What does 0^0 equal?

Quote: DJTeddyBear

Why can't you calculate the square root of a negative number?
Why will most calculators (including the app on my iPhone) allow cube roots of negatives?
What are the rules governing that?

Is zero a real number?

You can calculate the square root of a negative number, but there's a trick to it.
Consider the number as being positive, take the square root and append an "i" to it. Thus SQRT(-4) = 2i
Any engineer or mathematician will know what you mean. "i" stands for SQRT(-1). A number with an "i" attached is called an "imaginary" number. Numbers without an "i" are called real numbers. The sum of a real number and an imaginary number is called a complex number, example: 3 + 4i.

Advanced calculators such as the TI-83+ and TI-84+ can calculate with complex numbers. You will find a page on that here.

You can add, subtract, multiply, and divide complex numbers just like any algebraic binomial. When you get i² you replace it with -1 and when you get i³ you replace it with -i. i⁴ = +1.

Cube roots of negatives are no problem. Since the cube of a negative number is negative, so is a cube root.

In the sense of "real" or "complex," zero is a real number.

Here's a stupid math problem:

Using every numeral once and only "=" and one other math symbol, write a valid equation.

I'll post the solution after this spoiler pause



p

a

u

s

e

You've been warned.




23456789^0=1


As I said, a stupid amth problem. That's about my highest level of mathematical abstraction.

Now, what's the largest number that can be expressed using the numeral 1 four times and any number of any kind of symbol you want? example 11+11, 11^11, 11-11, 111-1, 111+1, 1+1+1+1, etc.

That one's even easier.

...there isn't one?

1111!
(1111!)!
((1111!)!)!
(((1111!)!)!)!
((((1111!)!)!)!)!
...

You can do that as many times as you like.

some kind of emoticon like cat eyes (here's cat and mouse).

Quote: puzzlenut

The Wizard said I could come here to see who's good at math. This generally would be regarded as a calculus limit problem.
0⁰ is not too meaningful in itself, but becomes tractible in the form lim x^x as x -> 0 which we'll take to mean approaching 0 as decreasing positive values of x

Your statement of the problem is not general enough. The problem is:

Find the lim f^{^g} as x -> 0 when it is true that f(x) -> 0 and g(x) -> 0 as x -> 0

f^{^g} = exp (g ln(f) ) = exp (ln(f)/ (1/g) )

lim f^{^g} as x -> 0 = lim exp( - g^2*f'/ (f*g') ) as x -> 0

In the simple case where f=g=x , the limit goes to 1.


The actual word on the poll should be "indeterminate" and not "undefined". The latter word means you can't find the answer, while the first word means it depends on how f and g are defined.

I still think you guys are missing the point, or just overlooking the point. Puzzlenut, you say from the start 0^0 is meaningless in itself. It's meaningless becuase 0^0 equals 0. Nothing is nothing. I'm guessing that's what you mean by 0^0 is meaningless. It's only going to be useful if you use a "cheat" or "shortcut" and make 0^0 equal 1. Then you can use 0^0 in your theorems and calculations. But, without the "cheat" 0^0 is still 0 or "meaningless". So, in other words, 0^0 on it's face value would be 0. But, it is also used as a mathematical "cheat" or "code" or "shortcut" where 0^0 is 1. So, if you take 0^0 at face value it would be 0. If you use it as a "cheat" or "shortcut" 0^0 may be referred to as equaling 1.

Getting back to the original question. The question is what does 0^0 equal. I would say with no other information given, we are being asked to take 0^0 at face value. Therefore, 0 would be the correct answer. If the question was what does 0^0 equal in this particular theorem or equation, that may have been different. But, at face value, 0^0 is 0 or "meaningless".

Quote: JyBrd0403

If the question was what does 0^0 equal in this particular theorem or equation, that may have been different. But, at face value, 0^0 is 0 or "meaningless".

There's plenty of meaning in 0.

Quote: JyBrd0403

I still think you guys are missing the point, or just overlooking the point. Puzzlenut, you say from the start 0^0 is meaningless in itself. It's meaningless becuase 0^0 equals 0. Nothing is nothing. (...)

Funny how people who miss the point usually insist on blaming the others.

