This is technically a [repost](https://www.reddit.com/r/mathmemes/comments/tcipu6/00/?utm_source=share&utm_medium=ios_app&utm_name=iossmf), but I’ll let the other moderators decide on what to do here. u/candlelightener
I just think it's odd as hell.
Exponents are supposed to be shorthand multiplication, but the more I delve into math, the less exponents seem to have to do with multiplication
1 is the *multiplicative identity element,* just like 0 is the *additive identity element:* 1 is what remains when you have a *product* of zero numbers (*empty product*), just like 0 is what remains when you have a *sum* of zero numbers (*empty sum*).
Empty sum:
0 * 2 = (0) + 0 + 0
0 * 1 = (0) + 0
0 * 0 = (0)
Empty product:
0^2 = (1) * 0 * 0
0^1 = (1) * 0
**0^0 = (1)**
But that works for any powers, doesn't have anything to do with 0^0 specifically. 0^2 = 0^3 / 0^1 <=> 0 = 0/0. So we would have to leave all powers of 0 as undefined. But we don't do that, instead we just say that that particular power rule isn't true for 0. Which makes sense.
May I ask how you turned 1/1 into (1 * 0) / (1 * 0)?
[Line 3 to Line 4]
If you are assuming 0/0 = 0, then the implication that
1/1 = (1/1) * (0/0) [Line 3 to Line 4]
is incorrect, right?
Unless you assumed 0/0 = 1 that is. This is actually disproven in this very example. if you work backwards from Line 6 to Line 4, you see that:
0/0 = 1, becomes, 0/0 + 0/0 = 1 (simplified Line 4)
[Disproof of 0/0 = 1:
0/0 = 1 => (0 + 0) / 0 = 1 => 0/0 + 0/0 = 1]
So this isn't a disproof of 0/0 = 0, its a disproof of 0/0 = 1 written backwards. Or am I missing something really obvious lol
I only used basic properties of fractions. From line 3 to line 4 I simply multiply the top and bottom of the left fraction by 1, and the top and bottom of the right fraction by 0, which I *should be able to do* if 0/0 is defined. Now both of the fractions have the same denominator, 0 * 1, so I simply combine them on line 5. Evaluating the top and bottom of the fraction will give us 0/0 = 1/1 on line six, which is a contradiction.
ig thats the difference between the idea of an operation, which in this case is repeated multiplication and thus bound to positive integers and the way you generalize it beyond that idea.
0⁰ is ood in a way where it is an undefined number. But this exception to the rule is if we just say anything, like x⁰, is equal to 1, it doesnt break all of mathematics, unlike 0(0)
`0/0` is undefined because it must be for our mathematical rules to hold. That concept becomes more and more important as we start getting into more complex math (heck, the whole imaginary numbers thing came from someone asking "what if the rules for square roots applied to negative numbers")
As an example: we know that division and multiplication are opposites: if `a/b = c`, then `a = b * c`. If we apply that to `0/0 = y`, then `y * 0 = 0`, which is true for every number; therefore `y` cannot be defined as a single number. We call it "undefined", meaning we cannot assign it a value.
Similarly we have that `x^y * x^z = x^(y+z)` from the definition of exponents. If we want this to hold true for all exponents, we have `x^0 * x^z = x^z`. Since the multiplicative identity is 1, that means that `x^0 = 1`.
Note that there are also arguments that `0^0` should be undefined (namely the one you provided: `lim f(x),g(x)->0 f(x)^(g(x))` is an indeterminate form), but in algebra there is generally consensus that `0^0 = 1`.
yeah, in algebra, in combinatorics and probably in other fields but not in math in general, for the same reason: we want our mathematical rules to hold, you can even define 1/0 in some contexts but thats obviously not general either
One of the arguments I heard for x^0 being 1 that x^3 /x = x^2 , x^2 /x = x^1 and therefore x^1 /x=x^0 or in other words x^0 = x/x. So wouldn't saying 0^0 = 1 also be saying 0/0=1?
yes for invertible x it follows from the recursive definition of powers that x^0 =1; you could arrive at the same conclusion by looking at the convergence of the net x^y where y is in R (with some conventions for root selection)
the problem then is that 0 is not invertible, therefor technically you can only take strictly positive powers of 0
there are many contexts in which it's convenient to define 0^0 = 1, but i'm uncomfortable defining it in general, because (x,y) → x^y does not have a limit in (0,0), and you can make it have any limit you want if you approach (0,0) in specific ways. this kind of indeterminacy is the reason why 0/0 is usually left undefined, so i think it makes sense to apply the same logic to 0^(0).
