Zero to the Zero Equals One #Shorts

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#Shorts

[xinhaiii]

Rare footage of Andy gaslighting his viewers

[mosatsoni]

We need to get Terrence Howard’s opinion on this 😂

[Soumya_]

0^0= 0^(1-1) = 0^1/ 0^1 = 0/0
Now if you think 0/0=1 then I have no problem with 0^0=1

[carultch]

0^0 is indeterminate, because you get conflicting answers depending on exactly how you approach it.

[timbeaumont]

Andy
Could you plot the graph of y=n^n
Where each value of n is a tenth of the previous value.
1
0.1
0.01
0.001
Etc
The closer you get to 0
The closer the result gets to 1

[fiction_enthusiast]

0⁰ is a very controversial question and can have numerous answers.

0^0 = 0^1-1 = 0 * (1/0) = undefined

However on simplifying some may say that 0 * something = 0 or 0/0=1

[CharlesB]

Andy casually starting up the old math war amongst his viewers/comments section. 😂

[westy]

It is indeterminate, what is shown here is not proof in any way.
In e.g. programming it is accepted as 1 for ease of use, in Set theory it is 1 because there is 1 empty set.
In algebra it is undefined because 0^0=0^(1-1)=0^1/0^1=0/0 which is inherently undefined.
The main limitation of such an algebraic definition is that using the same logic, 0^1=0^2/0^1=0/0, as such also undefined, given this it is again indeterminate, because depending on how you approach it, the expression becomes undefined, or defined as 1 or 0.

[greeklighter-countryball]

I want to know btw why 0 comes after solving the equation x^x=1:
x^x=1
Taking the natural log on both sides,
ln(x^x)=ln(1)
Laws of logarithms
x*ln(x)=0
Meaning that either *x=0* or ln(x)=0
So possible outcomes are *0*, 1 why is that?

[lunaropsv]

0^0 = 0^0
[ x^0 =1 ]
[ 0^x = 0 ] putting 0 in x,
Hence, 0 = 1.
Further, 1 = 1/0
Therefore, 1 = 0 = undefined

[lastchance]

So, it's defined as 1 so that anywhere it appears in a calculation, it doesn't negate products. Good explanation! Why didn't my math teacher intuit this?

[element]

The simplest common definition of exponentiation is as follows:
a^b s.t. a∈ℝ & b∈ℕ ≝ a^(b-1)·a s.t. a^1= a
However, we can also use
a^b s.t. a∈ℝ & b∈𝕎‎ ≝ a^(b-1)·a s.t. a^0 = 1
which produces all the same results but extends the definition to include zero. This definition should be the standard for the whole numbers

[templateorman]

The limit of x^x when x approaches 0 is 1. 0^0 is undefined, though in physics it is often interpreted as 1

[aleksanderszarat]

0⁰ = 0 because there are no numbers to multiple then

[DeemIsTaken]

Yes because when you have no candies that dont exist you have one of those candies

[Qermaq]

0^0 must be undetermined to some extent, whether indeterminant or undefined. I can make something converge to 0^0 in a lot of ways with a lot of limits. It's the height of arrogant hubris to assert that 0^0 = anything without some sort of specifically relevant calculus to back it up.

[Kevine]

this is not a proper explanation.
simply 0^0 is just undefined:
0³ = 0 • 0 • 0 = 0
0² = 0 • 0 • 0/0 = 0 • 0 • (any number) = 0
0¹ = 0 • 0/0 = 0 • (any number) = 0
0⁰ = 0/0 = undefined

[NinjaBear]

If you graph 0^x you would have an asymptote at 0 and have no answers as a negative integer. But with the straight line no matter how small your decimal is from the right hand side the limit is eventually going to hit 0 when x approached from the right side. X doesn't approach at all from the left side because for any negative integer, you will have a reciprocal that has a 0 in the denominator. Making the limit not existent because the left side doesn't have any direction. But looking from the right side, eventually we will get 0.

[totally_not_a_bot]

Exponentiation can be defined as "a to the b is 1 times b instances of a." In fact, I believe that is the commonly accepted definition. It's also how you would implement it computationally. So 0^0 is 1 times no zeros.

[lethal]

Okay, because once you get to the 60 = 4x3x5, then the logical next x value is 1
So the once we get to:
60=4x3x5 after squaring the 2 and then the 3 and 5 ^ 1. You get
60=4x3x5x1x1x1x1x1x1x1x1♾️ is always = to 60. So if you have 0 ^ 0 = 0 then the equation is 0=4×3×5×0.
N^0 is 0 until 0 becomes 1 of nothing