🧵 The debate about 0^0 makes no sense
Anonymous at Sun, 17 Nov 2024 21:03:25 UTC No. 16479730
Mathematicians post a video to youtube twice a year about how 0^0 is debated and controversial but it's literally so simple.
The function x^y is always 0 along the x-axis and always 1 along the y-axis.
When you zoom in around 0,0 the plotted surface gets increasingly fucked and disjoint, because 0,0 is a singularity. Why do so many people fall for the clickbait?
Anonymous at Sun, 17 Nov 2024 21:22:09 UTC No. 16479747
>>16479730
use case?
Anonymous at Sun, 17 Nov 2024 22:25:03 UTC No. 16479842
[math]0^0=1[/math] and anyone who disagrees is a retard
the fact that 0 is your base doesn't matter - it's quite literally the empty product
Anonymous at Sun, 17 Nov 2024 22:45:16 UTC No. 16479885
>>16479842
>i dont do anything
>i somehow get 1 by not doing anything
mathematicians are fucking RETARDED
Anonymous at Sun, 17 Nov 2024 22:51:33 UTC No. 16479896
>>16479747
there is none.
never in any real world scenario will you need to raise zero to the zeroth power nor will it matter what the "real" answer is. its just mathematicians pontificating about angles on the head of a pin.
Anonymous at Sun, 17 Nov 2024 23:04:06 UTC No. 16479918
>>16479842
This, nothing is being multiplied. In the expression x^0, x doesn't participate.
Anonymous at Sun, 17 Nov 2024 23:12:53 UTC No. 16479930
>>16479885
Not multiplying a number by anything is the same as multiplying it by 1. If I've got the number 5 and I just don't multiply it by another number, it doesn't suddenly turn into zero.
Anonymous at Sun, 17 Nov 2024 23:14:47 UTC No. 16479932
>>16479930
>If I've got the number 5 and I just don't multiply it by another number, it doesn't suddenly turn into zero.
ok then explain why it suddenly turns into 1.
Anonymous at Sun, 17 Nov 2024 23:17:59 UTC No. 16479937
>>16479932
NTA but 5 doesn't turn into 1, there's no 5. Nothing "turns into" anything
Anonymous at Sun, 17 Nov 2024 23:22:14 UTC No. 16479940
>>16479932
It doesn't, rather the absence of multiplication is the same as 1, just like the absence of addition is the same as 0.
If I start with nothing and I don't multiply it by anything, that's still nothing because 0×1 is zero. But 0^0 isn't necessarily starting with nothing. You can define the exponentiation to include an implicit 1 without contradicting anything about how it works in any other situation. And then using that definition, 0^0 can be interpreted as 1×0^0, or "What do you get when you take 1 and multiply it by zero zero times? Which of course is still 1.
Anonymous at Sun, 17 Nov 2024 23:36:30 UTC No. 16479957
>>16479940
but 1 multiplied by 0 = 0...
Anonymous at Sun, 17 Nov 2024 23:43:55 UTC No. 16479963
>>16479957
>1 multiplied by 0 = 0
"x mutliplied by 0" means you add x 0 times. Meaning that you don't add anything.
"x exponentiated by 0" means you multiply x 0 times. Meaning that you don't multiply anything.
Anonymous at Mon, 18 Nov 2024 02:03:04 UTC No. 16480099
>>16479932
I start off with 2.
I don't multiply it by 5.
What do I end up with?
DoctorGreen !DRgReeNusk at Mon, 18 Nov 2024 11:14:48 UTC No. 16480465
>>16479730
>0^0
this is just the big bang. why are all of you retarded?
Anonymous at Mon, 18 Nov 2024 11:31:51 UTC No. 16480483
>>16479730
>You can't divide by zero! It's undefined, math doesn't allow it!
>Black holes have real singularities
Makes you think
Anonymous at Mon, 18 Nov 2024 11:39:02 UTC No. 16480492
>>16479730
The exponential is only properly defined for something like 0^0 via analytic continuation, ie 0^0 = e^(0*ln0). The RHS is clearly undefined, but its right-sided limit evaluates to 1.
So 0^0 is undefined, but lim_a->0^+ a^a= 1.
Anonymous at Mon, 18 Nov 2024 12:07:22 UTC No. 16480538
>>16480492
The exponential is only properly defined for something like 0^1 via analytic continuation, ie 0^1 = e^(1*ln0). The RHS is clearly undefined, but its right-sided limit evaluates to 0.
So 0^1 is undefined, but lim_a->0^+ a^(1+a)= 0.
