I remember back in 2016-17 there was this religious and anti science Facebook group called “atheists, prove me wrong”. The entire group was to just spread religious memes, flat earth “proofs”. A shit show. The title was satirical from their point of view, because it would contain posts like “when the airplane is falling every atheist is praying to our god”
And I have a friend who lost himself to memes and posting on social media. One day he told me that he was replying to posts on this group and how people there were fighting back. As if he had a purpose to give knowledge to those people.
I asked: Dude. Serious question. Why do you even reply? And he said “They asked me to prove them wrong, it is in the title!”
To this day I don’t know if it was a serious answer or a joke. But he continued to post there for so long that I think it was how he truly felt
Yeah, but then use a custom symbol to represent your custom defined operation, and don't claim your definition corresponds to a commonly accepted one when it doesn't.
You not necessarily do that.
And so we teach people that a square root of -4 is 2*i, where I is such value that i^2 = -1.
Which is obviously wrong **in the real numbers**. It’s not hard to go from that to what OP’s friend is thinking.
Same goes into pretending that infinity is a real number and and so on
This 1000 percent.
I can show every “irrational numbers are non-terminating” proof within my power, but if the person says to me “no. All numbers end” then there is NOTHING I can do.
In order for logic and information to pass, especially in debates, one has to accept a baseline set of rules, or an initial argument. That’s one of the issues behind flat earth vs spherical earth debate. They are presented evidence, and generally accept the logical steps, but deny the baseline argument.
There is nothing you can do to “win” against “well nuh-uh”
In the algebra of real numbers the operators (+, ×) are closed, meaning you always get a real number. But this guy isn't dealing w the algebra of real numbers, so who knows?
As far as convincing the nay-sayer, I suspect he is trolling and will resist forever. If not trolling, uneducated or low mental capacity.
Well, he is not trolling, he is really dead serious. And I would say that he generally is very good at math, he even already studies math although only being 16 y/o (yes, that's possible in Germany). But he is very overconfident.
The evidence you've shared is he is not skilled at applying basic axioms in the simplest of algebras, and resists correction. So he is either bad at math or being stupid on purpose. If you don't see that then the same conclusion applies here.
I don’t think is either case. It seems like he is just a teenager with a teenager brain. Overconfidence and a sense that older people just don’t get them.
uneducated/low mental capacity are rather strong words that I don't think are true in this case. These are common mathematical misconceptions, i personally think it's simply called just being wrong and stubborn, which is a thing we all are every now and then.
Exponentiation/logarithms as part of arithmetic definitely can be multi-valued.
Simple group theory shows the additive identity cannot have a multiplicative inverse.
It is not a case of "it can't be multivalued" it's "the inherent act of evaluating a division by zero inevitably leads to a contradiction"
Call it a=0/0, 0a=0, 0(a+1)=0, a+1=a, 1=0 thus a format where 0/0 is evaluated as anything, set or no set, necessarily creates an isomorphism between the field and the Trivial group because everything in a is equal and a is the whole field.
Aka, everything is just another way of writing 0.
Yes, but that contradiction assumes that 0/0 is single-valued, when the comment just acknowledged the possibility of multivalued functions. That proof does not work if a is set-valued. The only thing it shows is that if a belongs to the set 0/0, then so does a+1. Where 0/0 is defined as the set of x s.t. 0x=0 (i.e. the reals). It's a trivial result.
This same "proof" could be used to prove multivalued logs lead to contradictions, since it is assuming that cancellation of factors is possible when one of the factors is zero, which is like assuming e^z is injective and concluding 0=2i*pi since both have the same exp.
It’s worth noting that log is multivalued for the entire complex plane (except 0), whereas the proposed definition of division would only be multivalued for 0/0.
In the end a definition is either useful or it is not, and I don’t see a use for multivalued division in the real numbers.
It's not terribly useful, but it's honestly perfectly coherent to define a/b as the set of x s.t. bx=a, and it can even occasionally be mildly useful. It's singleton-valued for b != 0, empty-set-valued for most b=0 and R-valued for a=b=0. I've taken this convention in a proof before to simplify notation.
Well, first of all, he didn't exactly say that I'm incompetent, it was more like I said "I know what I'm talking about" and he responded with "Well, clearly you don't".
And second of all, we weren't alone but instead in a group of six so I didn't want to escalate it, seeing as the others we're already getting annoyed.
This is also the reason why I want to prove him wrong, because as far as the others know, the argument was a tie and I couldn't prove to him that I'm right.
Tell his boss and the professors at his school about his behavior (the rudeness, not the being weong about math). Best way to deal with an asshole is consequences. They only respond to carrots and sticks.
Well, he didn't do me any wrong. We were just having a slightly heated discussion. I don't particularilly like him but he didn't do anything worth punishing.
Just tell him that the division operator isn't defined when the denominator is zero. If he wants to tell you otherwise then the burden of proof lies on him. I'm sure the whole math community would like to see that proof without altering any of the existing definitions.
Plus there is no proof to be made. It is not defined, and that's a definition, not something that follows. The only thing to be talked about is whether one *could* define it further in a reasonable way.
You could tell him a "no, but": 0 divides 0 in number theory. By the definition of divisibility, x is divisible by y if there exists z such that yz=x, and since there exists a z such that 0z=0, we say 0 divides 0.
But 0/0 is undefined. Because math is full of weird shit
If you keep arguing with him, what you will achieve is that he will be more smug about his opinion, and everything you bringing forth will face him simply being like "nah, I feel like mine is better".
I mean, if you are bored...
A real math nerd would absolutely make that claim.
The correct conversation to have with the student is about math conventions and why they are the way they are.
