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Anonymous Sun 29 Mar 22:46:14 2020 No.11513438 View Reply Original Report

Quoted By: >>11513459 >>11513496 >>11513513 >>11514074 >>11514089 >>11514096 >>11514184 >>11514296 >>11514315 >>11514395 >>11514412 >>11514692 >>11514763 >>11515453

0+0=0
0×0=0
0/0=Undefined
0^0=1
Why shouldn't 0/0=1? Alternatively, why shouldn't 0^0=0? Don't give me some fancy math degree mumbo jumbo. Don't give me books. Don't give me those gay equations math majors use to loophole their way through actual natural logic. I want the pure natural logic seen in the real world explained in a way a child could understand.

Anonymous

Anonymous Sun 29 Mar 2020 22:56:24 No.11513459 Report

Quoted By: >>11513462

>>11513438
Why did you ask this question if you don't want an answer to it?

Anonymous

Anonymous Sun 29 Mar 2020 22:57:26 No.11513462 Report

Quoted By: >>11513469

>>11513459
What's the point of an answer that is useless to me?

Anonymous

Anonymous Sun 29 Mar 2020 23:01:24 No.11513469 Report

Quoted By: >>11513483

>>11513462
Because an answer being useless to a useless person like you doesn't mean the answer is, in and of itself, useless.

Anonymous

Anonymous Sun 29 Mar 2020 23:06:39 No.11513483 Report

Quoted By: >>11514034

>>11513469
But it is. What's the point of an answer that is useless to me, the person asking for an answer? If you can't articulate your answer in a manner that is useful for the questioner, your answer is useless. It's as if you never answered the question.

Anonymous

Anonymous Sun 29 Mar 2020 23:13:41 No.11513496 Report

Quoted By: >>11513513 >>11514395

>>11513438
the values 0/0 and 0^0 are both indeterminate, not undefined.

For example, let x approach 0, then:
x/x = 1
x/0 = undefined
0/x = 0

x^x = 1
x^0 = 1
0^x = 0

Anonymous

Anonymous Sun 29 Mar 2020 23:21:14 No.11513513 Report

Quoted By: >>11514048 >>11514790

>>11513438
>Why shouldn't 0/0=1?
If we actually said 0/0=1, then we could say that since 0=0*0, we have 0/0=(0*0)/0=0*(0/0)=0*1=0. So 0/0 is undefined.
>Why shouldn't 0^0=0?
If we actually said 0^0=0, then we could say that since 0=0-0, 0^0=0^(0-0)=(0^0)/(0^0)=0/0, and again 0/0 is undefined.

>>11513496
>indeterminate, not undefined.
"Indeterminate forms" are just a special class of undefined expressions that arise in limits.

Anonymous

Anonymous Mon 30 Mar 2020 03:22:28 No.11514034 Report

Quoted By:

>>11513483
I could answer it, but you told us that any way to ACTUALLY answer the question is forbidden. So, since you clearly don't ACTUALLY want an answer, what's the point of asking?

Anonymous

Anonymous Mon 30 Mar 2020 03:26:40 No.11514048 Report

Quoted By:

>>11513513
>"Indeterminate forms" are just a special class of undefined expressions that arise in limits.
Not true. Indeterminate forms are a special class of discontinuities that arise in limits; they need not refer to an undefined expression.

Anonymous

Anonymous Mon 30 Mar 2020 03:41:37 No.11514074 Report

Quoted By:

>>11513438
>Why shouldn't 0/0=1? Alternatively, why shouldn't 0^0=0? Don't give me some fancy math degree mumbo jumbo. Don't give me books. Don't give me those gay equations math majors use to loophole their way through actual natural logic. I want the pure natural logic seen in the real world explained in a way a child could understand.

Simple. You've been listening to those who speak of no use. If math is the language of quantification, or "to count" then "zero" is literally not "of number". Nothing quantitative is being defined, therefore it lacks the qualification of being used in math or referred of as a number. It is at best the placeholder of number. It's defined by itself with circular reasoning basically, which means it it's not logical whatsoever.