Anonymous at Mon, 18 Nov 2024 12:09:42 UTC No. 16480541
>>16479963
So if you don't have anything and you don't multiply it by anything, you suddenly have 100% of a thing?
Anonymous at Mon, 18 Nov 2024 12:16:05 UTC No. 16480548
>>16480465
No, you are thinking of 0!.
Anonymous at Mon, 18 Nov 2024 12:37:54 UTC No. 16480578
>>16480541
if you have 3 and multiply it by nothing you have 3
if you have 3 and multiply it by 1 you have 3
multiplication by nothing is the same thing as multiplication by 1 qed
Anonymous at Mon, 18 Nov 2024 12:47:46 UTC No. 16480600
>>16480578
>multiplication by nothing is the same thing as multiplication by 1 qed
No, nothing is the additive element which is quantized as 0 and 0 multiplied by anything is always 0.
Adding nothing is the same thing as multiplying by 1, but not multiplying by nothing, multiplying by nothing is like adding the opposite number.
Anonymous at Mon, 18 Nov 2024 12:48:23 UTC No. 16480603
>>16480578
Zero is not the same thing as nothing
Anonymous at Mon, 18 Nov 2024 12:48:23 UTC No. 16480604
I'm following the lead you clue directly at now.
Anonymous at Mon, 18 Nov 2024 13:07:48 UTC No. 16480637
>>16480600
see >>16480603
"nothing" in the context of an operation means the identity of that operation. You don't delete a square by "doing nothing" to it
Anonymous at Mon, 18 Nov 2024 13:14:14 UTC No. 16480645
>>16480603
0 is the numerical equivalent of nothing since they are attempts to logically describe an additive identity.
Anonymous at Mon, 18 Nov 2024 13:15:39 UTC No. 16480646
>>16480637
You nullify a square by multiplying it by nothing since that is a multiplicative annihilator.
Anonymous at Mon, 18 Nov 2024 13:16:36 UTC No. 16480649
>>16480600
>have 3
>don't multiply it by anything
>suddenly don't have 3
explain
Anonymous at Mon, 18 Nov 2024 13:39:08 UTC No. 16480674
>>16480541
You don't "suddenly" have anything. The background of multiplication is 1. It's always there. If you don't put anything else on the background, all you have is the same thing you always had, 1.
Anonymous at Mon, 18 Nov 2024 13:39:38 UTC No. 16480675
>>16479730
>Scp-1471 is a big tiddy goth wolf temptress. D class and foundation staff alike have commented on it's lustful figure, several attempts were made to solicit intercourse by personnel prior to scp-1471's relocation to site 11 and revision of containment protocols. Relocation was prompted by event 1471-A.
Event 1471-A log.
This log transcribes events recorded by several CCTV cameras adjacent to and inside scp-1471's cell.
Dr ------ can be seen walking down the corridor towards scp-1471's cell, Dr ------ exchanges some words between the single guard on duty, Sgt ----, before passing him a brown envelope. Sgt ---- allows unlocks scp-1471's cell and permits Dr ------ to enter before closing the door behind him resuming his duties. Dr ------ immediately begins to communicate with scp-1471, the conversation suggests Dr ------ used prior authorized interactions with scp-1471 to arrange this meeting. Dr ------ makes several crude remarks on the attractiveness of scp-1471, which plays off these and attempts to ply Dr ------ for security information relevant to egressing the facility. A series of escalations initiated by Dr ------ result in both parties finding themselves in a state of undress. From this point onwards scp-1471's attempts to attain more information become more forceful, with questions overtly specific to methods of escaping the facility raised. Dr ------ and scp-1471 initiate coitous once Dr ------ reveals his access code to Entrance Zone gate.
Little of academic relevance is discussed during the following 3 minutes. This end period is marked by Dr ------'s exclamations of increasingly powerful contractions by scp-1471. It is moments after this that a sector wide lockdown is declared after Dr ------'s howls of confused ecstasy and agony are overheard. It is hypothesized that Dr ------'s genitals were obliterated around this time. MTF agents report to scp-1471's cell within 2 minutes, and are able to separate Dr ------ and scp-1471.
End log.
Anonymous at Mon, 18 Nov 2024 13:42:35 UTC No. 16480677
>>16480649
NTA but if you "have 3" and don't multiply it by anything, you still have 1 copy of 3. That's not 3^0, it's 3^1.
Anonymous at Mon, 18 Nov 2024 16:51:00 UTC No. 16480862
>>16480538
This is different because 1*ln0 is -infinity. The usual limit techniques for natural numbers still work. Whereas 0^0 is something that requires L’Hospital’s rule, ie a notion of differentiation. So it’s meaningless to talk about 0^0 in the context of non-differentiable functions.