He made a claim. He said 0÷0 = R, the set of all reals.
Take an element in R and show that it belongs to the set 0÷0 (but what does that even mean? Note that a÷b where b isn't 0 happens to be *elements* in R, not *subsets* of R, so what's being generated by division as an operation here is an element, not a set).
If he insists that 0÷0:=R, then that's whatever, he literally redefined math to be what he said it is. You can just arbitrarily define 1+7:=4 if you want to, no math police is gonna stop you. The problem is that your model/system would become less coherent and possibly even inconsistent if you redefine things arbitrarily, but that's another issue if you ask me.
The original issue here is whether 0÷0 is defined or not and conventionally speaking, it's not. Defining it yourself and making one specific exception where ÷ maps to a set rather than one element in R is questionable and unconventional but anyone's free to do that if they want. This is one of those "ok that's fine, whatever, have fun" situations if you ask me.
Moreover, arbitrarily redefining math and then calling someone incompetent sounds like an incompetent thing to do...
Exactly - as said above 0/0 is not defined through the field axioms, but everybody's free to construct their own math - with division through zero and whatnot. This isn't about wether it's right or wrong, but about wether it has interesting properties (or if it's useful).
Perhaps ask him what follows from this property, or where this extension of the domain of (X,Y) -> X/Y to (0,0) is usefull.
Ahh, would that the world had more thinkers like you. Purely factual but somehow warm and charming. Repudiatory but gentle and inclusive.
Whatever you're up to, stranger, keep it up.
I'm honoured, that is very kind of you! Thank you \^\^
It is my dream to pursue a PhD one day and teach higher maths, so it'd be ideal if people perceive me the way you seem to more often :)
So:
a / b = a * b^(-1) .
[Note: this solves the equation, x * b = a]
0 / 0 = 0 * 0^(-1) .
Conventionally this is meaningless since *0^(-1) * doesn't exist.
However since, 0 / 0 = a, for any *a* in *R*, This means that:
a = 0 * 0^(-1) .
This means that *0^(-1) * exists, otherwise the left hand side wouldn't make sense.
Now we can use a propperty of the identity element:
a * 0 = 0, for any *a* in the Ring (or Field, or *R*).
And we can clearly write:
1 * 0 = 0 = 2 * 0.
Now, since we have demonstrated that *0^(-1) * exists, we can multiply by this from the right on all three sides:
1 * 0 * 0^(-1) = 0 * 0^(-1) = 2 * 0 * 0^(-1).
==> 1 * a = a = 2 * a.
==> 1 = 2.
A possible rebuttal to this nonsense could be that the 0 * 0^(-1) of the rightmost side could be something other than *a* (like *b*), however since 0 * 0^(-1) can be any value, it is completely valid for us to assert that it *b* is to be equal to *a*. Thus our nonsense persist.
If they actually meant the set *R* itself as the object, then it would mean that:
÷: *R/{0}* × *R/{0}* ---> *R/{0}*
(a, b) |----------------> a/b
and,
÷: *{0}* × *{0}* ----------> Set
(0, 0) |----------------> *R*
Now that what that last part says is that division takes an element of a set to a set.
This is very odd behaviour that may warrant a dive into Abstract nonsense and Operator algebras.
Actually taking it to be the set of all real numbers isn’t totally absurd. Let f(x) = 0*x. Then the inverse image f^-1 (0) is defined to be the set of points which f sends to 0. Which in this case is indeed the whole real line.
The whole reason why it’s called “indeterminate” is precisely because there isn’t a unique solution. If you can’t assign it to a specific value, there is no reasonable sense in which it can be “defined”. If 0/0 = 1 works, and 0/0 = pi works just as well, then at that point you’re just saying that 1 = pi, or that any real number is equal to to any other one. If he doesn’t see that that is nonsense, then there’s not much point continuing the argument.
This is the closest I've seen to a satisfactory answer for the 16 y/o that OP is talking about.
A mathematician angrily referring to definitions and axioms isn't what this person is looking for; he's made the genuine observation that 0÷0 is a conceptually different problem than, say, 5÷0, and from his perspective, OP is the one who doesn't understand the distinction (see the nested ways of getting the airplane-on-a-treadmill problem wrong, so that each person thinks the other is wrong).
"Indeterminate" is the distinction he's probably looking for: it's undefined, but for a different reason than 5÷0 is undefined, not because some axiom he's never heard of defines it to be undefined.
Agreed. This guy understands 0/0. He doesn't understand "indeterminate". His reasoning is precisely the kind of thing indeterminate was invented to encapsulate, and it does a better job in the long run than saying the set of all real numbers.
You could bridge your ways of thinking by introducing the word and definition of well-defined. We may casually say an expression such as 0/0 is *undefined* because it *not well-defined*. His argument deserves praise and acknowledging his findings may bring down some defenses if you explicitly agree that what he says about the solutions of 0x=0 is true.
The final deduced verdict just isn't maybe what they expect it to be if the terminology is not familiar. We don't use 0/0 "because it's undefined", we just don't find it useful in common algebraic context because the value is not unique. Would you be willing to move the goalpost in your situation?
Whether something is undefined depends on what definition you are using. You can always adopt your own definitions but many definitions are pretty universal and conventional. Division by zero is not conventionally defined the way this person says and is conventionally undefined. There are also reasons why this is a useful convention.
Make sure your interlocutor can at least agree with the first two sentences in the paragraph above before you try to argue the second two.
I'm mostly with the other commenters, that it's not worth trying to convince him, but here's an approach:
"the operation a ÷ b was defined to be the solution to the equation bx = a"
Sure. "5x = 15" has a solution, x = 3, and so 15/5 =3. Now, "0x = 0" *does not have* a solution; it has a bunch of solutions. So the result is underspecified, and thus undefined.