Zero is not "nothing". You mix up a mathematical concept with an everyday-life concept. YOU miss the whole point. Sorry.

Quote: kubikulann

Zero is not "nothing".

If you don't know nothing, then you don't know anything.

Quote: kubikulann

Funny how people who miss the point usually insist on blaming the others.

Zero is not "nothing". You mix up a mathematical concept with an everyday-life concept. YOU miss the whole point. Sorry.

Funny ha ha
example
10 runners line up in a starting gate (like in horse racing) for a one-half mile run
(I like getting half done)

Start Gun goes off but only half the gates drop to allow 5 runners to start running.
"WAIT" they fire the gun again to stop those 5 runners.
we must start all over!

This is not fair!
crap!
shouts all the runners.

In the video room the director says
"rewind that tape of the race back to 3 seconds after the start to see who was in the lead at that point."
after that was done he says
"rewind it to 0 seconds"

but the tech says " no can do! 0 = nothing, so there was NO beginning of the race.
It did not exist because there was nothing at time 0"

next!

Well put 7Craps ...


Another example, I exist yet I have been insulted by being called a 0.

math guys and gals have specific definitions of events when using math

Is 0 a real number?
Yes
No
two very common replies to that question.

(OK. 3*3/3*3 = 9/9 = 1)
using the law of exponents
3^2 / 3^2 =
3^(2-2) =
3^0 = 1

if 3=x
x^0 = 1

how about when x = any real #?

what was the question again?

Being a LA Dodger fan (not a Cubs fan)
the waiting is the hardest part

Quote: kubikulann

Funny how people who miss the point usually insist on blaming the others.

Zero is not "nothing". You mix up a mathematical concept with an everyday-life concept. YOU miss the whole point. Sorry.

Well, Ann, I'm sure you believe that 0^0 is not meaningless either. You can make a post and let puzzlenut know he's mixing up everyday-life concepts and mathematical concepts. I'm sure he'll appreciate your wisdom on the subject.

I assumed Puzzlenut meant 0^0 was nothing or 0 when he stated it was meaningless. If he would like to clarify that or not is up to him.

Speaking of the Cubs, at the beginning of every game the score is nothing nothing, and there was nothing on the time clock at the end of the game when the Bears beat the steelers this weekend.

The correct response is all of the above.

Quote: JyBrd0403

I assumed Puzzlenut meant 0^0 was nothing or 0 when he stated it was meaningless. If he would like to clarify that or not is up to him.

I would be glad to clarify anything! I don't think it is profitable to analyze 0⁰ semantically because according to the law of subtraction of exponents anything to the zero power means that thing divided by itself, which is always 1. In the case of zero, however, it would mean 0/0. This asks the question "What number multiplied by the denominator equals the numerator" and in this case anything at all multiplied by zero is zero. I prefer 1. I have given a calculus derivation of this and the Wizard has given a numerical one.

Please, guys, do not provoke kubikulann. He is a teacher of mathematics and also is from Belgium, where people are more polite than they are here. He is too valuable a resource to risk losing.

Quote: puzzlenut

Please, guys, do not provoke kubikulann. He is a teacher of mathematics and also is from Belgium, where people are more polite than they are here. He is too valuable a resource to risk losing.

Whoaw! First time in my life I get such a compliment. I cannot but suppose this is ironic. And well-deserved if so.

Don't be fooled, people in Belgium are not more polite. Maybe we put pressure on other facets of the debate. Here, I am considered rude, actually, because when I attack issues and arguments my Belgian fellow countrymen immediately think you attack their person.

Quote: kubikulann

I cannot but suppose this is ironic. And well-deserved if so.

No, I am perfectly sincere. I have seen some vicious flaming in the Free Speech Zone, where posts cannot be flagged for deletion. That zone is now closed to new threads, however.

The people in this forum vary widely in their mathematical education. Many are merely gamblers and at least one (Teliot) has a Ph.D. in mathematics. If someone addresses you rudely I recommend either to ignore him or treat him as though he were a student. You see that I am trying to teach DJTeddybear complex arithmetic. I think he is worth the trouble because he is an ordained minister who officiates at weddings. Of course you have to plant your seeds in what you perceive to be fertile soil.

