What is vakue of 0 to the power of 0?

Question

Pl help me?

Answer 1

That's an indeterminate form.

Answer 2

Dear, Its so simple
Any number to the power zero is always 1
As 1^0=1
Hence 0^0=1

have a nice day.

Answer 3

1. anything to the power of 0 equals 1, even 0^0

proof:

a^(b+c) = a^b * a^c for any a,b,c (this is from the definition of power operation on the field of real numbers)

if c=0, then a^0 must be 1 for any a

this is algebra question, not calculus question, for all of you with indeterminate forms. 0^0 is an indeterminate form *in calculus* and *only* when you're trying to find a limit of a function (here no function is mentioned thus no limit can exist!). in algebra 0^0 is 1, which follows easily from the set of rules by which you define power operation and which power operation must fullfil.

avik, how do you come from 0^0 to 0/0? algebraicly you can't because 0^(0-0) = 0^0 * a^0, a being inverse element in R for 0 for operation of multiplication, which doesn't exist.

consider f(x,y) = x^y, here there is no limit at (0,0) (using two variables you can come from 0^0 to 0/0 - but then limit is undefined and not 1). so it all depends from what you get to 0^0 - this is what i ment by calculus or algebra question - what the context is. since poster didn't give any details for that i assumed algebraic context.

but you are right, it is indeterminate. whether we can define it as 1 depends on what the context is, and how much do we want to make our lives (and calculations :) easier

Answer 4

Indeterminate.

You cant say its 1. If you try and say that is so... prove it... you will end up with another indeterminate form i.e., 0/0.

You could define it as the limit of x^x as x -> 0 which is 1. However then you have to justify why that and not some other limit. Then your opening up a whole bag of worms by even defining it.

Answer 5

Everything to the power of 0 is 1
but if 0 to the power of 0 has no solution....

Answer 6

Anything to the power of 0 is 1.

Answer 7

Those of you giving "proofs" with logarithms should keep in mind that by your logic 0 is undefined. After all, 0=e^ln 0, but ln 0 is undefined, so 0 itself must also be undefined. Of course zero has a perfectly well-defined value, and so does 0^0, which is 1. Remember, b^x=e^(x ln b) is not the primary definition of exponentiation, but an extension of it to real numbers. Where the primary definition is applicable, as it is with any whole-number exponent, including zero, that definition takes precedence. And the primary definition is multiplication by the base n times. That is, b^3=b*b*b. b^2=b*b. b^1=b (and btw, those who mistook the primary definition as being multiplication of the number by itself n times are wrong: that would imply that b^1=b*b, since we are multilpying b by itself only once, but clearly that is wrong). And so b^0=*, where * represents the product of no numbers. Clearly, it doesn't matter what the base is, because the base _does not appear_ in the empty product. This tells us that b^0 must be a constant. Not a constant with a removable singularity at b=0, since that would mean that a number that doesn't appear in the empty product can influence whether it has a solution, which is absurd and contradictory. A constant. And since the only constant consistent with the laws of exponents is 1, b^0=1 for all b, and therefore 0^0=1. Q.E.D.

Edit: in response to avik r -

"x^0=1 when x is not 0."

Two things:

#1: argument from authority does not work when the consensus of professional mathematicians disagrees with you, and it certainly does not work when argued in opposition to a logical proof.

#2: Argument from anonymous authority never works.

"0^0=1 [as u say]
=>0/0=1

so what u say now?"

I say first that your use of the abbreviation 'u' is indicative of a faulty education, and second that your leap from 0^0 to 0/0 is logically unjustified. I assume that it's based on the idea that 0^0 = 0^(1-1) = 0^1/0^1 = 0^0, but if you consider that line of reasoning carefully, I believe you will find that it also implies 0^1 = 0^(2-1) = 0^2/0^1 = 0/0, and that would mean that 0 is undefined. Clearly, the rule b^(x-y)=b^x/b^y only works when division by b^y is actually defined. We are not giving exception to exponentiation here - consider the rule that x/x=1 - again, this only works when division by x is actually defined. But the fact that you can't naively apply exponent rules doesn't mean that 0^0 itself is undefined any more than 0^1 is undefined, because just as you don't have to compute 0^2/0^1 to figure out what 0^1 is, neither do you have to divide 0^1/0^1 to figure out what 0^0 is. The whole point of my post was to show that the value of 0^0 can be proved from first principles to be equal to the value of (anything else)^0, which is easy to find using the laws of exponents. Now, you are free to propose an alternate definition from which it does not follow that 0^0 = 1, but you cannot say that 0^0 is undefined as that expression is understood by the current body of accepted mathematics. If you still want to argue otherwise, at least try to do so using internally consistent arguments.

avik r (cotd):

This response is quite long, so I'll give you a brief summary - the fact that 0^0 is defined follows from the fact that x^0 is not defined as x^n/x^n, but rather as the product of no numbers. Below, I show that the product of no numbers may be defined without reference to exponentiation, and that this does imply that x^0=1 even when x=0. I then argue that the definition currently in use by most mathematicians is indeed the empty product, and discuss why mathematicians choose their definitions in the first place.