Anonymous

Anonymous Mon 30 Mar 2020 03:47:39 No.11514089 Report

Quoted By: >>11514106 >>11514399

>>11513438
>0^0=1
also undefined, but x^x has a nice limit so that gets used
https://www.wolframalpha.com/input/?i=0%5E0
http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiJ4XngiLCJjb2xvciI6IiMwMDAwMDAifSx7InR5cGUiOjEwMDAsIndpbmRvdyI6WyItMS4zMiIsIjEuOTMiLCItMC4zMjQ5OTk5OTk5OTk5OTk4NCIsIjEuNjc0OTk5OTk5OTk5OTk5OCJdfV0-

Anonymous

Anonymous Mon 30 Mar 2020 03:50:15 No.11514096 Report

Quoted By: >>11514106 >>11514373

>>11513438
>0/0=1
Division is reverse multiplication. You can rewrite a/b=x to be x*b=a. E.g. to find 2/2, find x so 2*x=2. x is obviously 1. For 0/0, find x so 0*x = 0. Since this is true for all x, we can't just pick one x to be "the answer" so we call it indeterminate since there's more than one answer.

>0^0=0
0^0 is undefined, not 1, because:
0 to the power of anything is 0
anything to the power of 0 is 1
In 0^0 the 2 definitions clash so we call it indeterminate since there's more than one answer.

Anonymous

Anonymous Mon 30 Mar 2020 03:56:04 No.11514102 Report

Quoted By:

https://www.google.com/search?q=y%3Dx%5Ex&oq=y%3Dx%5Ex

$\displaystyle <br />\lim_{x \rightarrow 0+} x^x=1<br />$

Anonymous

Anonymous Mon 30 Mar 2020 03:56:06 No.11514106 Report

Quoted By: >>11514115 >>11514121 >>11514399

>>11514089
>also undefined, but x^x has a nice limit so that gets used
It's exactly the opposite. 0^0 is well-defined, despite limits not being helpful.

>>11514096
No, this is not what being indeterminate means.

Anonymous

Anonymous Mon 30 Mar 2020 03:58:04 No.11514115 Report

Quoted By: >>11514126

>>11514106
>0^0 is well-defined
[citation needed]

Anonymous

Anonymous Mon 30 Mar 2020 04:01:03 No.11514121 Report

Quoted By: >>11514126

>>11514106
Since OP is gay and doesn't want "fancy math degree mumbo jumbo" I thought he couldn't handle the real definition of indeterminate form.

Anonymous

Anonymous Mon 30 Mar 2020 04:03:08 No.11514126 Report

Quoted By: >>11514275

>>11514121
That's fair.

>>11514115
What's that going to accomplish? I can cite you a source like https://www.askamathematician.com/2010/12/q-what-does-00-zero-raised-to-the-zeroth-power-equal-why-do-mathematicians-and-high-school-teachers-disagree/ , but you will inevitably be able to find a different source that disagrees. So sources are not going to be able to settle this.

Anonymous

Anonymous Mon 30 Mar 2020 04:29:16 No.11514184 Report

Quoted By: >>11514296

>>11513438
>0^1=0
>0^-1=infinity
>0^0=1
What kind of step function is this? And why does it intercept 1 on its way to infinity?

Anonymous

Anonymous Mon 30 Mar 2020 05:32:46 No.11514275 Report

Quoted By:

>>11514126
>What's that going to accomplish?
un-graduated anon detected

Anonymous

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Anonymous Mon 30 Mar 2020 05:43:14 No.11514296 Report

Quoted By: >>11514300

>>11514184
>>11513438
Wait a minute wait a minute, if $0^{(+n)}=0$ , and $0^0=1$ , and $0^{(-n)}=\pm\infty$ , then what is $0^{(\pm\infty)}$ ??

Anonymous

Anonymous Mon 30 Mar 2020 05:46:47 No.11514300 Report

Quoted By: >>11514310 >>11514405

>>11514296
all are undefined

Anonymous

Anonymous Mon 30 Mar 2020 05:56:26 No.11514310 Report

Quoted By: >>11514324

>>11514300
Don't fuck with me anon, $0^5$ is perfectly well-defined

Anonymous

Anonymous Mon 30 Mar 2020 05:59:26 No.11514315 Report

Quoted By: >>11514318

>>11513438
The following thread will likely enlighten you:
>>1511146

Anonymous

Anonymous Mon 30 Mar 2020 06:00:29 No.11514318 Report

Quoted By:

>>11514315
meant to link this thread:
>>11511146

Anonymous

Anonymous Mon 30 Mar 2020 06:05:34 No.11514324 Report

Quoted By: >>11514332

>>11514310
0^5 = 0^7/0^2
ok anon

Anonymous

Anonymous Mon 30 Mar 2020 06:11:57 No.11514332 Report

Quoted By:

>>11514324
Don't try anything funny on me you faggot, I'm onto you

Anonymous

Anonymous Mon 30 Mar 2020 06:38:00 No.11514373 Report

Quoted By: >>11514404 >>11515469

>>11514096
but is he wrong?