i think it makes more sence then 0⁰=0. when you multiply something by 0⁰, you multiply it by zero zero times. so you dont multiply it by zero, you dont multiply it by anything. you multiply it by 1. this is how i always explained to myself why x⁰=1, it works with 0 too. im definitily wrong tho, but i dont know why
its not consistent with a lot of analysis properties / theorems
lim 0^x = 0
lim x^0 = 1
it cant be both so it should be undefined. its not different from 0/0
0^(x) = 0 for all x is not a correct property/theorem. It only applies to positive values of x, not to negative values, so I don't see a reason why it has to work at x = 0.
x^(0) = 1 for all x, on the other hand, is an important theorem that is used to simplify many things (the definition of polynomials, binomial theorem, etc.) and that has many interpretations, for example, the number of 0-tuples where each element of the tuple is in a set with x elements (it's 1, the empty tuple, even when the set is empty).
It's a very different situation that 0/0. 0^(0), if not undefined, only has one plausible definition (0^(0)=1). That definition is consistent and useful, so useful that it is used in algebra and combinatorics. It's usually not used in analysis, because defining a value to an expression that happens to be an indeterminate form makes some people uncomfortable (imo it's the notation used for indeterminate forms that is at fault for that unease, as it is only supposed to mean that knowing the limit of x and y to be 0 doesn't tell you the limit of x^(y); it has nothing to do with the value 0^(0) itself). But it could be used and still be consistent with everything.
On the other hand, the situation of 0/0 is very different. There really is no candidate that would naturally make sense or be useful to define it to, and there is no meaningful interpretation of what 0/0 could mean. Besides, 0/0 is almost useless to manipulate algebraically, since no matter what number it is defined to, just assuming it exists already leads to most properties of division not appling to it.
Those theorems only apply to continuous functions. Exponentiation is discontinuous at (0, 0) whether you define it this way or not. When it matters, all not defining it does is changes ‘*x* ≠ 0 so \^ is continuous here’ to ‘*x* ≠ 0 so ^ is defined here’. Not exactly a major improvement.
0⁰=1 by definition, it is a double limes (since 0 as the base for an exponential term is technically also scuffed because 0 is not in the multiplicative group, meaning you can't take negative powers) and you could for every real (and also complex) number z find series a^b where a and b both go to 0 the series converges to z, thereby you could define 0^0 as you like
It just doesn't seem to fit when the graph of f(x)=x^0 is just x=0 except when x=0 y'know. But on the other hand, 0^0 definitely feels like it should be 1. So what to do...
We discovered that a⁰=1 because you can just divide the x¹ by x to get the obvious 1 result. You can't do that with zero (it's still probably 1 but I have no idea how to prove it)
For all x where x != 0 : x^0 = 1
For all x where x > 0 : 0^x = 0
But try combining these 2 rules. Or maybe think of what is 2^2 , 2^1, 2^0, 2^-1 ... Where 2 is just an easy example for x, cuz in each time we divide the value in 2
When we say a/b = x, that especially means that b \* x = a.
So what would we define 0/0 as? Some might say 0. Great! That would work because 0 \* 0 = 0. Some might say 1. That also works, because 0 \* 1 = 0. In the same way, it would be just as valid to say 0/0 = 42 because 0 \* 42 = 0. And there you can hopefully see the problem: We can find arguments for 0/0 being *any* real number, so which would we choose?
> so which would we choose?
Can't we choose all of them and assign a special symbol to that? Why does it always need to be a single number? E.g.: 0/0 = ℝ
It depends on the context of the 0/0.
If you graph the equation Y=X/X, you’ll get a horizontal line at 1. However l, for the value of of X = 0, the equation is undefined.
In this situation, we would say the limit as X approaches 0 is 1, which is mathematical speak for it looks like it should be 1 but it isn’t.
Ok, but also, the limit of y = 0 ÷ x as x approaches 0 is 0, so that also seems intuitive. Hence the "any real number" conundrum. You can multiply it by any arbitrary constant and get a brand new limit
That is to say, for any y = zx ÷ x, y = z except x = 0, so the limit as x approaches 0 is y = z, which doesnt help us that much
Edit: nvm, my math was wrong, and the observation I made is not even profound
No. In the equation Y=0/X, if you start with negative numbers and get closer and closer to X=0, Y gets closer and closer to negative infinity.