"how can we define the '0 times multiplication of any number x.' what does it mean in fact?"

We can define it quite simply as the product of no numbers, or the empty product. You can read about it on the wikipedia if you wish: http://en.wikipedia.org/wiki/Empty_product . Think of it this way: multiplication is fundamentally a binary operation, which takes two arguments and gives a single result. If you want to talk about, say, multiplying three numbers or one number or a multiset of any other size (we have to use multisets here, because the same number may appear more than once. If you are not familiar with multisets, consult the wikipedia: http://en.wikipedia.org/wiki/Multiset ), you have to expand the function. Expanding it to numbers larger than two can be done through function composition: for instance we define a*b*c as (a*b)*c. Of course there is the question of what order do we perform the function composition in, but since multiplication is commutative and associative, we don't have to actually answer that question, because it doesn't matter. But how do we define multiplication of one number - i.e. 5* ? In short, how do we define 5^1? You propose trying to use the laws of exponents to do this, and that actually works provided that the base is not zero. But what if it is? Your definition leaves 0^1 undefined, because (again) 0^2/0^1=0/0. Fortunately, it isn't necessary to make reference to exponents at all. We can define the multiplication of multisets of size 1 and 0 simply by using the properties of multiplication alone - specifically the fact that f(A∪B) = f({f(A), f(B)}), where f is the product function, A and B are multisets, and ∪ is the join (not the union - I tried to use the proper join symbol, but it won't show up here) of the two multisets. If that notation is too abstruse, I'm basically writing that a*b*c*d*e*f must equal (a*b*c) * (d*e*f), and the like. Under that definition, we can define the unary product by noting that a*b = (a*) * (b*), which is only possible if a*=a. In other words, the product of one number is the number itself. Similarly, the product of no numbers, which I've been writing *, must be 1, because A is the join of the multiset A and the empty multiset, and so (*) * (A*) = A*, which is only possible if *=1. By similar logic, we can show that applying any group operation to no numbers must yield the identity element for that group. Now note that x^0 is in all cases the product of x 0 times - in other words, x^0 is simply the product operation applied to the empty multiset. There is only one empty multiset, and so we cannot get a different result when x=0 than when x=some other number.

Basically the point of this very longwinded explanation is that you're trying to say that x^0 is defined as x^n/x^n, which if true would entail that 0^0 is undefined. What I'm trying to get across is that the function x^0 is defined most sensibly as the product of no numbers, in which case it entails that 0^0 is defined as 1. Now which definition is "right" is primarily a matter of convention, although in this case the convention on the side of defining 0^0=1 - see both the empty product link I gave you and the link to dr. math somewhere above. A similar argument may be had over whether 1 is a prime number. If you define a prime number as having exactly two factors, then it follows instantly that 1 is not prime, whereas if you define a prime number as having at most two factors, then 1 IS a prime number. This is not merely idle speculation: for much of the 19th century the latter definition was in use, and 1 was considered a prime number ( http://en.wikipedia.org/wiki/Prime_number#History_of_prime_numbers ). This followed trivially from the definition in use at the time. The definition was changed only because it was more useful to define primes in a way that excluded 1 ( http://mathforum.org/library/drmath/view/57058.html ). Similarly, it is more useful, and requires far fewer exceptions to established theorems, to define x^0 in terms of the empty product, precisely _because_ it allows 0^0 to be defined. Among other things, it is no longer required to state that there are exceptions to the binomial theorem, or that e^x= (n=0, ∞)∑(x^n/n!) except when x=0, or the like. That last one is particularly important: the exponential function is too fundamental to complex analysis to be arbitrarily restricted! Now, you may argue that _by the definition you give_ 0^0 is undefined, but you cannot argue that _by the definition used by most mathematicians_ 0^0 is undefined. And when people ask things like "is .99999... equal to 1" or "is 1 a prime number" or "what is 0^0" they are asking about the definition used by most mathematicians, which is indeed 0^0=1.

- Pascal

P.S. "please mention some reliable sources for your argument so that both of us can be benefited" - I'm a dead french mathematician. How much more reliable can you get?

Answer 8

Let's have some background:
e^x is a function from R the set of reals to R+, the set of positive reals. Its inverse function called "ln" or the natural log is therefore a function from R+ to R. Since 0 is not included in the domain of ln so ln (0) is not defined literally.
Now what do we mean by a^b. Its actual definition is that a^b:=e^(b*ln a). By this 0^0 would have been e^( 0 * ln 0). Since ln 0 is not defined so clearly 0^0 is also not defined.

Answer 9

It's 1 (one). Not only is anything to the power of 0 equal to 1, but the empty product and zero factorial are bothd defined as 1, so this follows.

Answer 10

hmmm, consider the function

f(x) = y = x^0, (for x not equal to zero)

except for the "special" case where x=0,
it is clearly a straight, horizontal line which
coincides with the line defined by the function

y = 1

hmmm again ...


"IF" we define: f(0) = 1, then lo and behold,

f(x) is a continuous function.

....

problem?

nope

[NOTE: there are other, cloudier ways to speak of
this, but the goal of mathematics is to eliminate
ambiguity ...... qed]

Answer 11

1

anything raised to 0 is 1