0^0 = 0^1 * 0^-1 = 0 * 1/0 = 0 * infinity = finite ~=1

1 / infinity = 0
1 / 0 = infiniity

Anonymous

Anonymous Mon 30 Mar 2020 07:00:33 No.11514395 Report

Quoted By: >>11514403

>>11513438
>why shouldn't 0^0=0?
There is exactly 1 function that maps a set of 0 elements to a set of 0-elements, hence 0^0 = 1.
>>11513496
Nope, 0^0 = 1.
>0^x = 0
That's not even true for negative x. On the other hand,
>x^0 = 1
Everywhere.

Anonymous

Anonymous Mon 30 Mar 2020 07:03:37 No.11514399 Report

Quoted By: >>11514401 >>11514584

>>11514089
It's not undefined, it's equal to 1.
>>11514106
There's even a good argument from limits for why 0^0 = 1, independent of the discrete/combinatorial reasons why it's true.

Namely, if $f$ and $g$ are analytic and $\lim_{t \rightarrow 0^+} f(t) = \lim_{t \rightarrow 0^+} g(t) = 0$ , then $\lim_{t \rightarrow 0} f(t)^{g(t)} = 1$ .

Anonymous

Anonymous Mon 30 Mar 2020 07:04:39 No.11514401 Report

Quoted By:

>>11514399
(cont.) There's also an even more general condition for the limit always being 1, but it's a more complicated one.

Anonymous

Anonymous Mon 30 Mar 2020 07:05:16 No.11514403 Report

Quoted By: >>11514407

>>11514395
>>x^0 = 1
>Everywhere.
not at x=0

Anonymous

Anonymous Mon 30 Mar 2020 07:05:42 No.11514404 Report

Quoted By: >>11514411

>>11514373
No, you just proved him wrong. There’s no reason “finite” should be 1, it can be anything, which is why 0^0 is undefined, not 0 or 1.

Anonymous

Anonymous Mon 30 Mar 2020 07:05:57 No.11514405 Report

Quoted By:

>>11514300
No, $0^0 = 1$ .

Anonymous

Anonymous Mon 30 Mar 2020 07:06:58 No.11514407 Report

Quoted By:

>>11514403
Yes, including at x=0, because 0^0 = 1.

Anonymous

Anonymous Mon 30 Mar 2020 07:08:00 No.11514411 Report

Quoted By:

>>11514404
No, 0^0 = 1 because there is 1 function from a 0-element set to a 0-element set, and because x^0 for any x is the empty product and therefore the identity element for multiplication, which is 1.

Anonymous

Anonymous Mon 30 Mar 2020 07:08:31 No.11514412 Report

Quoted By: >>11514421

>>11513438
0/0=?
What number (?) multiplied by 0 gives us 0? Absolutely any number, which is why it is undefined.
As to 0^0=1, mathematicians need it to be 1, instead of proving it; maybe 0^0 should never have been a thing to begin with.

Anonymous

Anonymous Mon 30 Mar 2020 07:16:49 No.11514421 Report

Quoted By: >>11514424 >>11515568 >>11516283

>>11514412
$0^0$ for a wide range of reasons:
- There is exactly 1 function from a set of 0 elements to a set of 0 elements.
- $x^0$ is an empty product and therefore the identity element of the monoid formed by the natural numbers under multiplication, namely 1.
- For any analytic functions $f, g$ , $f(t), g(t) \rightarrow 0$ as $t \rightarrow 0^+$ implies $f(t)^{g(t)} \rightarrow 1$ .
- It yields the right expression for a ton of combinatorial identities such as the binomial theorem, without having to create ugly special cases.
- $x^0$ is the identity element of the polynomial ring, whose evaluation homomorphism yields 1, so $x^0=1$ for all specializations of $x$ , be they natural, rational, real complex, or whatnot.
And so on.

