What is 0^0?

You heard me, is it 0 because 0*anything=0 or is it 1 because anything^0=1?

Prelude> 0^0
1
Prelude> 0**0
1.0

>>39
See >>38

This question pops up on /sci/ pretty regularly, and this is the last time I'm going to answer it.

To set the record straight the correct answer is "It depends". The Mathematics of numbers is basically a set of logical "models" (analogous to the mathematical models of physical situations) to things we intuitively know.

To clarify, the basic operations of arithmetic are addition, subtraction, exponentiation, multiplication, etc. are all intuitive for the nonzero finite numbers we know and love, but not always otherwise (for example, it is not clear what, if anything 0^0 should be), so the mathematical models can disagree.

I'll give the two main models and what happens. It is clear that many people are arguing based (possibly unknowingly) on one or the other model, but I must emphasize that these are models so there is no universal truth other than "it depends".

Modelling numbers in the cardinal sense (eg. seeing numbers as meaning the "size" of a set of distinct things) gives us the answer 0^0. Because in this model 0 is (usually) defined as {}, x^y is (usually) defined as the cardinality of the set of functions from y to x. Hence 0^0={}^{}=1 because there is only one map from {} to {} (namely {}, ie. the map that has an empty domain and codomain).

So 0^0=1. This result actually makes sense in a huge number of cases where counting is involved or numbers are treated in the discrete sense.

But let's model numbers in a different way now. The model above does not encompass negative, rational, or irrational numbers. Now, let's view numbers in the model often given in Analysis courses (one that does everything we expect and gives us all real numbers); that is, as the completion of the field of rationals (the set of rationals being defined as the quotient set of all ordered pairs of integers over the appropriate equivalence relation, and integers being defined in the usual way -- often on the definition given above). Usually, in this model we have the definition x^y=log (y exp x) (or ln, if you prefer that notation), where exp is defined in terms of its Taylor Series and log is defined as the inverse of exp. Now, in this case 0^0 is completely undefined and frankly meaningless; which is good, because if 0^0 is defined in this model then it would ruin much of Analysis!

So, it depends on your model -- and that's all there is to it.

>>43
>x^y=log (y exp x)
lol wut?

x^y = e^{y ln(x)}


In[1]:= e^(0 ln[0])

Out[1]= 1


In[2]:= Limit[x^0, x -> 0]

Out[2]= 1

In[4]:= Limit [0^y, y -> 0]

Out[4]= 0

\lim_{x \to 0}x^{\frac{\ln(k)}{\ln(x)} = k

For k = 0, the old 0^x trick works.

Thus, indeterminate.

X = 2+2

lim_{x \to 0}x^{\frac{\ln(k)}{\ln(x)}} = k

0^0=1

You are all wrong. 0^0 is a tasty sandwich. Not because it is logically arguable, but because it can be argued.

1, niggers

so whats 0/0

x/0 = 0
0/0 = 0

JESUS CHRIST GUYS

FINAL ANSWER:

DISCRETE MATHEMATICS—0^0=1
CONTINUOUS MATHEMATICS—0^0=¿

---END OF LINE---

0^0 is the face I made when I read this.

>>56
Even in anime, one's nose is not at the same altitude as one's eyes.

>>57
That's a mouth.

>>58
That's even moar anatomically incorrect.

>>55
That's a very discrete answer

hey fags

>>61
hey

>>59
He's raging probably.

1 >C