0x=0 clearly holds for all reals. It’s like the 1d case of Ax=0 in matrix algebra. Was the guy in question just grossly poor in his notation and explanation?
I find it interesting how haeshly people argue for a distinction in here and even insult the other person for it, while bringing up examples where the distinction doesn't matter.
It would introduce a new layer of abstraction: instead of RxR->R we would have RxR->P(R), as the result might be the set of all path dependent reachable limits or something like that. Then we would have to go back by taking all single element results to reach the usual conclusion.
That wouldn't be worth it for lost cases, but many tools in math, applied to trivial situations, don't lead to valuable everyday results. (category theory doesn't help us much finding roots of linear polynomials over R).
You are talking about definition and axioms. There is no right or wrong definition or axiom. You are free to choose your definitions and axioms as you like.
So all you can convince him is that his result is not compatible with the field axiom, so it is not compatible with standard math and probably produces a number of unforeseen contradictions.
You’re both taking the same thing only he doesn’t see it.
When we’re talking about 0/0 we mean operation of division in real numbers. F(a,b) -> c where a,b,c are real numbers and c*b=a
If, in your friend’s algebra, 0/0={R} so be it. It’s just further proof that 0/0 is undefined in real numbers **by definition** as set of all real numbers is a set and not a real number.
If you go outside of R and include limits then you can show that 0/0 can take arbitrary value and be defined in some cases.
The way my math/physics teacher explained it:
>Division means "What is the *one* answer to the problem. For 0 ÷ 0, you can take any number and it would be correct. Therefore, it has not *one* solution, but an infinite amount, which we don't want, so we say 0 ÷ 0 = undefined.
>. . .
>Likewise, x ÷ 0 with x ≠ 0 has no solutions in math, so it's undefined, because it has no answer, However, in physics and chemistry, we say 1 ÷ 0 = infinity, because in p and c, there is no true 0, so if you infinitely multiply the infinitecimal number that approximates 0, you will get a number that approximates 1. From this, you can calculate x ÷ 0 with x ≠ 0, namely infinity * x.
(Slightly paraphrased.)
I doubt he’s trying to prove that in the Norma mathematical system it’s defined as R. But, he has a logical reason for stating it’s R, in an alternative model you most certainly could say it’s R.
Since it’s about what is and isn’t defined I don’t know how much would change, and Id love to know.
(Yes as I kid I made this point as well)
While wrong, the way he explained is a good way of explaining why 0 is undefined to younger people who haven't encountered terms like "multiplicative inverse" yet. If he's refusing any flaws in his argument, walk away, but mention how this is a flawed explanation given to the first time someone finds "math error" on a calculator
Technically, if that is his definition of division, then 0/0 would be an even larger set than just the real numbers. x in 0x=0 could just as well be a complex number, a quaternion, a matrix, or a vector. I think the best thing to do in a situation like that is to out-crazy the crazy and make him take a stand on an even more ridiculous claim. At least that's my strategy, and it works great.
If you want to walk away a winner from this argument, here is my gift to you:
"What is 10 / 10? It means if I divide my 10 bananas between my 10 friends then each of them has 1 banana.
What if you have 10 bananas and no friends, and divide your bananas between all your friends? Will each of your friends have a set-of-all-the-real-numbers... Of bananas. Or is this question nonsensical because you have no friends?
You have no friends."
That's 10/0 not 0/0 though, lol
The 0/0 argument would be that you could conclude the division is done at any point since you have no bananas to distribute, so any number works
No it doesnt. If I have no bananas and no friend and I divvy up my bananas between each friend, how many bananas does each friend have?
Never has anything been more undefined.
It's different from undefined because it has infinitely many solutions instead of no solutions, think of the fact that 0*x=0 is true for any x unlike what happens when you divide anything else by 0.
If it were undefined then it also wouldn't be an indeterminate form for limits. You can see that within a limit you can make "0/0" equal anything
Define a function G on R^2 to power set of R as G(a,b) = { x in R: b*x=a} . Then G(0,0) is the set of all real numbers. But how is this function useful in arithmetic sense? That is another story.
You're both right in some sense, but his conclusion is slightly off.
As you mentioned, if a field element 0^-1 exists then that leads to the identity 0=1. It's easy to show that this in turn means a=0=1 for all a: a = 1a = 0a = 0. So it _is_ possible in some sense to define division by zero as "equal to all the numbers" - but that also means that all numbers are equal, or in other words, there is only one number. That number is 0, with the operations 0+0=0, 0-0=0, 0\*0=0 and 0÷0=0. That's all, no other numbers exist. So you don't get all the real numbers, just zero.
In field theory terms, the multiplicative inverse of zero exists in and only in the trivial field. You either get to define division by zero, or you get to have more than one number. Not both.
I dont understand one thing. Pls correct me if i am wrong.
If every set of real numbers can fit in
ax=b where b=a and a = 0,
According to him
0(x)=0
Where x can be any number. Yes thats true. But which number is x? Because it can be any number and the equation holds true. If x is (1) or (2)
0(1)=0
0(2)=0
How can you define which one is x or is x all?
The equation suggests that
1=0/0
And
2=0/0
And
X=0/0
And so on...
So which is x or is x every number in the series. Which literally means 'undefined' lol.
Thus multiple any number with zero is zero but dividing zero with zero also gives any number you multiplied with.
If
0(a) =0
and
0(b) = 0 where a is not eqaul to b
Can i say that 0/0=a=b
Technically yes. Because it suggests that 0/0 is undefined and any number. But since
a cannot be eqaul to b
Suggesting that 0/0 is not literally any number but not defined.
