**Clever student:**

I know!

= = = = .

Now we just plug in x=0, and we see that zero to the zero is one!

**Cleverer student:**

No, you’re wrong! You’re not allowed to divide by zero, which you did in the last step. This is how to do it:

= = = =

which is true since anything times 0 is 0. That means that

= .

**Cleverest student :**

That doesn’t work either, because if then

is

so your third step also involves dividing by zero which isn’t allowed! Instead, we can think about the function and see what happens as x>0 gets small. We have:

=

=

=

=

=

=

=

=

=

So, since = 1, that means that = 1.

**High School Teacher:**

Showing that approaches 1 as the positive value x gets arbitrarily close to zero does not prove that . The variable x having a value close to zero is different than it having a value of exactly zero. It turns out that is undefined. does not have a value.

**Calculus Teacher:**

For all , we have

.

Hence,

That is, as x gets arbitrarily close to (but remains positive), stays at .

On the other hand, for real numbers y such that , we have that

.

Hence,

That is, as y gets arbitrarily close to , stays at .

Therefore, we see that the function has a discontinuity at the point . In particular, when we approach (0,0) along the line with x=0 we get

but when we approach (0,0) along the line segment with y=0 and x>0 we get

.

Therefore, the value of is going to depend on the direction that we take the limit. This means that there is no way to define that will make the function continuous at the point .

**Mathematician: **Zero raised to the zero power is one. Why? Because mathematicians said so. No really, it’s true.

Let’s consider the problem of defining the function for positive integers y and x. There are a number of definitions that all give identical results. For example, one idea is to use for our definition:

:=

where the y is repeated x times. In that case, when x is one, the y is repeated just one time, so we get

= .

However, this definition extends quite naturally from the positive integers to the non-negative integers, so that when x is zero, y is repeated zero times, giving

=

which holds for any y. Hence, when y is zero, we have

.

Look, we’ve just proved that ! But this is only for one possible definition of . What if we used another definition? For example, suppose that we decide to define as

:= .

In words, that means that the value of is whatever approaches as the real number z gets smaller and smaller approaching the value x arbitrarily closely.

*[Clarification: *a reader asked how it is possible that we can use in our definition of , which seems to be recursive. The reason it is okay is because we are working here only with , and everyone agrees about what equals in this case. Essentially, we are using the known cases to construct a function that has a value for the more difficult x=0 and y=0 case.]

Interestingly, using this definition, we would have

= = =

Hence, we would find that rather than . Granted, this definition we’ve just used feels rather unnatural, but it does agree with the common sense notion of what means for all positive real numbers x and y, and it does preserve continuity of the function as we approach x=0 and y=0 along a certain line.

So which of these two definitions (if either of them) is right? What is *really*? Well, for x>0 and y>0 we know what we mean by . But when x=0 and y=0, the formula doesn’t have an obvious meaning. The value of is going to depend on our preferred choice of definition for what we mean by that statement, and our intuition about what means for positive values is not enough to conclude what it means for zero values.

But if this is the case, then how can mathematicians claim that ? Well, merely because it is useful to do so. Some very important formulas become less elegant to write down if we instead use or if we say that is undefined. For example, consider the binomial theorem, which says that:

=

where means the binomial coefficients.

Now, setting a=0 on both sides and assuming we get

= =

=

=

=

where, I’ve used that for k>0, and that . Now, it so happens that the right hand side has the magical factor . Hence, if we do not use then the binomial theorem (as written) does not hold when a=0 because then does not equal .

If mathematicians were to use , or to say that is undefined, then the binomial theorem would continue to hold (in some form), though not as written above. In that case though the theorem would be more complicated because it would have to handle the special case of the term corresponding to k=0. We gain elegance and simplicity by using .

There are some further reasons why using is preferable, but they boil down to that choice being more useful than the alternative choices, leading to simpler theorems, or feeling more “natural” to mathematicians. The choice is not “right”, it is merely nice.

0^0=1 because i^2=-1

http://oi57.tinypic.com/2qmzyvk.jpg

(1-1)^(1-1) view 😉

http://oi57.tinypic.com/v3d0rk.jpg

http://oi61.tinypic.com/zvbnv4.jpg

omg! 0-0^0=-1

http://oi62.tinypic.com/f3i6oz.jpg

the shape of the square root

http://oi61.tinypic.com/23tjms1.jpg

The argument is a construct for mathematical convenience.

It is not “true” for all abstracts. But it does work for most models which is why it is so widely used.

e.g.

If (in the form a^x)

1^1=1

and

1^0=1

then

x=1=0

which is not “true”

final countdown >>> 0^0=1^sqrt(e)

http://oi60.tinypic.com/2n1w5rs.jpg

peculiarity of oddity

http://oi57.tinypic.com/oiuaud.jpg

open design 😀

thanks for Britney 😉

http://oi57.tinypic.com/2v31xro.jpg

operating scheme:

