I tried to post this a couple times. The original title I was using was "0! = 1" which... I don't know, lemmy just gave up trying to post it when I hit the submit button. No errors, just a spinning circle like it was thinking about it and going to post it eventually.
Kinda funny.
24This meme is the mathiest of memes. I commend you.
6
This is right on the fucking edge of my knowledge of this stuff, but I fucking love this kind of a hot!
14Giving each character their own consistent operator is a great touch.
12I still can't wrap my head around why 0^0 = 1
If 0x0=0, and 0^0 is functionally identical to 0x0, why is it 1??
7A lot of people gave good responses. I think the best answer is that it depends on context. There are good reasons why it should be undefined, 1 or 0 depending on what you’re doing with it. Most people accept that it is 1, sometimes it makes more sense to be undefined or 0.
For an example for it to be 1, consider a combination lock on a briefcase. It has 4 digits, and each digit has 10 possible values. So there are 10^4 = 10,000 combinations. (Imagine the dials go from 0-9, so you have combos 0000-9999)
Now imagine the briefcase has no lock. How many ways can you can you arrange nothing? Just once.
The possible ways to arrange no objects in no possible ways is ∅ (empty set). So we have one possibility, empty set.
13Anything to the power of 0 is 1. For instance:
3^(-2) = 1/9
3^(-1) = 1/3
3^0 = ?
3^1 = 3
3^2 = 9
The pattern here is: Every time the exponent increments, the answer increases by a factor of 3. To get from 3^1 to to 3^2, you multiply 3 by 3 to get 9. Similarly, to get from 3^(-1) to 3^0, you multiply 1/3 by 3 to get 1.
This applies to exponentiation on any base, including zero (briefly checking a few examples, it seems to hold for all real numbers).
6it makes graphs look nicer.
however, 0^0 isn't 0x0, that would be 0^2. 0^1 is 0x1, anything^0 is... well, it's 1. afaik there isn't am equivalent mathematical expression to n^0, it's multiplying a number by itself -1x, or something equally mind melting.
4actually, I thought of a (maybe) helpful way to visualise this.
x^-n is equivalent to 1÷(x^n), so 10^-1 is one tenth, 10^-2 is one hundredth, so on. the number, x, appears in the equation n times.
you can view positive exponents as the inverse, (x^n)÷1. likewise, the number appears n times.
so what happens for x^0? well, zero is neither positive nor negative. and to maintain consistency, x must appear in the equation zero times. so what you're left with is 1÷1, regardless of what number you input as x.
4I'd imagine you want something defined recursively like multiplication
- ( 0x = 0 )
- ( xy = x(y-1)+ x ) ( y > 0 ).
So it needs to be
- ( x^0 = c ) (c is some constant)
- ( x^y = xx^{y-1} ) (( y > 0 ) (to see why, replace multiplication with exponentiation and addition with multiplication). So what could ( c ) be? Well, the recursive exponentiation definition we want refers to ( x^0 ) in ( x^1 ). ( x^1 ) must be ( x ) by the thing we wish to capture in the formalism (multiplication repeated a single time). So the proposed formalism has ( x = x^1 = xx^0 = xc ). So ( cx = x ) hence ( c = 1 ), the multiplicative identity. Anything else would leave exponentiation to a zeroth power undefined, require a special case for a zeroth power and make the base definition that of ( x^1 ), or violate the intuition that exponentiation is repeated multiplication.
On an unrelated note, it'd be nice if Lemmy had Mathjax. I just wrote all this on mobile with that assumption, and I'm not rewriting now that I know better.
3Indeed, my mind is melted.
1
00 is undefined, but often it's convenient to say it equals 1
4
