0^0 = 1 Prove me wrong. Pro tip: You can't.

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Because it's not wrong. One times zero, zero times is still one.

It's not wrong.

>>8114959
If it makes you feel any better, google calculator agrees with you even though you're wrong.

>>8114959
You're right.

x^3 = 1 * x * x * x
x^2 = 1 * x * x
x^1 = 1 * x
x^0 = 1

The empty product is one, because one is the multiplicative identity, just like the empty sum is zero, because zero is the additive identity.

>>8114959
Limit x->0+ of 0^x = 0

get rekt fuccboi

>>8115042
Nigga fuck you.

X^y can be written as x^2y/x^y, which gives is the subtractive property = x^y
So 0^0/0^0 gives us literally 0/0 which is an indetermination.

>>8115089
x^y = x^(y+1)/x^1
0^1 = 0^2/0^1
0^1 = 0/0

Whoops, I guess 0^1 is undefined too.

>>8115081
Sure, 0^x is discontinuous at 0. That doesn't mean it's undefined at 0.

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/thread

>>8115100
>uses arbitrary limit to argue 0^0 is defined as 1
>ignores arbitrary limit which disagrees
Fuccboi

>>8115114
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.

>>8115023
>1*0*0 =1
Ok

>>8115163
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

>>8115165
But 0^0 can be written as 0/0, and that's undefined.

>>8115208
So can 0^1, but that value's not undefined.

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>>8115163

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

>>8115297
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

>>8115358
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?

>>8115367
Nah, I just came here to prove OP wrong and I did.

>>8115375
But your "proof" also implies that 0^1 is undefined, which it's clearly not. This means that your proof must be wrong.

>>8115379
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)

>>8115398
a^2 = 1*(a*a)
a^1 = 1*(a)
a^0 = 1
a^-1 = 1/(a)
a^-2 = 1/(a*a)

>>8115466
a/a = 1 only if a != 0

>>8115594
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.

>>8114959
> he thinks math refers to universal truths
how embarrassing

>>8115697
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.

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

>>8114959
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

>>8115163
the zeroes cancel out

>>8115722
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.

>>8115722
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.

>>8115089
PEMDAS YOU IDIOT

>>8115160
>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

>>8116411
Perhaps you should try looking up the definition of "natural numbers", anon.

>>8116399
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.

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>>8117223
>implying 4 and 0.5 aren't natural numbers

>>8114959
anything ^0 is defined as 1.

i^0 = 1

You're not wrong.

>>8117243
>confusing rational with natural

>>8117243
Natural numbers are {1, 2, 3, 4, 5 ...} you dumbfuck.

>>8117243
Yes, that's exactly what I'm implying. I again suggest you look up the actual definition of the terms you're using.

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>>8114959

>>8117423
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.

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

>>8115223
0^1 = 0/1 not 0/0

>>8117596
0^1 = (0^2)/(0^1) = 0/0

>>8114959
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

>>8117603
just zero you moron

>>8117603
take a lap

>>8117855
is this bait? 0^-1 is never defined, that's the problem

>>8117855
0^-1 = 1/0 = undef

>>8117871
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.

>>8117826
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?

>>8117855
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

>>8118295
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?

>>8118304
Only idiots with useless pure math degrees give a shit about discontinuous functions.

>>8118315
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.

>>8118322
> transcendental functions

>>8118322
e^z or gtfo.

>>8118327
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.

>>8118351
What?

>>8115089
That's just as bad as
>multiply both sides by zero
>there is no solution!

>>8114959
[math] \lim_{x \to 0} x^{x} = 0 [/math]
I see no problem with this assertion.

>>8118375
x^x -> 1, I think you mean.

>>8117320
you forgot 0

>>8118358
> transcendental functions
> valid

>>8118397
So you don't agree that e^x is a valid function?

>>8118540
It isn't.

>>8118541
Why shouldn't it be?

>>8118548
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?

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>>8114959
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.

>>8115358
Division is a binary operator R x R/{0}->R
So your argument doesn't make sense.

>>8115108
>Academic licence

Buy it yourself, retard.

>>8118801
Read the thread. Discontinuity [math]\neq[/math] undefined

>>8118597
that reddit did it instead.