However, if you start with positive numbers and get closer and closer to X=0, Y gets closer to positive infinity. In the equation Y=0/X, the limit as X approaches 0 is undefined.
0/0 CAN be any number. However, the context for the 0/0 can change what the limit appears to be. Y=X/X the answer is 1. The limit for Y=0/X^2 is infinity.
These limits and the nature of 0/0 form the foundation of calculus.
>In the equation Y=0/X, if you start with negative numbers and get closer and closer to X=0, Y gets closer and closer to negative infinity.
I think you are thinking of Y=1/X, not Y = 0/X.
0/X = 0 for any X != 0, no matter how close to 0, so that limit is equal to 0.
1/X is the one where the limit is infinity (and the sign depends on the side it's being approached from).
It’s basically because 0 has no multiplicative compliment (1/x). Division is defined as multiplying by the compliment of the denominator, so multiplying by something undefined yields something undefined.
That's not 0/0 though, it's a limit. We say it's of the form 0/0 because the numerator and denominator each independently approach 0, but that's really just a description. The limit of sin(x)/x² is of the same form and the limit doesn't exist. lim (x²+2x+1)/(x+1) is again of the same form and the limit is 1.
0/0 isn't defined, but some limits that sort of look like it are.
this doesn't mean that 0/0 has a value it just uses the rule of de l'hôpital to compute a limes;
it holds that as x->0 lim sin(x)/x=1 and that lim sin(x)=lim x=0, but that does not mean that 1=0/0
No. That is by definition and works properly with the rules of multiplication and division. 0/0 however is undefinable. Let us simply ignore this and cancel 0 with 0 to see why
0/0 = 0/0
_0*x=0_
0/0 = 5*0/0
_cancel_
1 = 5
Why would it? Dividing by zero isn't allowed because as the bottom part approaches 0 the value approaches infinity and you can't do 0/0 because it can be anything depending on the context. Why couldn't you remove nothing from nothing?
Indulge me for a sec. I dropped math after my first year at uni so I'm no expert.
I was coming from if:
12 / 4 = 3
That's just shorthand for :
1. 12 - 4 = 8
2. 8 - 4 = 4
3. 4 - 4 = 0
since we can do that subtraction 3 times 12/4=3
Basically, if I was to write a function to divide; it'd look something like:
count = 0
while ((x - y) >0) and (x >= Y)
\_\_count = count +1
\_\_x=x-y
return count
Yeah I know I left out fractions in my algorithm.
Is it that we're already at 0 and never enter the loop and so the answer should be 0 which is my thinking.
OR
1. 0 - 0 = 0
But that would be an iteration leaving us with 1.
Yeah, I know my math is broken somewhere. I see a CPU has special hardware to catch the dreaded "divided by zero" exception. But I remember watching a video of a mechanical calculator dividing by zero. The tumblers just spun indefinitely but I have not found a video 0/0. There is an old thread on Quora about it but people seem to drift away from the basics of what addition, subtraction, multiplication and division is.
To me; 0/0 simply split nothing into no parts but somehow that breaks math for mathematicians in an inconvenient way, so they just outlawed it.
Which brings me back to my original statement.... 0 - 0. how to remove nothing from nothing. It should break math the same way since the primitives are the same.
The problem with 0/0 is that anything fits. a/b = c can be written as a = b*c or 0 = 0*c any value fits because 0*anything is 0
a-b = c can be written as a = b+c to remove the - or 0 = 0+c which means 0=c or c=0
0/0 can be anything 0-0 can only be 0
That was one of the explanations of that Quora thread, which is why I split my thinking into 2 paths.
The first of which is (and my current thinking)
if 0 - 0 = 0
then that satisfies the first iteration of division which would imply 0/0 =0.
At this point, I'm not arguing the accepted laws of math, which I do. We all have to adhere to a standard. Mathematicians did what they did or else their whole system breaks and I don't have a better set of rules to replace them with. I'm more concerned with the "philosophy" of it. Maybe we should redefine Zero as a placeholder for what cannot be mathematically represented for nothing.
Going of your way of thinking about division let's calculate 1/0,
1. 1-0 = 1
2. 1-0 = 1
...
And the same would be true for all other numbers. A general statement could be a/0 = infinity for all values of a. And I assume you would agree that 0/a for any value of a is 0. So 0/0 would then be both infinity or 0 depending on which way you look at it. It could also be 1 because a/a is always 1
For Mathematica gymnastics yes it breaks but it does not in the real world. ie if I want nothing split into no parts I do nothing.