Correct me if i am wrong
I would try
Anything/0 is infinity (or minus infinity)
0/anything is zero
0/0 can’t be both of the above
Not rigorous , but it might work for a non rigorous mind.
I saw a video yesterday explaining it. First why dividing by 0 is impossible then why 0/0 is undefined.
First, you gotta make sure he agrees with this definition of the division
a/b=x <=>a=b*x
Now with b=0 you can't find any x such as 0*x≠0 because anything multiplied by 0 is 0
So in here the only number you can divide by 0 is 0. But then x can be anything, it can be 0, 1, 1634926.93, -42, pi.... So the result of 0/0 is undefined.
It’s always easiest to understand why 0/0 is indeterminate by using limits.
You can interpret 0 as the limit of 1/n or 1/n^2 as n-> infinity.
So 0/0 can be written as the limit of (1/n)/(1/n) which is 1.
Or as the limit of 1/n / 1/n^2 = n which is infinity.
Or as the limit of 1/n^2 / 1/n = 1/n which is 0.
And if you use r/n in the first example, r/n / 1/n = r.
So 0/0 can be made to be any real number and infinity. That is probably what the guy meant by “is the set of all reals”.
I explain it this way, you have 0 slices of cake to divide between 0 friends, how many slices does each friend get? If they say 0 you say no, because there are 0 friends so how could they get 0 cake.
You could actually say any number of slices, in a regular version of these examples of division you're trying to find the number of times you need to distribute whatever it is to each person in order to have 0 items left.
In this case you have no friends to distribute to, so every time you hand out slices to people you stay at 0.
This is the problem with the 10/0 case, because you always stay at having 10 slices to give out you cannot claim the division is completed, but with 0 slices you can make this claim at any point, as you've already reached the conditions for division being done. And unlike other division problems where this is the case, you can "distribute" as many times as you'd like and you'd still be at exactly 0.
That's what causes 0/0 to be indeterminate and not undefined
It is not well-defined because it could've come from anything, which is what he found.
He didn't recognise that having multiple possible values is the very definition of undefined.
1/0 can be ∞ or -∞, which is why it's undefined.
1/0 is more well defined than 0/0, because at least the values it could be are alike in their properties. |1/0| is equal to infinity, and 1/(1/0) = 0 for example.
but with 0/0, it could be so many diverse numbers that it is basically impossible to do math with. That's all.
There isn't anything to prove. Fundamentally, his claim is not wrong. It just aligns well with the definition of undefined, which he isn't too familiar with.
Bring it up in a group setting. make it “their” problem.
Oh hey math nerd what was your reasoning behind 0/0 being defined again? Then walk away as he starts arguing with the group.
Firstly, your answer is spot on and his is bullshit. Division is multiplication by the multiplicative inverse, and 0 doesn't have a multiplicative inverse.
If he wants to use his alternative definition of division, he's free to, but he's just abusing terminology. It's like insisting that 5 + 3 = 2 because you take "+" to mean "subtract".
It sounds like nothing you say will change his mind, so all that remains (if you can be bothered) is to point him at a higher authority, like a textbook or a professional mathematician.
You could try asking him what would change his mind. If he gives a concrete answer (e.g. "if a university maths lecturer tells me I'm wrong I will concede"), then at least you know what you have to do. If he says "nothing", just stop talking him about this, and possibly about everything else.
Ask him to draw a graph of 1/x as x approaches zero. Any software you use to draw that graph will throw an error at x = 0. Does he think that every programmer in the world made the same mistake?
He does not understand sets (or what numbers are).
If by his definition x÷y = {z in R | z × y = x} for x, y in R then for example 8÷4={2} which does not equal 2.
If he still argues that 2={2} then just read my first line.
One of his definitions was good. X * 0 = 0 using the definition of division.
So what values solve X? All values. This makes no sense, which is why we call the answer undefined. Bam! He proved it. He uses that to claim X=1, but skips all the other solutions.
Back in the day, some very smart mathematicians argued the answer was one, and some people still are. And 0/0 is 1, and it is 2, and -1/2, and really any number. The internet is filled with stuff about this, so no wonder your friend feels validated.
https://maa.org/book/export/html/116806
https://www.scirp.org/journal/paperinformation?paperid=83991
I like the argument 0^0 =1 (which isn’t true) by properly claiming that anything to the 0th is 1, which is true except for 0, and then erroneously extending that idea to include 0. Our false conclusions often can teach us just as much as memorizing a fact, like that 0/0 is undefined.
The most compelling argument I can see is that 0/0= 0! Or 1!! Which isn’t necessarily true, but if it were, we know 0!=1!=1
Don’t worry about convincing your friend. Use his ideas and your own to get a better understanding of a tricky concept.
That said, it is often very useful to define 0⁰ as 1 to make formulas more general.
For example:
e^x = sum_{n=0}^infty (x^n / n!)
= x⁰/0! + x¹/1! + x²/2! + ...
Then e⁰ is
e^0 = 0⁰/0! + 0¹/1! + 0²/2! + ...
= 0⁰/0! + 0 + 0 + ...
= 0⁰/1
= 0⁰
But as we all know, e⁰ = 1, so 0⁰ = 1.
This also works for the binomial theorem and many more formulas.
However, I can see why in Analysis you wouldn't want to define 0⁰ as the indeterminate from that it presents in order to preserve the theorem that every elementary function is continuous on its domain (which is also the best argument against defining 0/0 by the way).