Quote: puzzlenut

I would be glad to clarify anything! I don't think it is profitable to analyze 0⁰ semantically because according to the law of subtraction of exponents anything to the zero power means that thing divided by itself, which is always 1. In the case of zero, however, it would mean 0/0. This asks the question "What number multiplied by the denominator equals the numerator" and in this case anything at all multiplied by zero is zero. I prefer 1. I have given a calculus derivation of this and the Wizard has given a numerical one.

Please, guys, do not provoke kubikulann. He is a teacher of mathematics and also is from Belgium, where people are more polite than they are here. He is too valuable a resource to risk losing.

Sorry, Kubikulann, I guess I should have used Kubik instead. My apologies.

Puzzlenut, it sounds like your saying that 0^0 can be answered as 0 or undefined or 1, depending on your preference. However, and maybe this would be better asked to the OP, but, the question was simply what does 0^0 equal. It was not qualified in any way. It was not asked as what is the calculus derivation of 0^0, or what is the numerical derivation of 0^0. So, while most mathematicians may automatically see 0^0 as being 1, it still must be noted that this is the calculus derivation, and/or numberical derivation, or whatever way you used to come up with 0^0 = 1. At face value 0^0 is still 0. Much as a common slang word such as "cool" would be known by everyone, but it still must be noted that this is a slang word. 0^0 does have the face value of 0, even if it is commonly used as 0^0 = 1, and therefore if a question is asked what does 0^0 equal, I believe, you would have to answer 0, since 0^0 = 1 must be qualified first.

Quote: JyBrd0403

Sorry, Kubikulann, I guess I should have used Kubik instead. My apologies.

:-) Actually the name is my young nephew's deformation of the name of Kubilay Khan, the Mongol emperor.

Mike asked me to say something here.

By convention, an "empty sum" is defined to be 0 and an "empty product" is defined to be 1. For example, 0! is an empty product, and it is defined to be 1. Similarly, 0^0 is an empty product. But so are 6^0 and 133^0. These conventions are used primarily because many theorems are then true in certain trivial cases. Special cases don't need to be singled out, thus making the expression of the theorem more complex. These conventions also assist in creating recursive definitions, as well as aiding base cases for inductive arguments. But these are conventions and not definitions. A definition assumes there is something meaningful there to begin with. But, if you want to use "definition" then who am I to deny you that pleasure.

For those who don't know me and may wonder why Mike asked for my input, I have a Ph.D. in mathematics and was a professor of mathematics / computer science for 26 years.

A similar puzzle,where the value actually exists and is not a matter of convention or definition: let i = sqrt(-1). What is i^i ?

Quote: DoubleOrNothing

The correct response is all of the above.

As much as it is none of the above.

Quote: teliot

Mike asked me to say something here.

By convention, an "empty sum" is defined to be 0 and an "empty product" is defined to be 1. For example, 0! is an empty product, and it is defined to be 1. Similarly, 0^0 is an empty product. But so are 6^0 and 133^0. These conventions are used primarily because many theorems are then true in certain trivial cases. Special cases don't need to be singled out, thus making the expression of the theorem more complex. These conventions also assist in creating recursive definitions, as well as aiding base cases for inductive arguments. But these are conventions and not definitions. A definition assumes there is something meaningful there to begin with. But, if you want to use "definition" then who am I to deny you that pleasure.

Well, now I find myself just trying to win an argument here, because I understand what you guys are saying, and I think you understand what I'm saying, so I'm kinda just nitpicking to win an argument now. However, and again, this might be better asked to the OP, but, I think the "root" or "definition" of what 0^0 equals is what the spirit of the original question was.

Once again 0^0 = 1 is qualified as being used by convention. I think in that case the question would need to be asked as, "By convention, what does 0^0 equal?". I understand that's not how mathematicians would talk to each other, they wouldn't qualify 0^0, and say "by convention", 0^0 equals 1, but I am nitpicking to show that you can clarify the subject, that it has a "face value" or "root" definition.