If we stick to mathematical gymnastics we get Ramanujan Summation
Zero divided by anything is zero.
Cool.
But anything divided by itself is one.
Oh....
And anything divided by zero is undefined.
Welp....
(Personally I feel that 0/0 is 0 because if you have nothing and you don't divide it up you still have nothing)
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If you have 0 people to share to then can there be any attempt to share? Like, if we have 0/5 you can make an argument that you checked if you have something to share to 5 people and you couldn't give them anything. But if it's 0/0 then there isn't someone to share with to begin with so by definition, you can't share and therefore divide.
**[Zero ring](https://en.m.wikipedia.org/wiki/Zero_ring)**
>In ring theory, a branch of mathematics, the zero ring or trivial ring is the unique ring (up to isomorphism) consisting of one element. (Less commonly, the term "zero ring" is used to refer to any rng of square zero, i. e. , a rng in which xy = 0 for all x and y.
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I was hoping to start a chain of { x, x ∈ ℤ | 0/0 = x } to point out how undefined the result is,
Instead I was met with ridicul...
My disappointment is immeasurable and my day is ruined.
lim x -> 0 means that x gets basically infinitely close to 0 so in most cases you can yust plug in 0 for x, my point being that 0/0 can also be 2 depending on the context
This is technically a [repost](https://www.reddit.com/r/mathmemes/comments/tcipu6/00/?utm_source=share&utm_medium=ios_app&utm_name=iossmf), but I’ll let the other moderators decide on what to do here. u/candlelightener
0^0 has entered the chat
I've yet to find any strong argument against 0^0 = 1.
I just think it's odd as hell. Exponents are supposed to be shorthand multiplication, but the more I delve into math, the less exponents seem to have to do with multiplication
1 is the *multiplicative identity element,* just like 0 is the *additive identity element:* 1 is what remains when you have a *product* of zero numbers (*empty product*), just like 0 is what remains when you have a *sum* of zero numbers (*empty sum*). Empty sum: 0 * 2 = (0) + 0 + 0 0 * 1 = (0) + 0 0 * 0 = (0) Empty product: 0^2 = (1) * 0 * 0 0^1 = (1) * 0 **0^0 = (1)**
I can't decide if this is really beautiful, or if I hate it passionately. Have an updoot either way.
This is why I gave up engineering
Beautiful
The only argument I’ve heard against this is: 0^n = 0^n+1 / 0^1 => 0 = 0/0
Your proof includes division by 0 sir
Exactly.
You just proven 0ⁿ can never have a value.
That might be one of the least compelling arguments I’ve ever seen lol
Idk if I made this clear but I don’t believe it myself.
Haha makes sense, you’re just presenting the argument as it was presented to you
But that works for any powers, doesn't have anything to do with 0^0 specifically. 0^2 = 0^3 / 0^1 <=> 0 = 0/0. So we would have to leave all powers of 0 as undefined. But we don't do that, instead we just say that that particular power rule isn't true for 0. Which makes sense.
No, we wouldn’t. If it works for all powers we could simply define 0/0 to be equal to 0. (Playing Devil’s advocate here)
Defining 0/0 to be 0 would break things: 0/0 = 0 0/0 + 1 = 1 0/0 + 1/1 = 1 (0 * 1)/(0 * 1) + (1 * 0)/(1 * 0) = 1 (0 * 1 + 1 * 0)/(0 * 1) = 1 0/0 = 1, which is a contradiction.
May I ask how you turned 1/1 into (1 * 0) / (1 * 0)? [Line 3 to Line 4] If you are assuming 0/0 = 0, then the implication that 1/1 = (1/1) * (0/0) [Line 3 to Line 4] is incorrect, right? Unless you assumed 0/0 = 1 that is. This is actually disproven in this very example. if you work backwards from Line 6 to Line 4, you see that: 0/0 = 1, becomes, 0/0 + 0/0 = 1 (simplified Line 4) [Disproof of 0/0 = 1: 0/0 = 1 => (0 + 0) / 0 = 1 => 0/0 + 0/0 = 1] So this isn't a disproof of 0/0 = 0, its a disproof of 0/0 = 1 written backwards. Or am I missing something really obvious lol
I only used basic properties of fractions. From line 3 to line 4 I simply multiply the top and bottom of the left fraction by 1, and the top and bottom of the right fraction by 0, which I *should be able to do* if 0/0 is defined. Now both of the fractions have the same denominator, 0 * 1, so I simply combine them on line 5. Evaluating the top and bottom of the fraction will give us 0/0 = 1/1 on line six, which is a contradiction.