Anonymous at Mon, 18 Nov 2024 17:38:32 UTC No. 16480932
>>16480677
it's [math]3^1 \times x^0[/math] for arbitrary[math]x[/math]
Anonymous at Mon, 18 Nov 2024 17:57:31 UTC No. 16480959
>>16480932
Sure, you could describe that post that way, too.
What anon seems to be missing is that x is arbitrary for any x^0 because no x is actually involved anywhere in what that operation describes.
Anonymous at Tue, 19 Nov 2024 05:16:20 UTC No. 16481764
>>16480483
Black holes don't have real singularities you idiotic pop soience consoomer. They encompass finite volumes with finite mass.
Anonymous at Tue, 19 Nov 2024 06:00:37 UTC No. 16481784
it's definitely 1 but I actually forget why as it's been a long time since high school. But, negative exponents are between 0 and 1, an exponent of 1 is the number, and so an exponent of 0 being 1 makes sense.
Anonymous at Tue, 19 Nov 2024 06:14:20 UTC No. 16481790
Anonymous at Tue, 19 Nov 2024 07:53:52 UTC No. 16481872
>>16479842
Bullshit. (You need to be 18+ to post on 4chan. Are you?) [math]x^0=x^{(1-1)}=\frac{x}{x}[/ma
Anonymous at Tue, 19 Nov 2024 08:42:18 UTC No. 16481904
>>16480674
>The background of multiplication is 1. It's always there.
Then why isn't 0*x=1?
>If you don't put anything else on the background,
They did, though, they put 0 in the background and the foreground.
Anonymous at Tue, 19 Nov 2024 08:43:58 UTC No. 16481906
>>16480959
Are you blind or just being disingenuous? x is set to 0 in the 0^0 operation being described.
Anonymous at Tue, 19 Nov 2024 09:38:35 UTC No. 16481957
my 0^0 identifies as female
Anonymous at Tue, 19 Nov 2024 09:40:59 UTC No. 16481962
If 0/0 = 1 then 0 is a prime number
Anonymous at Tue, 19 Nov 2024 12:47:35 UTC No. 16482182
>>16481906
It's set to 0 and, like any other x in the operation x^0, doesn't participate anywhere in what the operation describes.
Anonymous at Tue, 19 Nov 2024 12:51:08 UTC No. 16482188
>>16481904
>Then why isn't 0*x=1?
Because the background of addition is 0.
>they put 0 in the background and the foreground
No, they put nothing anywhere. x^0 means that no multiplication is happening.
Anonymous at Tue, 19 Nov 2024 13:10:52 UTC No. 16482223
>>16481872
It's common for people who fail at latex to also fail at understanding the difference between continuity and single operations.
Anonymous at Tue, 19 Nov 2024 13:16:56 UTC No. 16482232
>>16479963
so if you multiply it 0 times doesn't x stay the same? If you multiply 100 dollars 0 times you have $100
Anonymous at Tue, 19 Nov 2024 13:20:31 UTC No. 16482237
>>16482188
No addition is in that formula, though, its a product.
Anonymous at Tue, 19 Nov 2024 13:22:45 UTC No. 16482243
>>16482232
If you have $100, you've multiplied $100 by 1, not by 0.
Anonymous at Tue, 19 Nov 2024 13:23:45 UTC No. 16482246
>>16482237
That's what a product is, it tells you how many copies of something are added.
Anonymous at Tue, 19 Nov 2024 13:30:52 UTC No. 16482260
>>16482246
But you said no multiplication is happening if you multiply 0 times, so if there is no multiplication, there is no addition either by that logic.
Anonymous at Tue, 19 Nov 2024 13:34:01 UTC No. 16482266
>>16482260
Right, x*0 means no addition is happening. All you ever have is the background 0.
Anonymous at Wed, 20 Nov 2024 06:40:52 UTC No. 16485968
>>16482266
That is the opposite of what was said >>16480674
Anonymous at Wed, 20 Nov 2024 07:06:15 UTC No. 16485981
>>16481790
Division by zero is undefined for real numbers ,so I don’t think this really proves your point.
0 raised to the zero power seems like a matter of convention.
The interpretation of b^0 as an empty product assigns it the value 1.
The combinatorial interpretation of b^0 is the number of 0-tuples of elements from a b-element set; there is exactly one 0-tuple.
The set-theoretic interpretation of b^0 is the number of functions from the empty set to a b-element set; there is exactly one such function, namely, the empty function.
