0^0 = 1

/thread

>uses arbitrary limit to argue 0^0 is defined as 1
>ignores arbitrary limit which disagrees
Fuccboi

You don't have to use limits to determine 0^0 = 1. If you just define exponentiation on the natural numbers as repeated multiplication, 0^0 must be 1 because the empty product is always 1. Or if you prefer set theory, 0^0 is 1 because there's one function from the empty set to itself. Neither of these rely on arbitrary limits.

>1*0*0 =1
Ok

Not 0 times as in "*0", 0 times as in "repeated 0 times", i.e. the empty string "".

0^2 = 0*0
0^1 = 0
0^0 = [empty product] = 1

But 0^0 can be written as 0/0, and that's undefined.

So can 0^1, but that value's not undefined.

Are you fucking retarded? As in not multiplying one by zero at all, hence "zero times" and thus leaving one.

>ITT: Handwaving

0^0 is undefined for the same reason 0^-1 is undefined

But 0^0 = 1 follows intuitively from discrete definitions of exponentiation and causes no inconsistencies, while no definition of 0^-1 is natural or free of inconsistencies. If 0! can be defined, why shouldn't 0^0 be?

a^1 = a
a^-1 = 1/a
(a^n)(a^m) = a^(n+m)
(a^1)(a^-1) = a^0 = a*(1/a) = a/a = 1

0^0 = 0/0 = undefined

checkmate atheists

Good job, you just proved that 0^x is undefined for all x. Example:

a^2 = a*a
a^-1 = 1/a
(a^n)(a^m) = a^(n+m)
(a^2)(a^-1) = a^1 = (a*a)*(1/a) = (a*a)/a = a

0^1 = 0/0 = undefined

Care to try again?

Nah, I just came here to prove OP wrong and I did.

But your "proof" also implies that 0^1 is undefined, which it's clearly not. This means that your proof must be wrong.

whoops, good call bro. back to the drawing board.

a^2 = a*a
a^1 = a
a^0 = a/a
a^-1 = a/(a*a)
a^-2 = a/(a*a*a)

a^2 = 1*(a*a)
a^1 = 1*(a)
a^0 = 1
a^-1 = 1/(a)
a^-2 = 1/(a*a)

a/a = 1 only if a != 0

I wasn't simplifying your definition of a^n. I was giving an alternative definition which is more natural and which allows 0^0 to be defined without contradiction.

> he thinks math refers to universal truths
how embarrassing

Well, I can at least say that "0^0 = 1" is consistent with standard arithmetic, but that "0^0 = x" is not consistent with standard arithmetic for any other value of x. It is of course possible to define x^y so that it has a hole at (0,0), but this is an artificial and silly definition.

You can't prove it's wrong/right (yet). There's no real waterproof mathematical evidence but the consensus indeed is 1.

cs.uwaterloo.ca/~alopez-o/math-faq/mathtext/node14.html

lim of x->0 of x^0 is 1,
lim of x ->0 of 0^x is 0
at the same point these two contradict each other, a contradiction shows that the result is undefined
as it is with lim of x->0 of 1/x it can be positive or negative infinity it is undefined therefore this is also undefined in the same way

the zeroes cancel out

The value of the LIMIT at a point says absolutely nothing about the value of the actual function at that point. The limits of x^0 and 0^x disagree because x^y is discontinuous at (0,0), not because it is undefined at (0,0).

Given that 0^0 = 1, it's still true that the limit as x -> 0 of f(x) = 0^x is 0. But AT 0, f(0) = 1. Yeah, f(x) is discontinuous; so what? It's discontinuous whether 0^0 = 1 or 0^0 is undefined.

Example:
lim as x -> 0 of floor(x^2) is 0,
lim of x -> 0 of floor(-x^2) is -1
at the same point these two contradict each other, a contradiction that shows the result is undefined

But wait, floor(0) = 0. It's not undefined at all, it's just discontinuous at x = 0, much like 0^x.

PEMDAS YOU IDIOT

>If you just define exponentiation on the natural numbers as repeated multiplication
But that's not how it's defined. Otherwise 4^0.5 would be meaningless. F U C C B O I

Perhaps you should try looking up the definition of "natural numbers", user.

PEMDAS doesn't say shit about what the ascii x^2y means, that's his own shitty notation, (x^2)y or x^(2y) would have been appropriate. Or laytek.

>implying 4 and 0.5 aren't natural numbers

anything ^0 is defined as 1.

i^0 = 1

You're not wrong.

>confusing rational with natural

Natural numbers are {1, 2, 3, 4, 5 ...} you dumbfuck.

Yes, that's exactly what I'm implying. I again suggest you look up the actual definition of the terms you're using.

...

So? All that does is prove x^y is discontinuous at (0,0), which is irrelevant to the actual value 0^0 = 1.

0^0=1 because thats how we define it. Graphical arguments are shit and are made by math plebeians.

>let's define an operation on the natural numbers....
>...using limits from real analysis
when did you realize Veeky Forums has zero aesthetics?

0^1 = 0/1 not 0/0

0^1 = (0^2)/(0^1) = 0/0

I've seen 0^0 defined as 0 in a textbook because it simplified a formula.

defining 0^0 makes exponentiation non associative
it's kind of ugly
>0^0 * (0^1 * 0^-1) != (0^0 * 0^1) * 0^-1

just zero you moron

take a lap

is this bait? 0^-1 is never defined, that's the problem

0^-1 = 1/0 = undef

Yes, that was my point. The fact you can rewrite 0^1 to be 0/0 is not enough to prove that 0^1 is undefined. Therefore the fact that you can rewrite 0^0 to be 0/0 shouldn't be enough to prove that 0^0 is undefined either.

Really? I'm surprised there are any formulas (other than the trivial 0^x) which get simplified when 0^0 = 0. Do you remember what it was?

Using 0^-1 as though it were a valid expression is what makes exponentiation non-associative in your example.

do you motherfuckers even l'hoptal

You do realize the limit of f(x) as x -> c has absolutely no relation to the actual value of f(c), right? And that bringing up techniques to evaluate limits is therefore completely irrelevant to whether 0^0 = 1?

Only idiots with useless pure math degrees give a shit about discontinuous functions.

f(x,y) = x^y is discontinuous at (0,0), whether or not 0^0 is defined. I don't think exponentiation is entirely useless.

> transcendental functions

e^z or gtfo.

And? The fact that functions like e^x, ln(x), and sin(x) are all transcendental doesn't stop them from being valid and useful.

What?

That's just as bad as
>multiply both sides by zero
>there is no solution!

[math] \lim_{x \to 0} x^{x} = 0 [/math]
I see no problem with this assertion.

x^x -> 1, I think you mean.

you forgot 0

> transcendental functions
> valid

So you don't agree that e^x is a valid function?

It isn't.

Why shouldn't it be?

It's the product of an infinite process.
Infinite processes don't exist.
ergo exp(x) doesn't exist.

Imagine if the resolution of the continuous and discrete proofs of 0^0=1 comes from some autists on a Taiwanese water buffalo husbandry forum. What would the history books say?

Undefined. Consider the following equations:

[eqn]\lim_{x\rightarrow0^+} 0^x=0[/eqn]

[eqn]\lim_{x\rightarrow0^+} x^x=1[/eqn]

Both are of the form [math]0^0[/math], and thus a definite value can't be assigned to [math]0^0[/math] without specifying some context.

Division is a binary operator R x R/{0}->R
So your argument doesn't make sense.

>Academic licence

Buy it yourself, retard.

Read the thread. Discontinuity [math]\neq[/math] undefined

that reddit did it instead.