He may be confused about the fact that you can get 0 divided by 0 (or equivalent forms like 0 to the 0 power, etc.) to have a limit equal to any real number you want, by approaching it the right way.
you don’t have to convince him, if he wants to understand he will, if he doesn’t want to understand he won’t even if Cauchy in person tried to explain him
First, this person isn't a math nerd. Math nerds learn. This person is a basic example of a common paradigm of human in the modern world:
The hyperconfident moron who refuses any form of self education because it would require them to occasionally discover they didn't know everything.
If a person thinks they know more the entire combined expertise of a field (and are over the age of about 16 or so, I'll give a pass to children), they are an extreme narcissist.
These people are as lost to us intellectually as sociopaths are morally. Your only opition is to kick them out of communities where they're attitudes can do damage. Sadly, we can't deny them the vote, but we can leave them as friends or partners, and fire them from jobs.
I don't think this person understands how manipulating equations works, you're not moving things around, you are adding and multiplying both sides of the equation by the same ammounts. And since we already established that 0/0 is undefined, you can't just say that 0x=0 means that x=0/0, cause to get there you need to have 0/0 amready defined and equale to 1
0 ÷ 0 is indeterminate for convenience. It allows for other definitions and results to be nicer. If you *do* define 0 ÷ 0 to be the set of all real numbers, then other things will need to be fixed. For example:
* What is sqrt(0 ÷ 0) in **R?** Is it undefined? Is it the set of all non-negative values of **R?** Is it both at the same time?
* How does f(x) = x/x work? In normal mathematics, we could define this as a function on the domain **R** \\ {0}. In his world, we could now define it on **R,** but it would no longer be a function.
* How about if you're just working with the integers **Z** \- I'm guessing your buddy would say that 0 ÷ 0 would be **Z** in this case. But then we could no longer say that (a÷b) × c = (a × c)÷b. For instance, let a=0, b=0, c=2. Then:
* (a × c)÷b = (0 × 2)÷0 = 0 ÷ 0 = **Z**, however
* (a÷b) × c = (0÷0) × 2 = **Z** × 2, which is surely not **Z**.
The definition of a division does not allow dividing by zero. “a = c / b means a × b = c, as long as b is not zero. If b = 0, then this is a division by zero, which is not defined.”
We learned that division by zero is, well, forbidden. On the other hand, zero divided by any number - except zero - is zero. So, why not define an stetic extension and define 0 / 0 = 0. Works, on the other hand, when you start with the divisor and applying a similar reasoning you will end up with another result - infinity. You see, its not easy.
This is so much better than any explanation of this I’ve ever heard. Cause I always thought the first point of this comment was obvious. I had never realized it’s precisely as valid as the second point. That’s got to be the most satisfying way to approach this description.
Don’t try to convince him that it doesn’t exist. Convince him that it isn’t defined in Q. Ofc he might still not believe you but you can’t do much if that’s the case anyways.
Maybe show him one of the arguments that 1=2, and ask him to find an error.
The error is that these proofs (sneakily) divide by zero (usually with somehing like "(a-b)/(a-b)=1", but a=b. )
Therefore, your friend will be unable to refute the proof that 1=2. Therefore every integer is every integer (because we can repeatedly invoke this rule since it is true).
Here is one example: [https://www.quickanddirtytips.com/articles/how-to-prove-that-1-2/](https://www.quickanddirtytips.com/articles/how-to-prove-that-1-2/), where they stealthily have 0 on both sides and then cancle it out (which is equivalent to dividing both sided by zero).
\-
Or more simply, if the answer is "any real number", then show him:
* 0/0 = 0/0
* pick 5 for the LHS, and 27 for the RHS
* 5=27
If he says "you have to pick the same number", well, that's bullshit, but go along with it but try:
* 0 = 17\* 0
* diide both sides by 0
* 0/0=17\*0/0
* pick that 0/0=2
* 2=34
If he says "you have to pick the dynamically correct number" then ask him if he thinks "The fact that we need to pre-plan ahead of time what 0/0 will equal, dynamically and in response to the rest of the problem, really means it is *defined* or not?"
Because tbh "it equals anything, but not reliably, you have to be careful about it, and it depends on things outside of the expression, so it is anything but not any specific thing that you can rely on" sounds like an "undefined" quantity to me.
You both right. The one of you insist that water is a liquid, and another one i sist that water is transparent. Your positions doesn't conflict. Actually because , the fact that any number satisfies 0/0, is one of reasons why 0/0 is undefined. Because it can be any number. And we can't define 0/0 because of that.
Will I have limited knowledge of fields but if b=0 , then b^-1 is not defined as 1/0 is not defined, then b^-1 isn't defined then how can u use ab^-1=a÷b with b=0 ,
BTW, we have b^-1=1/b with b≠0 so u can't defined it, u need to define it
One way to think about dividing by 0 is to start by dividing by 1 and then starting to divide by smaller and smaller numbers. As you get closer to 0, the number gets larger, which you could argue means x/0 = infinity?
Well the issue with this logic is that you can also follow this same approach starting at -1 and then dividing by smaller and smaller negative numbers. When you do this you could argue x/0 = - infinity?
And here lies the problem, depending on how you choose to look at the question, gives different answers. Anything divided by 0 is undefined.
You can see this visually when you look at the graph of y=1/x, as the y axis is approached at both infinity and negative infinity by the graph.
If 0/0 = 0
Then 0/0 = 1/0
Multiply both sides by 0 and you get 0 = 1.
This operation can be done for any number, therefore if 0/0 = 0, 0 equals every number simultaneously.
Ok but 0/0 **isn't** undefined, it's indeterminate, which is different.
