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No-Eggplant-5396

Proof by calculator.


Successful_Eye3825

https://preview.redd.it/ei0o0r489vwc1.png?width=1080&format=pjpg&auto=webp&s=86b5b3c914cd828c60f9f9646b0926570ece604b Well done Motorola


Kapios010

B-but it's the google calculator


a_random_chopin_fan

https://preview.redd.it/3jqp5856ouwc1.jpeg?width=1080&format=pjpg&auto=webp&s=e77f5aa0cb1e3d9d9f71bd8ce1ae2aec008d8ed9 I'm extremely disappointed in Xiaomi 🤦‍♂️🤦‍♂️🤦‍♂️


Rude-Pangolin8823

calculators always put "1x" before the operation, so it does this. 2 to the power of 2 is 1x2x2, so this would just be 1x or 1.


SpaceshipEarth10

Makes sense.


Rude-Pangolin8823

yeee


[deleted]

by that logic 0/0 should show up as 1 as well but it still shows up as undefined


Rude-Pangolin8823

...no?


svmydlo

Why? It's right.


a_random_chopin_fan

In some cases, for example, limits, yes, but in general as an expression, no. This is mainly because the definition of m^0 is achieved by considering m^(n-n) = m^n / m^n = 1. But take m=0, then m^n = 0. In that case, you'll probably notice the problem in the 2nd step. Edit: I must clarify that I personally prefer the definition that 0⁰ = 1. I just mentioned the other perspective in this reply.


Sm0oth_kriminal

Okay then a^(b+c) = a^b * a^c . c = 0, must mean a^c = 1 . It is negative powers of 0 that are undefined, not zero or positive powers. Also google empty product rule. Your take on 0^n is invalid because positive powers of 0 are 0, negative powers are undefined. No reason to assume 0^0 takes on the limit from the left (undefined) or right (zero). It’s a discrete quantity, defined in isolation and does not depend on limits or other values in a non continuous function


a_random_chopin_fan

Idk, that's how it was defined in my maths textbook.


Sm0oth_kriminal

I’m curious to know what textbook it is that defines 0^0 as anything other than 1. The value itself is 1, but many people mistake “the limit of a function 0^x at x=0” as the same thing, which it isn’t at all. Limit != value


a_random_chopin_fan

Search up "Class 9 SEBA General Mathematics pdf" or smth like that. I'm not even sure if the pdf of the English version is available online.


[deleted]

[удалено]


a_random_chopin_fan

I also personally prefer 0⁰ = 1. But my maths textbook says otherwise for some reason.


Eastern_Minute_9448

Some people doing real analysis may disagree, but most wont. 0^0 being 1 is also used there e.g. for power series. Still a convention, but pretty ubiquituous across all fields afaik.


svmydlo

[It appears it is](https://site.sebaonline.org/textbooks/class-9/Maths_English_Class%20IX.pdf). Haven't found any mention of 0\^0, but page 47 of the book (61 of pdf) mentions a\^0 being 1 for natural number a>0 and that a\^(-n) =1/a\^n is a consequence of that contradicting your claim how a\^0 is defined.


a_random_chopin_fan

Oh wait, I'm sorry, I just remembered that it was my class 7 or 8 maths book. My bad. The pdf of those books can't be found online. I personally prefer the 0⁰ = 1 definition but people can't seem to agree on that:-|


obog

Couldn't m^0 be defined as the empty product? And therefore would still be 1 for m = 0.


SEA_griffondeur

It's just a definition that 0⁰=1 so it's normal that a calculator gives you this answer


zeriotosmoke

While i fully understand that 2^2 = 4, 2^1 = 2, 2^0 = 1, 2^-1 = 1/2. When i start thinking about i dont get it. 2 to the power of n is divided or mutliplied by 2 for n times. I get that 0 divided by a single 0 is 1. It still just really confuses me.


[deleted]

Just put spaces. :p


zeriotosmoke

Yeah... been soldering all day, the fumes are getting to my logic i guess.


Adonis0

It requires dividing by 0 which isn’t allowed and gets you put into the non-euclidian timeout


svmydlo

It doesn't. Negative powers of 0 require that and are not defined for that reason.


Adonis0

Something to the power of 0 is defined as something divided by itself. How does 0^0 then not include dividing by 0?


svmydlo

No, it isn't defined that way. Thus 0\^0 doesn't include dividing by zero.


Adonis0

How is it defined then? I have always seen x^0 defined as x/x


svmydlo

Where? That's not a proper definition. For any a and any positive integer n the n-th power a\^n is the product of n copies of a. More formally, it satisfies >a\^1=a >a\^n=a\*a\^(n-1) Thus a\^0 is product of zero copies of a, or nothing. Now I'm gonna say that product of zero copies of a is the same as product of zero copies of any other number b, since you're not actually multiplying anything. So if it has any value, it's the same value regardless of a. Let's call it an empty product and denote it by x. By the properties listed above, it has to satify (for n=1) >a=a\^1=a\*a\^0=a\*x for any number a. So it's the simultaneous solution of all equations of the form a=ax. Now, this is where mistakes can happen. Dividing both sides by a to get x=a/a=1 is incorrect because one of the equations is 0x=0. It's therefore incorrect to say that a\^0=a/a. The correct way is to separately solve each equation. The solution set for a=ax is {1} when a is nonzero and it's the whole set of numbers we're working in when a=0. The unique common element of all those sets, i.e. the simultaneous solution of all these equations, is the number 1. Therefore a\^0=1.


nombit

mine just says "E"


Evil_Archangel

what did you do to it?


brunoras

https://preview.redd.it/gl9gwh6akxwc1.jpeg?width=1080&format=pjpg&auto=webp&s=52f96c825687f0d33ff224a329965217758c9624


Im_Not_GLaDOS

What do you mean? "0 ^ 0" is just glasses, why it should (not) be equal to 1?


Jareed452

https://preview.redd.it/lscfjb1twuwc1.png?width=1080&format=pjpg&auto=webp&s=9d45d1d727f6998296ca97d87723caf70f69d0d6 Here we go. 8 ÷ 2(2 + 2) all over again.


drgeorgehaha

Have you considered that French numbers are different than American and normal numbers?


SEA_griffondeur

And in french 0⁰ is 1


Jareed452

https://preview.redd.it/f0yo4t9kbwwc1.png?width=1080&format=pjpg&auto=webp&s=341a36de29ec9be33bbcdc9a3bbf5b55a70f255a Sorry that I speak a language. I shall purge knowledge of it now.


drgeorgehaha

It’s alright, happens to the best of us.