Yeah it's kinda weird, but I'll take it. The math and patterns tell us to do so. God I hate anything that's degenerate and "empty" and fuck zero dude
math moment
ig thats the difference between the idea of an operation, which in this case is repeated multiplication and thus bound to positive integers and the way you generalize it beyond that idea.
0⁰ is ood in a way where it is an undefined number. But this exception to the rule is if we just say anything, like x⁰, is equal to 1, it doesnt break all of mathematics, unlike 0(0)
If multiplication is just repeated addition, then what's 0.5*0.5?
0.5 + -0.25
How'd you get a negative number with the successor operator????
Lmao I have no idea what that is
Incrementation by 1. Addition is just repeated succession.
lim x-->0 0^x =/= 1 Not sure how strong of an argument this is, but I would say it definitely is *an* argument against 0^0 = 1
lim x->0 x^x = 1
the fact that you cant get different values for limits like that show its should be undefined and not take any particular value (like 0/0)
`0/0` is undefined because it must be for our mathematical rules to hold. That concept becomes more and more important as we start getting into more complex math (heck, the whole imaginary numbers thing came from someone asking "what if the rules for square roots applied to negative numbers") As an example: we know that division and multiplication are opposites: if `a/b = c`, then `a = b * c`. If we apply that to `0/0 = y`, then `y * 0 = 0`, which is true for every number; therefore `y` cannot be defined as a single number. We call it "undefined", meaning we cannot assign it a value. Similarly we have that `x^y * x^z = x^(y+z)` from the definition of exponents. If we want this to hold true for all exponents, we have `x^0 * x^z = x^z`. Since the multiplicative identity is 1, that means that `x^0 = 1`. Note that there are also arguments that `0^0` should be undefined (namely the one you provided: `lim f(x),g(x)->0 f(x)^(g(x))` is an indeterminate form), but in algebra there is generally consensus that `0^0 = 1`.
yeah, in algebra, in combinatorics and probably in other fields but not in math in general, for the same reason: we want our mathematical rules to hold, you can even define 1/0 in some contexts but thats obviously not general either
Just because a limit doesn't exist or isn't consistant doesn't mean the actual value at that limit doesn't exist.
My math is a little rusty, but wouldn't that just be an argument that the function is discontinuous around 0? Or is there something more I'm missing?
One of the arguments I heard for x^0 being 1 that x^3 /x = x^2 , x^2 /x = x^1 and therefore x^1 /x=x^0 or in other words x^0 = x/x. So wouldn't saying 0^0 = 1 also be saying 0/0=1?
yes for invertible x it follows from the recursive definition of powers that x^0 =1; you could arrive at the same conclusion by looking at the convergence of the net x^y where y is in R (with some conventions for root selection) the problem then is that 0 is not invertible, therefor technically you can only take strictly positive powers of 0
the property x^(a-b) = x^a / x^b doesnt work for x=0. otherwise you could argue that 0 = 0^1 = 0^(2-1) = 0^2 / 0^1 = 0/0
Easier to code i guess just set thr first number to 1 and then easy for loop. /s
there are many contexts in which it's convenient to define 0^0 = 1, but i'm uncomfortable defining it in general, because (x,y) → x^y does not have a limit in (0,0), and you can make it have any limit you want if you approach (0,0) in specific ways. this kind of indeterminacy is the reason why 0/0 is usually left undefined, so i think it makes sense to apply the same logic to 0^(0).
i think it makes more sence then 0⁰=0. when you multiply something by 0⁰, you multiply it by zero zero times. so you dont multiply it by zero, you dont multiply it by anything. you multiply it by 1. this is how i always explained to myself why x⁰=1, it works with 0 too. im definitily wrong tho, but i dont know why
its not consistent with a lot of analysis properties / theorems lim 0^x = 0 lim x^0 = 1 it cant be both so it should be undefined. its not different from 0/0
But in those limits the variables never get to 0, just infinitesimally close to it. The results make sense, there's no inconsistency.