I took the question as being the "face value" or "root" or "definition" of 0^0. I think that was the spirit of the question, and if indeed that is the spirit of the question, the answer, I still believe, would be 0. It seems, Teliot, you would agree with that, if indeed the "definition" of 0^0, was the spirit of the question.

http://www.askamathematician.com/2010/12/q-what-does-00-zero-raised-to-the-zeroth-power-equal-why-do-mathematicians-and-high-school-teachers-disagree/

It's defined as or taken to be one for convention. But not a really definitive expression.

deleted

Quote: Ibeatyouraces

My new answer, 0^0 = impossible calculation.

I would accept that. But, in the same breath, go on to challenge the impossible.

Quote: teliot

What is i^i ?

Wolfam Alpha says i^i = e^(-π/2)
I don't know how pi and e get involved.

Sorry to be absent from the thread I started, but G2E week is always very hectic.

Eliot and I had breakfast this morning and he was kind enough to work this problem with me. In part, Eliot reminded me that, by definition:

a^x = e^x*ln(a)

This allows us to define values of a^x for non-integer values of x.

Using this formula for 0⁰ we get e^0*ln(0).

ln(0) does not exist. It is an absurd question.

I hate to put words in Eliot's mouth, but I think he would say that since ln(0) does not exist, so doesn't 0^0.

However, I don't think Eliot has a problem with ... assigning ... a value of 1 to 0^0, in order to express other formulas, like the binomial theorem, without having to state exceptions for the case of 0.

So, maybe I could have asked the question better. A lot of the arguing seems to be semantic in nature.

My apologies in advance if I misquoted Eliot in any way.

Quote: Wizard

I hate to put words in Eliot's mouth, but I think he would say that since ln(0) does not exist, so doesn't 0^0.

However, I don't think Eliot has a problem with ... assigning ... a value of 1 to 0^0, in order to express other formulas, like the binomial theorem, without having to state exceptions for the case of 0.

Well, you guys lost me with the math again. :) If it ever gets back to 0 multiplied by itself 0 times I'm still going with 0.

The whole point is that these expressions can be taken at "face value". I don't know what the big college words are , but I do know that there is a way to say that 0^0 taken at "face value" is 0. Especially since I already know that 5^5 taken at "face value" is just asking you to take 5 and multiply it by itself 5 times. So, somewhere in some brilliant college kids mind there is a way to express with big words nobody understands, that 0^0 taken at "face value" is asking you to take 0 and multiply it by itself 0 times, and that you have to use a "cheat" or mathematical convention of some sort or another to get 0^0 = 1.

If I were to ask someone to write down .999... on a piece of paper, and they kept writing down 1 on the paper, and saying they did actually write .999... because .999... is 1 and 1 is .999... and so therefore they actually wrote down 1. There is still a way to express to this person that you want him to write down .999... and not 1. It's not just uh oh were all lost now, there's just no way to express to someone that you literally want them to write down .999... and not 1. The person could, of course, be a smart ass and just continue to write down 1, and insist that he is writing down .999... because .999... is literally 1. But, the point I'm trying to make is that the person who keeps writing down 1 after you've expressed that you literally want him to write down .999... is in the WRONG! .999... does have a literal value and a literal definition. Now, again I don't know what the big college words are to get the person to actually write down .999..., but, if they keep insisting that they are actually, literally writing .999... down on the paper, they are going to find themselves with a one way ticket to a funny farm.

Again the point is these expressions can be taken at "face value". I know mathematicians use mathematical conventions, and algebraic derivations, and etc. etc. But, those are still just "cheats" or "code" or qualified answers. At "face value" 0^0 = 0, and .999... = .999... And the whole point is that once you recognize that, then you can clarify what you actually mean to express by 0^0 or .999... whether you want to take 0^0 at "face value" or by mathematical convention. .999... literally or by mathematical convention 1. If you can't do that I have no idea how you can get someone to literally write down .999... except by accident, and that seems pretty lame for college guys to not be able to get someone to literally write down .999...


Almost forgot about the quote. If 0^0 doesn't exist, then I believe 0^0 would be 0. And, just as a side note, 0^10 (or 0, multiplied by 0, 10 times) is 0 on the calculator.

Quote: JyBrd0403

If 0^0 doesn't exist, then I believe 0^0 would be 0.

Non-existence is nothingness. Strictly speaking, zero isn't nothing or none. Zero to itself has form, and hence exists. If only as a big nag which is driven by a puny existence.