0 rotate

0 plus

0 minus

http://oi62.tinypic.com/988q6c.jpg

13^2=12^2+5^2

http://oi59.tinypic.com/sm5z83.jpg

negative curvature rescues

http://oi60.tinypic.com/hsn7e0.jpg

strange tangent to displace the variable y

http://oi60.tinypic.com/33e1yzb.jpg

four-dimensional uncertainty with four metrics

http://oi57.tinypic.com/zyg6tg.jpg

pi e “gravity”

http://oi61.tinypic.com/34r7ibb.jpg

Better not ask how this idea

http://oi60.tinypic.com/s2d6bd.jpg

smells like aluminum

http://www.green-planet-solar-energy.com/images/aluminium-bohr.gif

pi2e square

http://oi59.tinypic.com/ege7f5.jpg

0^0 is indeterminate quantity : proof

let 0^0= k ; taking log both side

0loge 0=loge k

0(-infinite)= loge k

hense proof no value of exist

oh no! e number is strange

http://oi58.tinypic.com/apa0s9.jpg

calculate the neutral element

http://oi60.tinypic.com/qxufbs.jpg

Zero to zero was in need of dimensions 28297

Zero Complex only 13

harmonize the needs of quaternions referenced to prime numbers

3 neutral elements “0” , “1” , “i”

The operating matrix

http://oi62.tinypic.com/2nhmcxt.jpg

Rao KD,

Limits are irrelevant. Consider lim(h->0) (1+h)^(1/h). This is an indeterminate form of the type 1^infinity. Clearly 1^infinity = 1, not e. By changing the above to lim(h->0) (1+2/h)^h we get e^2. So limits are not unique, not equal to the correct answer of 1, and therefore are of no help. In fact, we already know that the indeterminate forms produce different answers in different cases, which is why we have, for example, L’Hopital’s rule for the 0/0 cases. (Which is why they’re called “indeterminate”.) So you can’t determine 0^0 via limits any more than you can 0/0.

As for 3^7 > 7^3 and such: You say that if x,y>2, the exponent rules (gives the larger result). The actual boundary is e, not 2. Example: 2.1 ^ 2.2 = 5.1154… whereas 2.2 ^ 2.1 = 5.2370…

where you want something dumb to count nothing

+0=(-0)^2 😉

http://oi59.tinypic.com/10eo5t1.jpg

tangens(0)-tangens(0i)

cute puzzle

http://oi62.tinypic.com/2ef5csw.jpg

each features a zero changes e -> sqrt(2)

eats action by replacing a part number

http://oi62.tinypic.com/sni8hk.jpg

circumvent the uncertainty of the logarithm

already knows where so many dimensions

http://oi62.tinypic.com/2e5ryx4.jpg

Welcome to the Matrix 8)

http://oi60.tinypic.com/25g3w36.jpg

(0/i)^(0/i) & (0i^2)^(0i^2) 😉

http://oi61.tinypic.com/1syvrm.jpg

total derivate x^x => {v}* [1+log(xx’),1+tg(x/x’)]

http://oi61.tinypic.com/11r8b2t.jpg

beast tamed 😀

no real, only imaginary

http://oi57.tinypic.com/241uwlv.jpg

something great in anything -> 0/sin(0)

http://oi57.tinypic.com/2q32fld.jpg

Finally calculations

http://oi58.tinypic.com/2qs1ik5.jpg

7 Card Stud Poker vs zero 😉

http://oi57.tinypic.com/k2enwy.jpg

zero space formula ^-^

http://oi62.tinypic.com/281eiv5.jpg

Kate Ryan rulez!

kyrbis rocks 😉

Zero is usually measured zero unusual

http://oi62.tinypic.com/33ytd1u.jpg

no comment 😀

http://oi59.tinypic.com/250pahu.jpg

to battle physicists 😉

http://oi58.tinypic.com/25h1v2x.jpg

trident for the white queen

no i dont know meybe 😕

http://oi59.tinypic.com/330fdxz.jpg

no torus is tangent and its behavior

“all for you”

http://oi57.tinypic.com/9uvv5w.jpg

((cos(z)+sin(z))/(cos(z)-sin(z)))^((cos(z)+sin(z))/(cos(z)-sin(z))) =?=1

http://oi59.tinypic.com/16ackzs.jpg

eye-opener 😉

http://oi62.tinypic.com/3179myq.jpg

You need a more sophisticated approach:

https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

I am of the view that 0^0 =0 for the reason that 0^n=0. 0^n=0 x 0 x 0 x 0x…….n times. You are multiplying NOTHING so you get NOTHING. In the case of 0^0, you get nothing and you do not even dare to multiply, you get NOTHING.

0^0 = 1. Every math book I’ve seen that gives the following formula,

e^x = Sum(n=0->oo) x^n/n!

says that it is good all real numbers. I have never seen an exception given. Now, for this formula to work for x=0 you have to have 0^0=1. So you have to choose: Either 0^0=1 and the formula is good for all numbers, or 0^0!=1 and the formula for e^x above is good for all numbers except zero. You can’t have it both ways.

You also need 0^0=1 for other formula that use the capital sigma notation for polynomials (e.g., the binomial theorem) or infinite series (like the one above) to work.

On top of this there’s the argument that y^x is the number of ways to map x objects to y objects, which gives 0^0=1. And Euler says so. And limits are no good because lim(h->0) (1+h)^1/h is e, not 1, as 1^oo would be. And lim(h->0) (1+2h)^1/h is e^2 — yet another answer! And then there’s also the empty product bit.

0^0 = 1.

I agree, 1. Because e=2.71828, defined by 1/O! + 1/1! + 1/2! … which

resolves to 2.71828, not 1.71828. That’s my 2¢.

Probably many of you were taught that any number to the 0th power = 1 and that 0 to any power = 0

The question here is weather or not zero is a number.

This is better than Chebyshev polynomials 😀

may explain something

https://en.wikipedia.org/wiki/Quadrifolium

i have not understand the answer of this i want specific answer

i think zero means nothing

so in both cases zero means nothing

so 0^0=0