Strikes me as similar to why we define 1 to be not be prime - it screws up things like the fundamental theorem of arithmetic if we don’t.
Anonymous at Wed, 20 Nov 2024 07:58:52 UTC No. 16486005
>>16485981
>interpretations of b^0
How about interpretation of 0^a? For each a>0 0^a=0 so lim (a->0+) 0^a is also equal to 0.
Anonymous at Wed, 20 Nov 2024 08:05:03 UTC No. 16486010
The definition of exponentiation with a natural number exponent is
[eqn]z^n = \begin{cases} 1 & n=0 \\ z \cdot z^{n-1} & n > 0 \end{cases}[/eqn]
So [math]0^0 = 1[/math].
Anonymous at Wed, 20 Nov 2024 08:06:32 UTC No. 16486011
>>16486010
What if zero is not a natural number?
Anonymous at Wed, 20 Nov 2024 08:08:40 UTC No. 16486012
>>16486011
Ex falso quodlibet
Anonymous at Wed, 20 Nov 2024 09:22:57 UTC No. 16486047
>>16486020
>pseud
>lmao
You've got to get back to /pol/, stray election tourist.
Anonymous at Wed, 20 Nov 2024 10:37:01 UTC No. 16486101
>for any x, 0^x = 0
>0^0 = 1 in most programming languages
programmers are the pseuds
Anonymous at Wed, 20 Nov 2024 10:51:59 UTC No. 16486121
>>16486101
>for any x, 0^x = 0
Consider x=-1 or any other non-positive x.
Anonymous at Wed, 20 Nov 2024 10:54:22 UTC No. 16486125
>>16486121
yes I have considered that these cases are undefined and experts agree with me
Anonymous at Wed, 20 Nov 2024 14:19:24 UTC No. 16486384
>>16486010
[eqn]
z^{n} = \left\{ \begin{array}{cl}
z & : \ n = 1 \\
\frac{z^{n+1}}{z} & : \ n < 1
\end{array} \right.
[/eqn]
Anonymous at Wed, 20 Nov 2024 14:23:49 UTC No. 16486389
>>16486005
The limit existing from the right does not guarantee the function equals the value of the limit.
Consider a piece wise function that maps x to x squared if x does not equal 1 and maps x to a billion if x equals 1. Here the limit as x approaches 1 exists and equals 1 ,yet the function itself does not equal 1 at x=1.
I reiterate that 0 raised to 0 is defined to be 1 by convention. For example, the power rule of differentiation is only valid for n=1 and x=0 if 0 raised to the zero equals 1.
Anonymous at Wed, 20 Nov 2024 14:25:48 UTC No. 16486391
>>16485968
It's the same.
For ^0
>You don't "suddenly" have anything. The background of multiplication is 1. It's always there. If you don't put anything else on the background, all you have is the same thing you always had, 1.
For *0
>You don't "suddenly" have anything. The background of addition is 0. It's always there. If you don't put anything else on the background, all you have is the same thing you always had, 0.
Anonymous at Wed, 20 Nov 2024 22:01:08 UTC No. 16487052
Anonymous at Fri, 22 Nov 2024 06:13:45 UTC No. 16488871
>>16486391
No, its not the same you were saying, you just said multiplication and addition are the same thing. >>16482246, so why would they have different "backgrounds"?
Anonymous at Fri, 22 Nov 2024 07:50:44 UTC No. 16488936
>>16480492
>its right-sided limit evaluates to 1
since when?!?
Anonymous at Fri, 22 Nov 2024 14:58:01 UTC No. 16489207
>>16488871
You're just being a brat lol.
Anonymous at Sun, 24 Nov 2024 06:12:09 UTC No. 16491347
>>16480492
>>16488936
the limit of x^x includes x^x sooo...
Anonymous at Mon, 25 Nov 2024 17:40:37 UTC No. 16493040
>>16491347
the limit of x^x is not a set, it doesn't include anything.
Anonymous at Mon, 25 Nov 2024 17:53:57 UTC No. 16493061
>>16479730
What debate is there? It's an indeterminate form. The value of this expression depends on how you got there, i.e. taking a limit.
I feel like this "debate" only exists on the wrong side of the Dunning-Kruger curve. Maybe don't comment on things you don't understand?
Anonymous at Mon, 25 Nov 2024 18:04:26 UTC No. 16493076
>>16493061
>The value of this expression depends on how you got there, i.e. taking a limit.
No, the value of a generalized function x^y as x and y both approach 0 depends on that.
The value of the simple expression 0^0 is simply and always 1.