Anonymous

Anonymous Mon 30 Mar 2020 07:17:51 No.11514424 Report

Quoted By:

>>11514421
* $0^0=1$ for a wide range of reasons

Anonymous

Anonymous Mon 30 Mar 2020 08:03:41 No.11514584 Report

Quoted By: >>11514622

>>11514399
>Namely, if $f$ and $g$ are analytic and $\lim_{t \rightarrow 0^+} f(t) = \lim_{t \rightarrow 0^+} g(t) = 0$ , then $\lim_{t \rightarrow 0} f(t)^{g(t)} = 1$ .
No it's not. If $f(t) = 0$ and $g(t) = t$ , then $\lim_{t \rightarrow 0^+} f(t)^{g(t)} = 0$ .

Anonymous

Anonymous Mon 30 Mar 2020 08:31:06 No.11514622 Report

Quoted By: >>11514637 >>11515579

>>11514584
Sorry, forgot that $f$ cannot be the everywhere zero function.

Anonymous

Anonymous Mon 30 Mar 2020 08:38:33 No.11514637 Report

Quoted By: >>11514668 >>11514674

>>11514622
Huh, I did not know that. Interesting.

Anonymous

Anonymous Mon 30 Mar 2020 08:48:45 No.11514668 Report

Quoted By:

>>11514637
There are also independent or more general conditions (i.e. beyond the analyticity condition). See https://www.jstor.org/stable/3595845.

Anonymous

Anonymous Mon 30 Mar 2020 08:54:27 No.11514674 Report

Quoted By:

Anonymous

Anonymous Mon 30 Mar 2020 09:08:16 No.11514692 Report

Quoted By: >>11514705

>>11513438
>I want the pure natural logic seen in the real world explained in a way a child could understand.
it leads to nonsense
1 = 0/0 = (0*2)/0 = 2*0/0 = 2

Anonymous

Anonymous Mon 30 Mar 2020 09:16:40 No.11514705 Report

Quoted By:

>>11514692
Pretty much this, the question is, do you want
>muh x/x is 1!!!!11!!
or
>*brain whirs*
>adjusts glasses
>fixes tie
>hmmm, the commutative and associative properties of multiplication are quite useful, if I do say so myself
Sorry, but you can only pick one.

Anonymous

Anonymous Mon 30 Mar 2020 10:05:55 No.11514763 Report

Quoted By:

>>11513438
Some children (OP possibly not included) may understand that 0/0 is meaningless: "if we have no cake and we cut it into pieces to divide evenly between no children, then how much cake will each child get?"

Anonymous

Anonymous Mon 30 Mar 2020 10:29:49 No.11514790 Report

Quoted By:

>>11513513
/thread

Anonymous

Anonymous Mon 30 Mar 2020 16:55:32 No.11515453 Report

Quoted By:

>>11513438
>0×0=0

I don't know about that one chief.

Anonymous

Anonymous Mon 30 Mar 2020 17:02:39 No.11515469 Report

Quoted By: >>11515570

>>11514373
inf isn't a number so stuff like
1 / infinity = 0
1 / 0 = infiniity
doesn't work, 0*inf is undefined

Anonymous

Anonymous Mon 30 Mar 2020 17:47:13 No.11515568 Report

Quoted By: >>11515579 >>11515950 >>11515955 >>11515965 >>11516094

>>11514421
> There is exactly 1 function from a set of 0 elements to a set of 0 elements.
Fine to use this if you're defining exponentiation strictly over the finite cardinals.
> $x^0$ is an empty product and therefore the identity element of the monoid formed by the natural numbers under multiplication, namely 1.
It's only defined that way for convenience, and generally not for x=0. Also, the multiplicative monoid of natural numbers? Maybe the most useless object in algebra?
>For any analytic functions $f,g, f(t),g(t)\to 0 as t\to 0^+ implies f(t)^{g(t)}\to 1.$
$f(t)=0, g(t)=t.$
>It yields the right expression for a ton of combinatorial identities such as the binomial theorem, without having to create ugly special cases.
What? Where does the binomial theorem contain exponentiation?
> $x^0$ is the identity element of the polynomial ring, whose evaluation homomorphism yields 1, so $x^0=1$ for all specializations of x, be they natural, rational, real complex, or whatnot.
And so on.
x is an indeterminate in the polynomial ring, it is not a variable over any number system.