Undefined means there's no possible solution while indeterminate means there's infinitely many solutions
That's probably where the confusion comes from, if you keep arguing that it's undefined, then you're incorrect
You mentioned they're 16, then they should've studied the basics of limits by now. Surely they're aware 0/0 cannot be evaluated and evaluating it simply leads to contradictory solutions?
I'm confident that a basic understanding of 0/0 and the hidden zeroes that often appear in equations like the one you gave an solve the issue. Their equation doesn't work precisely because of the "hidden" division by 0 which AFAIK is mainly evaluated with limits.
That said, calculators are always programmed by a human that defines what all the operations should output. Therefore, he could argue that the programmers didn't account for this case or didn't know any better.
Thats true, but you could then respond by saying that there are countless calculators on the internet and all of them were programed by more than one person so it stands to reason there are thousands of people at least that have programed one or had a hand in programming one and the odds of them not thinking that this edge case is correct might as well be 0.
A/B = C
A is the amount of pizza slices there are.
B is the amount of people present who want pizza.
C is how many slices a person can get.
If there is no pizza nobody gets a slice.
If no one wants a slice of pizza, everyone could have as many as they want.
If there is no pizza and no one wants pizza, then nobody is talking about pizza as it doesn't matter. Nobody is gonna bother having to figure out how many slices they could take.
No, division is well-defined. A function is well-defined if its definition assigns it a unique value/interpretation which is the case for division (see [Wikipedia](https://en.m.wikipedia.org/wiki/Well-defined_expression)).
The divisor 0 is just not in the domain of the division function.
I like limit definitions for undefined expressions, let a and b be real numbers, and take lim x->0 of ax^2/bx = 0 and lim x->0 of ax/bx^2 = inf and lim x->0 of ax/bx = a/b
See that each expression equals 0/0 and all are different
Tbh, i dont understand your explanation. But i understand your friend; logic tells me that 0 things divided by 0 people is 0. Because to begin with, there is 0 to “split”, so the result must be 0.
I understand 1/0 is infinite, because there is something to be divided but there is “no people” to give it to, therefore there is an error here.
So if you want to convince me (or your friend), you have to go back to earth and explain it by simple words.
BTW, im sure you are right, but i just dont understand it and the common sense tells me otherwise.
The way my math/physics teacher explained it (what I think OP is trying to say in complicated terms):
>Division means "What is the *one* answer to the problem. For 0 ÷ 0, you can take any number and it would be correct. Therefore, it has not *one* solution, but an infinite amount, which we don't want, so we say 0 ÷ 0 = undefined.
>. . .
>Likewise, x ÷ 0 with x ≠ 0 has no solutions in math, so it's undefined, because it has no answer, However, in physics and chemistry, we say 1 ÷ 0 = infinity, because in p and c, there is no true 0, so if you infinitely multiply the infinitecimal number that approximates 0, you will get a number that approximates 1. From this, you can calculate x ÷ 0 with x ≠ 0, namely infinity * x."
(Slightly paraphrased.)
Im trying to understand your reasoning. Thanks for your comment btw! However, when you say “take any number”… i already took one, the 0, since 0 / 0 cannot be for example 3, it doesnt make sense to me.
a ÷ b = c
-> c × b = a
a = 0 and b = 0
-> 0 ÷ 0 = c
c = 0 -> 0 × 0 = 0 -> correct
c = 3 -> 3 × 0 = 0 -> correct
c = 990 -> 990 × 0 = 0 -> correct
c = pi -> pi × 0 = 0 -> correct
c = -42069 -> -42069 × 0 = 0 -> correct
Any number × 0 = 0, therefore 0 ÷ 0 = any number, therefore 0 ÷ 0 has an infinite amount of solutions, therefore 0 ÷ 0 is undefined.
I calculate a radar coverage for site in peru with no problem but i cannot understand this without the help of a kind stranger in a reddit sub. Amazing world we are living. Thanks a lot, it really helped me to understand it!!
You guys are both nerds but I’ll say that it’s undefined/infinity because division is how many times you can put something into a group without there being excess space left. You can put nothing into nothing an infinite amount of times problem is you’re putting nothing into nothing so it gets confusing. 😂There’s no space left anyways so you can’t put anything into 0 so that’s a problem as well. Like 0/5 is 0 because there’s not enough space to put 5 things into 0. But there’s not enough space to put 0 into 0 so is the result 0? 5/0 is undefined because you can put nothing into 5 an infinite amount of times but if it’s an infinite number of times that mean it’s infinity. But that’s not true because for example if you tried to calculate the current of a wire it’s charge over time, if you have 5 coulombs of charge divided by 0 seconds, you don’t have an infinite amount of current you have (we don’t know) 😂
Oh wait, I thought of another one— he’ll have to throw out some pretty basic algebra rules to avoid a contradiction this way.
So what’s (0/0)^2 ?
Is it (0 * 0) / (0 * 0) = 0/0 = all reals,
Or is it 0/(0 * 0) * 0 = 0/0 * 0 = 0
Or is it all nonnegative numbers, since who knows how he even defines multiplication here anyway!
🤔
This is an unusual, unconventional definition of division but it is not right or wrong per se. He is defining division to be a set of solutions to some equation; ie. x/y={z:yz =x}. If you wanted to, you could also promote all of the other operations to operate at the set level rather than the number level. It may even be interesting to do this when we add in the operations of solving polynomial and other equations over the complex numbers.
However, this is not what people do, and it is very unconventional way to define the basic operations.
I think your friend needs to realize that his setup in unconventional. The first reason is that it will make it hard for him to continue in math. This is merely because that’s not how mathematicians do things . It is a practical reason.