Anonymous at Tue, 26 Nov 2024 23:01:12 UTC No. 16494601
>>16486384
bruh this shii don't make no damn sense
🗑️ Anonymous at Wed, 27 Nov 2024 22:04:45 UTC No. 16495527
0^0 = thrembo you gorilla niggers take your fucking meds
Anonymous at Thu, 28 Nov 2024 00:20:48 UTC No. 16495658
>>16493061
What does one gain from saying "Dunning-Kruger"? Like legitimately what value do you add to yourself, this thread, or the world?
I'm sure the rest of your post is just fine and informative but what is the point of the last half?
Is it to make you feel better? Make you feel superior? What is wrong with people commenting incorrectly on things? Is it not a net benefit for figuring out the correct things? Does it not make others who would not have spoken, speak the correct things? Is it not an exercise that helps people figure the right things out? Should people never learn an instrument because they can't play it perfectly?
What are these rules that you judge others on saying they shouldn't do this and that like you're some moral authority? You have never spoken about something incorrectly i guess.
Do you not want people to not explore ideas ever? Play around with things in their head and share what they find? Go on and correct them, or not but it's really cringe when people are offended by these things.
Anonymous at Thu, 28 Nov 2024 15:56:52 UTC No. 16496238
>>16479842
It’s undefined.
Anonymous at Thu, 28 Nov 2024 15:57:58 UTC No. 16496240
>>16479963
>"x exponentiated by 0" means you multiply x 0 times. Meaning that you don't multiply anything.
Then 00 must equal 0, according to your logic.
Anonymous at Thu, 28 Nov 2024 16:04:11 UTC No. 16496251
>>16496240
>Then 00 must equal 0
Correct, 00 (or any x0) means that no addition is happening, so all you ever have is the background 0.
Anonymous at Thu, 28 Nov 2024 16:05:22 UTC No. 16496255
>>16496238
False.
Anonymous at Thu, 28 Nov 2024 16:30:10 UTC No. 16496280
>>16479932
[math]anything^0[/math] expands to fuck all, not zero
we are allowed to multiply it by 1 without changing anything
1 times fuck all is just 1 - remember, the fuck all is not zero
Anonymous at Thu, 28 Nov 2024 16:51:11 UTC No. 16496300
>>16496280
By PEMDAS, [math] anything^{0} [/math] expands to [math] anythin [/math]
Anonymous at Thu, 28 Nov 2024 19:04:10 UTC No. 16496448
>>16493061
Of course the fellow who brings up Dunning-Kruger is the one falling victim to it, kek.
Anonymous at Fri, 29 Nov 2024 05:25:47 UTC No. 16496890
>>16496251
Eh, 4chins hates the fact that my keyboard layout allows me to input maths directly. It was supposed to say 0 to the power of 0.
Anonymous at Fri, 29 Nov 2024 13:45:24 UTC No. 16497164
>>16496890
So then [math] 0^{0} [/math] ( or any [math] x^{0} [/math] ) means that no multiplication is happening, so all you ever have is the background 1.
Anonymous at Fri, 29 Nov 2024 14:30:08 UTC No. 16497184
>>16497164
>4
There is no background 1 here, though.
Anonymous at Fri, 29 Nov 2024 15:40:12 UTC No. 16497217
>>16497184
What's this supposed to be?
>4
Anonymous at Sun, 1 Dec 2024 10:51:30 UTC No. 16499107
>>16479842
fpbp
Anonymous at Sun, 1 Dec 2024 11:10:23 UTC No. 16499117
>>16499107
why did the retard bump a page 11 thread?
Anonymous at Wed, 4 Dec 2024 02:13:01 UTC No. 16501727
X/∞=0
X/0=∞
∞*0=X
∞*0=0/0
0/0=X
X can be any real number.
further evidence: a line segment has a finite length X and is made up of an infinite number of points whose length equals 0. this holds true for a line segment of any definite length.
Anonymous at Thu, 5 Dec 2024 21:22:24 UTC No. 16504204
>>16494601
wrong
Anonymous at Fri, 6 Dec 2024 02:49:31 UTC No. 16504448
>>16480483
Math allows it, just not in every algebraic structure
Anonymous at Fri, 6 Dec 2024 02:54:43 UTC No. 16504451
>>16481790
/thread
Anonymous at Fri, 6 Dec 2024 02:56:11 UTC No. 16504454
>>16481962
It’s not, that’s just its continuous extension.
Anonymous at Sat, 7 Dec 2024 00:55:48 UTC No. 16505360
>>16479842
fpbp
Anonymous at Mon, 9 Dec 2024 01:57:45 UTC No. 16507248