Okay, so I agree that it works nicely if we define exponentiation over the finite cardinals, but that's about it.
Incidentally, I'm happy to be proven wrong, but I think I provided fairly strong points.

Anonymous

Anonymous Mon 30 Mar 2020 17:48:29 No.11515570 Report

Quoted By: >>11515590

>>11515469
>1 / infinity = 0
>1 / 0 = infiniity
>doesn't work,
They absolutely do work, in the extended reals.

Anonymous

Anonymous Mon 30 Mar 2020 17:51:49 No.11515579 Report

Quoted By: >>11516069

>>11515568
Sorry, I didn't see this >>11514622.
Regardless, doesn't that restriction weaken the whole idea? Because it means that there is then a discontinuity?

Anonymous

Anonymous Mon 30 Mar 2020 17:55:31 No.11515590 Report

Quoted By: >>11515612

>>11515570
if my aunt had a dick, she'd be my uncle

Anonymous

Anonymous Mon 30 Mar 2020 18:03:55 No.11515612 Report

Quoted By:

>>11515590
We can identify the 1-sphere with $\mathbb{R}\cup\{\infty\}$ and the complex projective line with $\mathbb{C}\cup\{\infty\},$ and define the usual operations on these spaces, and it makes sense to do so.

Anonymous

Anonymous Mon 30 Mar 2020 20:08:05 No.11515950 Report

Quoted By: >>11516010

>>11515568
>strictly over the finite cardinals
It works for infinite cardinals too. It's also true for ordinals too.
https://en.wikipedia.org/wiki/Ordinal_arithmetic#Exponentiation

Anonymous

Anonymous Mon 30 Mar 2020 20:10:17 No.11515955 Report

Quoted By: >>11515958 >>11516010

>>11515568
>What? Where does the binomial theorem contain exponentiation?
$(1+x)^n = \sum_{k=0}^n \binom{n,k} x^k$ is not valid unless $0^0=1$ .

Anonymous

Anonymous Mon 30 Mar 2020 20:11:18 No.11515958 Report

Quoted By:

>>11515955
* $(1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k$ is not valid unless $0^0=1$
for $x=0$

Anonymous

Anonymous Mon 30 Mar 2020 20:12:53 No.11515965 Report

Quoted By: >>11516010

>>11515568
>x is an indeterminate in the polynomial ring, it is not a variable over any number system.
Evaluation homomorphism.
See https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero#Polynomials_and_power_series
https://en.wikipedia.org/wiki/Polynomial_ring#Polynomial_evaluation
https://proofwiki.org/wiki/Definition:Polynomial_Evaluation_Homomorphism

Anonymous

Anonymous Mon 30 Mar 2020 20:25:58 No.11516010 Report

Quoted By:

>>11515950
>>11515955
>>11515965
Thanks so much! It is very instructive!

Anonymous

Anonymous Mon 30 Mar 2020 20:47:33 No.11516069 Report

Quoted By: >>11516074

>>11515579
There is always necessarily a discontinuity (if any type of path is allowed). The question is whether one can still assign a natural value at the discontinuity.

Anonymous

Anonymous Mon 30 Mar 2020 20:48:48 No.11516074 Report

Quoted By:

>>11516069
...either by a different set of arguments (e.g. the combinatorial arguments) or by restricting to some natural class of paths.

Anonymous

Anonymous Mon 30 Mar 2020 20:54:08 No.11516094 Report

Quoted By:

>>11515568
>so I agree that it works nicely if we define exponentiation over the finite cardinals
The thing is, 0 *is* a finite cardinal. Since any definition in the continuous setting ought to agree with the values that were determined in the discrete setting (since it’s supposed to *extend* it), it follows that in the continuous setting it also “ought” to be true that 0^0 = 1.

Anonymous

Anonymous Mon 30 Mar 2020 21:57:36 No.11516283 Report

Quoted By: >>11516305

>>11514421
If $0^0 = 1$ and $0^0 = 0/0$ (as $x^0 = x/x$
Therefore, $0/0 = 1$

Anonymous

Anonymous Mon 30 Mar 2020 22:03:46 No.11516305 Report

Quoted By:

>>11516283
x^0 is not defined as x/x; it is defined as being equal to 1. 0/0 is still an indeterminate form.

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