Index Laws (2 of 2: Zero power)

[@9TailsExar]

Well~~~ 0^4=0 0^3=0 0^2=0 0^1=0 0^0=1. WTF, MATH! YOU DRUNK! GO HOME AND SLEEEEEEEEP! But then... 0 says: Imma no change no matter of powah (well except myself)! -1 says: Omae wa mo shindeiru 0 says: NANI?! -1 : 0^-1= 1/0

[@lux6230]

But What about 0 to the 0 power? If you're going down (using the method of dividing by the base) would it be 1 or undefined? Both numerator and denominator are zeros, so which would prevail? Maybe I'm digging too deep into this. QwQ

[@yossefbudagov8748]

Zero to the power of zero is undefined because it can be either zero(zero to the power of anything is zero) or one(anything to the power of zero is one)

[@davidepasero7032]

Actually it's one, look at the video of eddie woo, or try this: 0.00...001^0.00...001 the more you get close to 0, with those numbers, the more you close to 1

[@yossefbudagov8748]

@@davidepasero7032 it's not one. Mathematicians all around the world defined it to be undefined.

[@davidepasero7032]

@@yossefbudagov8748 actually in some areas of mathematics it' considered to be 1 (for example algebra) and in some other areas like analysis it's considered to be undefined, and, if you search in google 0 to the power of 0 the result is 1

[@lucca7716]

yossef budagov « they defined it to be undefined »

[@heisenberg1601]

Well all the math teachers I've asked these past two years (we learnt that thing in 8th grade) wouldn't explain it to me and would say iTs a dEFinItiOn so thank you

[@davidwright8432]

last tiny bit at the end - 0^0 - is a bit weird, since it's (formally) 0^1/0^1 = help! Division by zero is a BIG no-no! But in terms of the pattern established ... um ... formally, yes! If you could do it. which you can't!

[@9TailsExar]

Well - you can write it like (a-x)^1/(a-x)^1 where a=x. but if you forget about it for second - you get (a-x)/(a-x)=1

[@brunobatista6568]

Your explanation is very good, congratulation, but I think you should improve your organization on the blackboard.

[@brunobatista6568]

@@kangolklm5490 Just an advice.

[@justabunga1]

That’s only true as long as the base is not 0 since 0^0 is indeterminate. To make things clear, it should be any non-zero number raised to 0 power is always equal to 1.

[@TrollAxeThrower]

is this high-school level?

[@nataliajimenez5973]

Yes

[@fhsid4669]

畢格力

[@zeboo8301]

ive had many math teachers but Eddie is amazing

[@tomr3499]

Hey can someone help me I have a question If I have r to the power of 3 divided by r to the power of 2 what do I do it’s not r to the power of 1 is it just r

[@JohnSmith-rf1tx]

"just r" IS r to the power of 1. Writing it without the "power of 1" is shorthand.

[@oscarguinane2184]

Clever

[@sebastianr.r4688]

first XD

[@bottlebrush4824]

Grow up

[@pelasgeuspelasgeus4634]

Pathetic tricks as usual. The exponent tells you how many times to use the number in a multiplication. That's the definition and there's no debate. So, how can any number raised to 0 be anything else other than 0?

[@AndresFirte]

Nice question! Observe these results: A: 4⁴ = 256 B: 4³ = 64 C: 4² = 16 Notice how to go from C to B, you can multiply by 4. And to go from B to A you can also multiply by 4. Therefore: to go up, you multiply by 4 But notice that to go **down** you can divide by 4. So if we have this: A: 4⁴ = 256 B: 4³ = 64 C: 4² = 16 D: 4¹ = ? E: 4⁰= ? And we want to go down, from C to D, we can divide by 4. We get 16/4 = 4. So that means that 4¹ = 4. Which seems right Now, if we want to go from D to E we can also divide by 4 again: we get 4/4 = 1. Therefore, the pattern tells us that 4⁰ = 1. Ultimately, this is just a convention: A⁰ = 1 because we decided to define it that way, because that’s a definition that allows us to have formulas like A^n × A^m = A^(n+m). And it also fits in nicely to the behavior of some patterns like the ones I mentioned in the other explanations. We defined it this way because it’s useful and completes patterns in many parts of mathematics and programming. For example, try to define exponents in a recursive way: 4^n = (4^(n-1))×4. If you program a function that does this, you need to define a base case. If your base case is at the exponent 0, then you must define the result as 1, otherwise it won’t work. If you define the base case at -1, then you must define the result as 1/4 (and consequently 4^0 = 1) for it to work, etc. 🟡🟡In other words: you’re kinda right! The initial definition for exponents is only unambiguous for positive natural numbers. It is a function from positive naturals (domain) to the natural numbers (codomain). If we want to extend it to a larger domain in such a way that we preserve the properties of the function, the only way to so it is to define 4^0 = 1. And that’s what we did: we extended the domain. Because it’s useful.

[@pelasgeuspelasgeus4634]

@@AndresFirte As usual you oversimplify stuff and pretend that definition is not that important.

[@AndresFirte]

@@pelasgeuspelasgeus4634 as usual? What do you mean? And, on the contrary, my reply focused on the definition, which is where the misconception could be: the original definition for exponents was a function from the positive naturals to the naturals. But we generalized that definition to a larger domain, and as a consequence, 4^0 = 1

[@pelasgeuspelasgeus4634]

@@AndresFirte Do you need to "generalize" to show that when you divide a number with itself the result will always be 1? Also, by exponent properties definition x^m/x^n=x^(m-n) only for different m, n. Yes or no?

[@AndresFirte]

@@pelasgeuspelasgeus4634 no, you don’t need to generalize for that. I said that we generalize the definition so it is applicable to the integer numbers instead of only the positive integers. And the formula you mentioned works even when m = n if we’re using the definition of exponents for integers. And I’m still confused why you said “as usual”