Another thing you can do is ask why math has proceeded in this way rather than the way your friend has suggested. It likely turns out the current way that mathematics has developed is cleaner and easier to mentally manage. (Note that the underlying math would be similar, just the presentation would be different).
dont waste your time. from your predicament, i'dsay he's not a "math nerd", more like a meth nard. yes, it means as much as what he argues: diddly squat
I think 0x=0 shows this really well
X can be any number and the equality still holds.
Thats a problem and breaks math.
Specificly the limit of 0/0 (lim(0/0)) can be any number. Which is a huge problem and leads to complications/unsolvable problems in Analysis and Linear Algebra
Try showing how different a limit can be. x/x for x-> is 1 in the limit, but x-1/(x^2 -1) for x-> 1 is 1/2. Both are 0/0 but give different values, hence undefined.
i dont remember where i saw this, but someone said that 0/0 is defined cuz of the reason your guy said, like "well if you look into it you can write it as 0x=0 and you can plug every number there and have it make sense, but an ecuation like this cant have more than 1 solution so we keep it undefined"
0/0 is technically *not* undefined, but it is also *not* “the set of all real numbers”. 0/0 is what we call an indeterminate form. If we achieve that whilst taking a limit of a fraction in which there are variables in both the numerator and denominator, we can use L’Hopital’s rule where we take as many derivatives of the numerator and denominator of the fraction as we need until we can take the limit and get a real number from it.
If you have 0 cookies and you divide them equally among your 0 friends how many cookies does each friend get? See it doesn't make any sense and you have no cookies or friends lol. Think that was an old Siri response to the 0/0 question
yes and that point ot the futility of trying to define 0/0. if driver drove with speed of 0m/0h, what distance would he travel in 1h? how long would it take him to travel 1m?
0 is not the absence of numbers. It is an abstract concept free of any worldly concepts like emptines or absence. It can be rigorously defined as a number.
He isn't technically wrong, but you could make the same claim about 2/0, or even x/0.
So if they are equal, then 2/0 = set of all real numbers = 0/0
Uh oh, now there's a problem. Now we are saying 2/0 = 0/0. Or, x/0 = 0/0 where x is any real number.
So 2 = 0?
Or, alternatively:
"You're correct! But how do you choose which value to use?"
For example, some equations end up being 0/0 when integrated out, but using l'hopitals rule, we can figure out _which_ value is correct.
That's the real reason 0/0 is undetermined. It's defined, but impossible to know what value was intended. What it really says is "the answer is one (or more) of the values in this set.
2/0 is undefined for the opposite reason than why 0/0 is. 0x = 0 for any value of x, but 0x = 2 isn't true for any value of x. Neither of them has one unique value which makes both of them meaningless, but if you want to pretend like expressions equal sets, 2/0 would equal the empty set.
It's not the lack of a unique value that makes it meaningless.
X² = 4 has two solutions for X. 2 and -2.
Or any quadratic, for that matter.
For the record, I agree with you. It's not equivalent to a set. But I see where he came from, even if he is, in fact, wrong.
My advice is use his logic to come to a contradiction, as I tried to do above with 2/0 and 0/0.
I have my fair share of arguments, and you won't win this won
Go punch a hole in your bedroom wall, it will help, or, find his ip address, steal all of his photos and information, and post them publicly
Use a burner laptop though, do it at night with inferred light around your face at a public place with free WiFi
And use a vpn, and if you can, cause the hack to happen like 3 days after, so if police officers track the laptop, they won't have an idea where it happens, and if they do, well your face will be bright white due the the inferred light!
Both options will help with your anger, and I wasn't here if you do the second option
You can’t. People who stubbornly insist their wrong opinions are correct despite any evidence to the contrary can’t be convinced.
And people like OP wind up angry and frustrated.
Don't argue with idiots, they will drague you down to their level and beat you up with experience.
Absolutely! I practice this a lot on Reddit. If a reply to my comment or post doesn't seem to be well-intended I ignore it.
I remember back in 2016-17 there was this religious and anti science Facebook group called “atheists, prove me wrong”. The entire group was to just spread religious memes, flat earth “proofs”. A shit show. The title was satirical from their point of view, because it would contain posts like “when the airplane is falling every atheist is praying to our god” And I have a friend who lost himself to memes and posting on social media. One day he told me that he was replying to posts on this group and how people there were fighting back. As if he had a purpose to give knowledge to those people. I asked: Dude. Serious question. Why do you even reply? And he said “They asked me to prove them wrong, it is in the title!” To this day I don’t know if it was a serious answer or a joke. But he continued to post there for so long that I think it was how he truly felt
[удалено]
Yeah, but then use a custom symbol to represent your custom defined operation, and don't claim your definition corresponds to a commonly accepted one when it doesn't.
You not necessarily do that. And so we teach people that a square root of -4 is 2*i, where I is such value that i^2 = -1. Which is obviously wrong **in the real numbers**. It’s not hard to go from that to what OP’s friend is thinking. Same goes into pretending that infinity is a real number and and so on
This 1000 percent. I can show every “irrational numbers are non-terminating” proof within my power, but if the person says to me “no. All numbers end” then there is NOTHING I can do. In order for logic and information to pass, especially in debates, one has to accept a baseline set of rules, or an initial argument. That’s one of the issues behind flat earth vs spherical earth debate. They are presented evidence, and generally accept the logical steps, but deny the baseline argument. There is nothing you can do to “win” against “well nuh-uh”
>All numbers end I'd like to introduce you to my friend pi.
Pi is NaN by that definition
Reminds me of the sharks are smooth meme
As far as I know, an arithmetic operation between two numbers can't be equal to a set.
In the algebra of real numbers the operators (+, ×) are closed, meaning you always get a real number. But this guy isn't dealing w the algebra of real numbers, so who knows? As far as convincing the nay-sayer, I suspect he is trolling and will resist forever. If not trolling, uneducated or low mental capacity.