Ask how zero can be zero to itself.

Quote: Wizard

Using this formula for 0⁰ we get e^0*ln(0).

http://www.math.fsu.edu/~bellenot/class/f99/cal1/indeterminate.pdf

Quote: teliot

A similar puzzle, where the value actually exists and is not a matter of convention or definition: let i = sqrt(-1). What is i^i ?

Hint : Euler's equation.
e^ix = cos x + i sin x

Begin by finding the value of ln i. Then find e^(i ln i).

Such strangeness of the numbers hints at a universal format with no need to choose one convention over another. Eg, where -0 and +0 can be held non-identical, so that o = root(-0) is merely a suspended (-) result of the sign. A further celebration of zero, as o^o in the process of zero.

The mathematics as a single logical dimension of an ordered space-time.

Quote: kubikulann

Hint : Euler's equation.
e^ix = cos x + i sin x.

Speaking of Euler's equation, we can use it to show that e^pi*i=-1.

My question: Is the case of x=pi (I wish I could make a nice looking pi symbol) a degenerative form in that equation?

What I really want to know is does e^pi*i equal a strong, firm, no room for debate, -1? Or is it more like the 0^0=1, where some of us *ahem* assign it that value to make other laws of mathematics hold water, but nobody can prove it really does.

Quote: Wizard

I wish I could make a nice looking pi symbol

With Num Lock on, hold down Alt and type 227 on the numeric keypad to get the lowercase pi symbol (π). It doesn't look so good in the default forum font (Tahoma). I think Times New Roman has a decent-looking pi symbol. So:

[font=Times New Roman]Alt-227 [/font]

gets you π

You might also want to use a size of 16 to make it better match the font size of Tahoma:

[font=Times New Roman][size=16]Alt-227 [/size][/font]

gets you π

Quote: JB

With Num Lock on, hold down Alt and type 227 ...

π

Thanks!

e^πi=-1

Quote: Wizard

Quote: kubikulann

Hint : Euler's equation.
e^ix = cos x + i sin x.

Speaking of Euler's equation, we can use it to show that e^pi*i=-1.

My question: Is the case of x=pi (I wish I could make a nice looking pi symbol) a degenerative form in that equation?

What I really want to know is does e^pi*i equal a strong, firm, no room for debate, -1? Or is it more like the 0^0=1, where some of us *ahem* assign it that value to make other laws of mathematics hold water, but nobody can prove it really does.

I think this is worthy of being broken off into its own thread; it's going to get missed at the tail end of this one. I'd be interested in attempting to follow the reasoning.

In my non-mathematical opinion, since both i and -1 are imaginary numbers (assuming, anyway, that the square of an imaginary number is itself an imaginary number; feel free to correct that assumption), and as such mathematical abstractions, I don't know how by definition you could derive a firm value. OTOH, there is an imaginary number on either side of the = sign, so at least theoretically, the imaginary values could be cancelled out in some configuration that would leave a firm value, so I suppose it could be done. Not that I know how.

0^0






Babs darling, this is not the mathematical symbol for a bra.

Quote: beachbumbabs

I think this is worthy of being broken off into its own thread; it's going to get missed at the tail end of this one. I'd be interested in attempting to follow the reasoning.

So do I. Please resume discussing this topic here.

Quote: Buzzard

0^0





Babs darling, this is not the mathematical symbol for a bra.

Kinda looks like a dunce hat between two Princess Leia cinnabons to me....so I can relate when it comes to abstract mathematical concepts. It's pretty fun when all the boys patronize me for making a doofus comment. I love it when it's Big Brains on Parade Day; I learn a lot. No?

Now if the caret went the OTHER way, a fair argument for major cleavage could be made....0v0....

0^0 = eyeglasses

But it seems to me why 0^0 =1 is because its te number of null sets being counted. Therefore there is one null-set. Its very similar (IMHO) to a universe-in-a-container, one needs proof of the container, followed by the rhetoric of what happens when a speed-of-light object stikes the container wall. So it is with 0^0... we need an extra lable (in my example a "set") or we need to introduce a constant "e" to rectify our rationale. Unfortunatly, the latter makes 0^n discontinuous. Thus my rationale linguified "If you have "nothing", anything done to it that does not introduce an exterior quantity or quantification, does not alter "nothing". Conversly, "When "nothing" has a quantity applied, or is quantified, nothing is no longer nothing".