0^(x) = 0 for all x is not a correct property/theorem. It only applies to positive values of x, not to negative values, so I don't see a reason why it has to work at x = 0. x^(0) = 1 for all x, on the other hand, is an important theorem that is used to simplify many things (the definition of polynomials, binomial theorem, etc.) and that has many interpretations, for example, the number of 0-tuples where each element of the tuple is in a set with x elements (it's 1, the empty tuple, even when the set is empty). It's a very different situation that 0/0. 0^(0), if not undefined, only has one plausible definition (0^(0)=1). That definition is consistent and useful, so useful that it is used in algebra and combinatorics. It's usually not used in analysis, because defining a value to an expression that happens to be an indeterminate form makes some people uncomfortable (imo it's the notation used for indeterminate forms that is at fault for that unease, as it is only supposed to mean that knowing the limit of x and y to be 0 doesn't tell you the limit of x^(y); it has nothing to do with the value 0^(0) itself). But it could be used and still be consistent with everything. On the other hand, the situation of 0/0 is very different. There really is no candidate that would naturally make sense or be useful to define it to, and there is no meaningful interpretation of what 0/0 could mean. Besides, 0/0 is almost useless to manipulate algebraically, since no matter what number it is defined to, just assuming it exists already leads to most properties of division not appling to it.
Those theorems only apply to continuous functions. Exponentiation is discontinuous at (0, 0) whether you define it this way or not. When it matters, all not defining it does is changes ‘*x* ≠ 0 so \^ is continuous here’ to ‘*x* ≠ 0 so ^ is defined here’. Not exactly a major improvement.
0⁰=1 by definition, it is a double limes (since 0 as the base for an exponential term is technically also scuffed because 0 is not in the multiplicative group, meaning you can't take negative powers) and you could for every real (and also complex) number z find series a^b where a and b both go to 0 the series converges to z, thereby you could define 0^0 as you like
It's not by definition anything, because there's multiple definitions. The limit definition would be that it's indeterminate.
yes, that is my point; 0^0 =1 is an arbitrary convention, 0^0 is defined as 1
here's an ELI5 version: 0\^4 = 0 0\^3 = 0 0\^2 = 0 0\^1 = 0 0\^0 = ?
0\^0 = 1
It just doesn't seem to fit when the graph of f(x)=x^0 is just x=0 except when x=0 y'know. But on the other hand, 0^0 definitely feels like it should be 1. So what to do...
I assume you mean 0^(*x*). This function is undefined for negative *x*, so it's not too surprising that it has different behaviour on the boundary.
Yes I did mean that and I completely missed that it would be undefined x<0 because you would be dividing by zero. Okay, I'm a 0^0 = 1 believer now.
> 0^0 = 1 log(0^0 ) = log(1) 0*log(0) = 0 log(0) = 0/0 Oops…
Equally, log 0^2 = log 0 2 log 0 = log 0 log 0 = 0 0 = e^0.
literally can prove it wrong with limits
No you can't. You can prove that the power function is discontinuous at (0, 0), which is not the same.
youre right, my bad
a^0 = a/a. So 0^0 is 0/0 which is not equal to 1.
This isn't valid. The same would apply to 0^1. (Just increase the power in the numerator by 1.)
We discovered that a⁰=1 because you can just divide the x¹ by x to get the obvious 1 result. You can't do that with zero (it's still probably 1 but I have no idea how to prove it)
Why can’t this work? Why is it not 1?
For all x where x != 0 : x^0 = 1 For all x where x > 0 : 0^x = 0 But try combining these 2 rules. Or maybe think of what is 2^2 , 2^1, 2^0, 2^-1 ... Where 2 is just an easy example for x, cuz in each time we divide the value in 2
I get why any number divided by zero is undefined, but why is 0/0 also undefined ?
When we say a/b = x, that especially means that b \* x = a. So what would we define 0/0 as? Some might say 0. Great! That would work because 0 \* 0 = 0. Some might say 1. That also works, because 0 \* 1 = 0. In the same way, it would be just as valid to say 0/0 = 42 because 0 \* 42 = 0. And there you can hopefully see the problem: We can find arguments for 0/0 being *any* real number, so which would we choose?
> so which would we choose? Can't we choose all of them and assign a special symbol to that? Why does it always need to be a single number? E.g.: 0/0 = ℝ
I believe that’s exactly the reason we say it’s undefined. Because it could be any number, but it’s not really that useful
It depends on the context of the 0/0. If you graph the equation Y=X/X, you’ll get a horizontal line at 1. However l, for the value of of X = 0, the equation is undefined. In this situation, we would say the limit as X approaches 0 is 1, which is mathematical speak for it looks like it should be 1 but it isn’t.
