0^0=1 Fun

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[katsura_grey]

Now I can say 0^0 = 1 with confidence! Much appreciated✌️

[bilaalahmed]

If we had to cancel the 0s, why didn't we just cancel the zeroes from the first place when we had 0/0,

[kruksog]

What about looking at the left and right limits of the functions 0^x and x^0 around the point x=0? For x^0, coming from the right side, the limit appears to be 1 (i.e. 0^0=1) but coming from the left side of 0^x, it looks like the limit is 0 (i.e. 0^0 = 0).

Im trying to parrot the argument I got as to why 0^0 is undefined from calculus a long time ago, so please forgive me if I didn't get it all right, but I think you can see what I'm getting at.

I don't have a dog in the fight. Just trying to understand more fully. I was always taught it was undefined when I got my math degree.

[IamExeller]

In the last, we have a hidden division by zero when we eliminate zeros out.

[thedoublehelix]

Actually the logic here is flawed. I do believe 0^0 = 1, but you can leave it undefined. 0^0 is more of a definition than a theorem.

[allozovsky]

Just for the reference (not supposed to be used as an argument):

C/C++*
printf("%g", pow(0, 0));
cout << pow(0, 0) << endl;
Output: 1

Python*
>>> [0**0, pow(0, 0)]
Output: [1, 1]

Haskell*
> [0^0, 0^^0, 0**0]
Output: [1.0, 1.0, 1.0]

Java*
System.out.println(Math.pow(0, 0));
Output: 1.0

And the same goes for Mathcad, Matlab, Maple and so on.

[BloodyDuke]

With the last thesis, it would also mean 0/0=1 which is something I would not agree to 😅 so I guess I will stick to 'undefined'.

[ShorTBreak]

a to the b is: the number of application from a set of cardinal b to a set of cardinal a
Then 0 to the 0 is the number of application from the empty set to the empty set, there is only 1 application from empty set to empty set which is the identity application

[attackhelicopteriscool]

we got brainrot mathematic before gta 6 😂

[linebreaker]

I actually found a total solution to division by zero. But I am not sure how will I tell this to world. If I share it with people in a YouTube video would it be a loss in means of Copyrights? Because I do not want any other people to claim copyright on my invention and I actually do not want to claim copyright on people for this math invention because I think it is common wealth. Please give me legal advice.

[JPK]

-Guys this video is a joke-

-The whole point of 0^0 being undefined in analysis is that you can show it equals anything. 0, 1, whatever. That's what being undefined means to keep the system consistent.-

OK so I'm dead wrong about the reason 0^0 is undefined. However, the following statement still holds: The simple fact is that undefinedness is not a property that holds across an equals sign. No one argues that 0/0 is undefined - but *it is reasonable to expect that any definition of this would have the property that all sequences approaching 0/0 should have the limiting value we have defined for 0/0. If we define 0/0=1 then, sure, we have lim x->0 x/x = 1 like we want, but also lim x->0 2x/x=2 and this also is a sequence approaching 0/0. Therefore we can't associate 0/0 with a value, making it meaningful to leave it undefined. In this vein, a proof showing 0/0 converges to 1 or something doesn't mean 1 is undefined like 0/0 is. The undefinedness meta property doesn't transfer across an equals sign, and we shouldn't be using it as an argument against leaving 0^0 undefined.*

-Saying 0^0=1 means that 1 = 0^0 = lim x->0 0^x = 0 so 1=0. The logic is sound, and so the conclusion being false implies the premise is as well. Therefore 0^0≠1.-

Edit: ok this last paragraph uses a seriously flawed argument. See the replies for a better discussion.

[renefernandez]

To say that 0^2=0^(3-1) is in fact incorrect.
Lets assume whe have a certain number x at power a (x^a) and we have two certain numbers b and c, such that a=b-c. Then we can say that x^a= (x^a)*1 and we can make 1= (x^c)/(x^c), if and only if x≠0, then because a=b-c we can say that b=a+c
And we can say that

Now to the why x≠0, using the earlier proof to go from 0^2=0^(3-1) we would need to multiply by 1, but here is where we cannot say that 1=0/0 because 0/0 is undefined, therefore we cant reach the right side so 0^2 ≠ 0^(3-1)

[kairostimeYT]

Limits exist for a reason. You can define several functions such that your functions approach this value. You'll clearly see that the value is different for different functions. So undefined limits. Also the argument in the video, though it's just a joke, is trash lol

[nexa]

law of exponents don't work with base 0.

[korayacar]

This only makes sense in a discrete context. In a continuous context, however, 0^0 is undefined, as the repeated multiplication definition stops working, and for the new exponential definition, 0^0 is a limiting case for a limiting case, which means its definition involves a bivariable limit.

Here's the simplest case of such a bivariable limit that unequivocally equals 0^0, and yet simply doesn't exist:

lim(a, b-->0) a^b = 0^0
= lim(a-->0)lim(b-->0) a^b = 1
= lim(b-->0)lim(a-->0) a^b = 0

This is a contradiction to the well-definition of 0^0 in the positive real exponential, and as such, 0^0 is an indeterminate form, just like 0/0.

Here's also a rigorous argument for why 0^0=1 for discrete exponentiation:

A^B can be thought of calling a repeated multiplication function with B arguments, all of which are A. Given that, here's how to think about 0^0 by continuing the pattern that, in the continuous limit, would have yielded 0:

0^3 = *(0, 0, 0) := 0*0*0 = 0
0^2 = *(0, 0) := 0*0 = 0
0^1 = *(0)

Here, the multiplication function takes fewer arguments than it's defined for, so a consistent definition is therefore needed. However, since the multiplication function can be thought of scaling one argument by the other, a simple extension of this is that *(A) = A for all A. This means that *(0)=0 as well.

How about 0^0? Here, the multiplication function doesn't even take an argument. For all A, A^0=*(), a consrant, for which a definition is now required. Since 1^A=1 not only for all positive integer A, but for even all complex A, it's safe to say that 0^0=*()=1^0=1. QED.

[kylefox]

They started with 0^2 and ended back at 0^2 lol

[shohamsen]

0^2=0 just from the property of multiplication. Your argument that 0^2=0^(3-1), while correct does not translate to 0^3/0, cause that is an undefined operation. Also u cannot cancel zero. If 0/0=x, then we can rewrite it as 0*x=0. If i cancel 0 from both sides, I get x=1. Anyway, there are better arguments to show why 0^0=1 that makes sense for discrete math and algebra.

[heatedturtle]

Objection. My calculator says Overflow error

[zgnytpg]

Please, one tell me, is this proof true mathematically?

[danteeightsix]

0 isn't a number. It's a placeholder or nothing.