So 0^0 is a logical argument for the existence of multiple dimensions, by extrapolation. Really interesting path to omniopsism, 98. Yes, I made that word up. It means what I want to say, as an all-inclusive philosophy that grants the concept there is existence beyond our knowledge but not beyond our reasoned imagination. If there's a real word for that, I don't know it.

Even fractional dimensions. And it might be that the definition of 0 is i^-3+i^3.

Quote: teliot

A similar puzzle,where the value actually exists and is not a matter of convention or definition: let i = sqrt(-1). What is i^i ?

If exp(i*PI)=-1=i*i then exp(i*PI/2)=i then i^i = exp(i*PI/2)^i = exp(-PI/2) approx 0.2078795764

Can you explain the fault in Bernoulli's 1727 argument that ln(–1) = 0, since 0 = ln(1) = ln(-1*-1) = 2ln(-1) ?

Hello Everybody,


i've been gone for a while and i'm sorry to bring this (old) post up again but the reason is the fact that the correct answer would be "indeterminate" (not undefined).

I voted "other" and here is the explanation:
together with 0^0 all other following cases are called indeterminate: 0/0, infinity/infinity(dont have the symbols :-), infinity - infinity, 0*infinity, infinity^0 and 1^infinity.

They are called this way because for example 0/0 symbolises a ratio of any 2 series that converge to zero but for which we cannot reach to any conclusion about the limit of the ratio. It is similar for 0^0.

Just wanted to clear this up, but i don't think i can offer other explanations, i remembered this from my university years ...

I would use a set theoretic definition as a "first principles" definition of exponentiation. So, how many functions map the empty set to the empty set? The answer is 1 (a function is just a set, which can be empty).

So, I'm going with 1.

Quote: AxiomOfChoice

I would use a set theoretic definition as a "first principles" definition of exponentiation. So, how many functions map the empty set to the empty set? The answer is 1 (a function is just a set, which can be empty).

So, I'm going with 1.

Haha, i don't know what to answer, all i say is that all those cases are already defined as "indeterminate". So if you want to change existing mathematics then yeah..

Quote: edward

Haha, i don't know what to answer, all i say is that all those cases are already defined as "undetermined". So if you want to change existing mathematics then yeah..

Defined where? I have never seen such a definition for 0^0. By the standard set theoretic definition of exponentiation, the answer is 1.

Do you get a different answer if you start with the Peano axioms and define exponentiation accordingly? I've actually never done that; only used set theory to define whole numbers, followed by defining functions, followed by defining exponentiation.

Quote: AxiomOfChoice

Defined where? I have never seen such a definition for 0^0. By the standard set theoretic definition of exponentiation, the answer is 1.

Do you get a different answer if you start with the Peano axioms and define exponentiation accordingly? I've actually never done that; only used set theory to define whole numbers, followed by defining functions, followed by defining exponentiation.

I've read just now a reply from MathExtremist in this thread (page 7), which i skipped somehow until now. Apparently some textbooks give it as indeterminate others as 1.
And that it should be 1 ... can't argue with that.

My question about the Peano axioms was stupid and poorly formed, anyway.

I guess, I can't think of a reasonable definition of exponentiation where this would be undefined. You can always choose to make it undefined by convention, but it's not a special case, so I don't see why you would do that. It's just 1.

FWIW, I just checked, and when I type it into my calculator, I get 1. Are there really conventions that make it undefined? That seems very strange.

Quote: edward

I've read just now a reply from MathExtremist in this thread (page 7), which i skipped somehow until now. Apparently some textbooks give it as indeterminate others as 1.

For the simplest problem as x -> 0+ the limit of x^x =1. As x -> 0- the function x^x does not exist,

But for the more general problem the limit only equals one if and only if f and g are locally given by convergent power series. at c, and f is positive for x sufficiently close (but not equal) to c.


So it is customary to say "indeterminate" because there are cases that exist where the limit approaches something other than 1. But most of the time the conditions are satisfied and the limit goes to 1.

===========
The indeterminate form 0/0 is much more likely to go to different constants than 0^0.