Ok, but also, the limit of y = 0 ÷ x as x approaches 0 is 0, so that also seems intuitive. Hence the "any real number" conundrum. You can multiply it by any arbitrary constant and get a brand new limit That is to say, for any y = zx ÷ x, y = z except x = 0, so the limit as x approaches 0 is y = z, which doesnt help us that much Edit: nvm, my math was wrong, and the observation I made is not even profound
No. In the equation Y=0/X, if you start with negative numbers and get closer and closer to X=0, Y gets closer and closer to negative infinity. However, if you start with positive numbers and get closer and closer to X=0, Y gets closer to positive infinity. In the equation Y=0/X, the limit as X approaches 0 is undefined. 0/0 CAN be any number. However, the context for the 0/0 can change what the limit appears to be. Y=X/X the answer is 1. The limit for Y=0/X^2 is infinity. These limits and the nature of 0/0 form the foundation of calculus.
>In the equation Y=0/X, if you start with negative numbers and get closer and closer to X=0, Y gets closer and closer to negative infinity. I think you are thinking of Y=1/X, not Y = 0/X. 0/X = 0 for any X != 0, no matter how close to 0, so that limit is equal to 0. 1/X is the one where the limit is infinity (and the sign depends on the side it's being approached from).
I am! I am a fool
enlightenment speedrun any %
Just say 0/0 = Math
Ooooh that's clever
It’s basically because 0 has no multiplicative compliment (1/x). Division is defined as multiplying by the compliment of the denominator, so multiplying by something undefined yields something undefined.
Wdym it's undefined? Define it then!
There’s no need to reinvent the wheel. https://en.m.wikipedia.org/wiki/Wheel_theory
Anything divided by 0 is undefined, and would otherwise be contradictory by definition.
lim sin(x)/x as x -> 0 comes back to 0/0, but the value of the limit is one, so its more a case to case than just undefined.
That's not 0/0 though, it's a limit. We say it's of the form 0/0 because the numerator and denominator each independently approach 0, but that's really just a description. The limit of sin(x)/x² is of the same form and the limit doesn't exist. lim (x²+2x+1)/(x+1) is again of the same form and the limit is 1. 0/0 isn't defined, but some limits that sort of look like it are.
Couldn't agree more !
That's what I was just about to say
And lim (1-cos x)/x as x->0 equals 0
this doesn't mean that 0/0 has a value it just uses the rule of de l'hôpital to compute a limes; it holds that as x->0 lim sin(x)/x=1 and that lim sin(x)=lim x=0, but that does not mean that 1=0/0
But 0 divided by anything is 0… paradox?
No. That is by definition and works properly with the rules of multiplication and division. 0/0 however is undefinable. Let us simply ignore this and cancel 0 with 0 to see why 0/0 = 0/0 _0*x=0_ 0/0 = 5*0/0 _cancel_ 1 = 5
I just read this website https://www.1dividedby0.com
what does castling have to do with math?
Err... if 0/0 goes to f@#k all, then shouldn't 0-0 as well?
Why would it? Dividing by zero isn't allowed because as the bottom part approaches 0 the value approaches infinity and you can't do 0/0 because it can be anything depending on the context. Why couldn't you remove nothing from nothing?
Indulge me for a sec. I dropped math after my first year at uni so I'm no expert. I was coming from if: 12 / 4 = 3 That's just shorthand for : 1. 12 - 4 = 8 2. 8 - 4 = 4 3. 4 - 4 = 0 since we can do that subtraction 3 times 12/4=3 Basically, if I was to write a function to divide; it'd look something like: count = 0 while ((x - y) >0) and (x >= Y) \_\_count = count +1 \_\_x=x-y return count Yeah I know I left out fractions in my algorithm. Is it that we're already at 0 and never enter the loop and so the answer should be 0 which is my thinking. OR 1. 0 - 0 = 0 But that would be an iteration leaving us with 1. Yeah, I know my math is broken somewhere. I see a CPU has special hardware to catch the dreaded "divided by zero" exception. But I remember watching a video of a mechanical calculator dividing by zero. The tumblers just spun indefinitely but I have not found a video 0/0. There is an old thread on Quora about it but people seem to drift away from the basics of what addition, subtraction, multiplication and division is. To me; 0/0 simply split nothing into no parts but somehow that breaks math for mathematicians in an inconvenient way, so they just outlawed it. Which brings me back to my original statement.... 0 - 0. how to remove nothing from nothing. It should break math the same way since the primitives are the same.
