(m/n)}=(x^{(1/n)})^m$, not $x^{(m/n)}=(x^m)^{(1/n)}$.

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