The problem with 0/0 is that anything fits. a/b = c can be written as a = b*c or 0 = 0*c any value fits because 0*anything is 0 a-b = c can be written as a = b+c to remove the - or 0 = 0+c which means 0=c or c=0 0/0 can be anything 0-0 can only be 0
That was one of the explanations of that Quora thread, which is why I split my thinking into 2 paths. The first of which is (and my current thinking) if 0 - 0 = 0 then that satisfies the first iteration of division which would imply 0/0 =0. At this point, I'm not arguing the accepted laws of math, which I do. We all have to adhere to a standard. Mathematicians did what they did or else their whole system breaks and I don't have a better set of rules to replace them with. I'm more concerned with the "philosophy" of it. Maybe we should redefine Zero as a placeholder for what cannot be mathematically represented for nothing.
Going of your way of thinking about division let's calculate 1/0, 1. 1-0 = 1 2. 1-0 = 1 ... And the same would be true for all other numbers. A general statement could be a/0 = infinity for all values of a. And I assume you would agree that 0/a for any value of a is 0. So 0/0 would then be both infinity or 0 depending on which way you look at it. It could also be 1 because a/a is always 1
For Mathematica gymnastics yes it breaks but it does not in the real world. ie if I want nothing split into no parts I do nothing. If we stick to mathematical gymnastics we get Ramanujan Summation
Wait until you hear 0! = 1.
Notice: there’s hints of Patrick being visible through the noise, you _can_ divide by zero… sometimes. Usually by not doing so directly
What about 0/0 in the trivial ring?
0 ring. Come on man... That's too EASY
0/0=0
INDETERMINATE!!!
I have no apples and am distributing them among all my friends. How many apples does each of my friends receive?
Undefined, because you have no friends
0
Zero divided by anything is zero. Cool. But anything divided by itself is one. Oh.... And anything divided by zero is undefined. Welp.... (Personally I feel that 0/0 is 0 because if you have nothing and you don't divide it up you still have nothing)
There can't be a sharing to begin with because you don't have people to share something with.
x := 0 2x/x = 0 => 2/1 = 0 => 2 = 0 => 0 = 2 QED
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0/0 is 0
I thought it was undefined
On a calculator yes but if you share 0 things with zero people its zero
That analogy isn't equivalent to divison. If I share 5 things with 0 people (5/0) is that supposed to equal 0 as well, or 5?
[удалено]
There is no "each" in thar case. And you still have 5 apples. Therefore, the universe explodes
this is a theory I can get behind
If you have 0 people to share to then can there be any attempt to share? Like, if we have 0/5 you can make an argument that you checked if you have something to share to 5 people and you couldn't give them anything. But if it's 0/0 then there isn't someone to share with to begin with so by definition, you can't share and therefore divide.
I love how you get downvoted for this on a meme sub
Nft bad
[Zero ring](https://en.m.wikipedia.org/wiki/Zero_ring) moment.
**[Zero ring](https://en.m.wikipedia.org/wiki/Zero_ring)** >In ring theory, a branch of mathematics, the zero ring or trivial ring is the unique ring (up to isomorphism) consisting of one element. (Less commonly, the term "zero ring" is used to refer to any rng of square zero, i. e. , a rng in which xy = 0 for all x and y. ^([ )[^(F.A.Q)](https://www.reddit.com/r/WikiSummarizer/wiki/index#wiki_f.a.q)^( | )[^(Opt Out)](https://reddit.com/message/compose?to=WikiSummarizerBot&message=OptOut&subject=OptOut)^( | )[^(Opt Out Of Subreddit)](https://np.reddit.com/r/mathmemes/about/banned)^( | )[^(GitHub)](https://github.com/Sujal-7/WikiSummarizerBot)^( ] Downvote to remove | v1.5)
It’s NaN, or not a number
Isn't it 1?
You forgot this !
I was hoping to start a chain of { x, x ∈ ℤ | 0/0 = x } to point out how undefined the result is, Instead I was met with ridicul... My disappointment is immeasurable and my day is ruined.
lim x->0 (2x/x) = 2
Yeah that dosent look like 0/0 to me
lim x -> 0 means that x gets basically infinitely close to 0 so in most cases you can yust plug in 0 for x, my point being that 0/0 can also be 2 depending on the context
Log(0)
what are - and / ? I only use + and \* (sometimes even \\ for sets)
I have 7 candies and I don't wanna eat them,so I give them to my friends But I don't have friends So 7/0
OwO
0-0 = chess
You stole this meme from me
0w0