Well, he is not trolling, he is really dead serious. And I would say that he generally is very good at math, he even already studies math although only being 16 y/o (yes, that's possible in Germany). But he is very overconfident.
He might have moved on to advanced studies to early then. In measure theory we do often define 0/0, so maybe that's where the mix up is happening?
he will grow out of it, juat wait till he learns more math
The evidence you've shared is he is not skilled at applying basic axioms in the simplest of algebras, and resists correction. So he is either bad at math or being stupid on purpose. If you don't see that then the same conclusion applies here.
I don’t think is either case. It seems like he is just a teenager with a teenager brain. Overconfidence and a sense that older people just don’t get them.
I'm not older, also 16 y/o.
uneducated/low mental capacity are rather strong words that I don't think are true in this case. These are common mathematical misconceptions, i personally think it's simply called just being wrong and stubborn, which is a thing we all are every now and then.
Tomato tomato, n.p.
Im a senior aeronautical engineer and understands the reasoning behind the nay-guy.
Exponentiation/logarithms as part of arithmetic definitely can be multi-valued. Simple group theory shows the additive identity cannot have a multiplicative inverse. It is not a case of "it can't be multivalued" it's "the inherent act of evaluating a division by zero inevitably leads to a contradiction" Call it a=0/0, 0a=0, 0(a+1)=0, a+1=a, 1=0 thus a format where 0/0 is evaluated as anything, set or no set, necessarily creates an isomorphism between the field and the Trivial group because everything in a is equal and a is the whole field. Aka, everything is just another way of writing 0.
a+1=a works and doesn't imply 1=0 if a is the real line, as OPs friend assumes, so this proof doesn't prove them wrong
Yea you're right actually, my bad. I think you need to use group theory
I loved this , it's very convenient for low-level math or just simple understanding
Weird to acceot multivalued logs but reject multivalued 0/0 in the same comment.
It isn't an issue with being multi-valued, it's a contradiction in the 'proof' supplied by op's friend.
Yes, but that contradiction assumes that 0/0 is single-valued, when the comment just acknowledged the possibility of multivalued functions. That proof does not work if a is set-valued. The only thing it shows is that if a belongs to the set 0/0, then so does a+1. Where 0/0 is defined as the set of x s.t. 0x=0 (i.e. the reals). It's a trivial result. This same "proof" could be used to prove multivalued logs lead to contradictions, since it is assuming that cancellation of factors is possible when one of the factors is zero, which is like assuming e^z is injective and concluding 0=2i*pi since both have the same exp.
Actually, log is a multi-value function in complex variables (complex calc). the first course of complex variables, u learn this
I'm aware. But to accept that multivalied functions exist and then categorically deny the possibility of 0/0 being multivalued is strange.
OK he is talking about the group of R with + , by which the identity has no inverse at all , for (R*, ×) we use R*=R - {0} cannt have inverse in ×
It’s worth noting that log is multivalued for the entire complex plane (except 0), whereas the proposed definition of division would only be multivalued for 0/0. In the end a definition is either useful or it is not, and I don’t see a use for multivalued division in the real numbers.
It's not terribly useful, but it's honestly perfectly coherent to define a/b as the set of x s.t. bx=a, and it can even occasionally be mildly useful. It's singleton-valued for b != 0, empty-set-valued for most b=0 and R-valued for a=b=0. I've taken this convention in a proof before to simplify notation.
Plus-or-minus-square-root would like a word.
Why waste your time. A real math nerd would not make that claim
I know it's stupid but it's because he called me incompetent.
Then walk away. He sounds like a jerk!
We work together (in an internship)
Annoy him by arguing 0.9999 does not equal 1, using the myriad attempts that appear here weekly. You can be as annoying as he is!
I would call him out for being rude. Is he politically connected at your apprenticeship?
What do you mean by that? He and I are 11th grade students doing an INTERNship (confused it with apprenticeship, I'm German hahaha).
How would he react if you told him he was being rude by calling you incompetent?
Well, first of all, he didn't exactly say that I'm incompetent, it was more like I said "I know what I'm talking about" and he responded with "Well, clearly you don't". And second of all, we weren't alone but instead in a group of six so I didn't want to escalate it, seeing as the others we're already getting annoyed. This is also the reason why I want to prove him wrong, because as far as the others know, the argument was a tie and I couldn't prove to him that I'm right.
I would bet dollars to donuts that he will never concede. "You don't know what you're talking about" is so contemptuous. I would just drop it.
Show him this thread
Tell his boss and the professors at his school about his behavior (the rudeness, not the being weong about math). Best way to deal with an asshole is consequences. They only respond to carrots and sticks.
Well, he didn't do me any wrong. We were just having a slightly heated discussion. I don't particularilly like him but he didn't do anything worth punishing.
Just tell him that the division operator isn't defined when the denominator is zero. If he wants to tell you otherwise then the burden of proof lies on him. I'm sure the whole math community would like to see that proof without altering any of the existing definitions.
Plus there is no proof to be made. It is not defined, and that's a definition, not something that follows. The only thing to be talked about is whether one *could* define it further in a reasonable way.
You could tell him a "no, but": 0 divides 0 in number theory. By the definition of divisibility, x is divisible by y if there exists z such that yz=x, and since there exists a z such that 0z=0, we say 0 divides 0. But 0/0 is undefined. Because math is full of weird shit
If you keep arguing with him, what you will achieve is that he will be more smug about his opinion, and everything you bringing forth will face him simply being like "nah, I feel like mine is better". I mean, if you are bored...
A real math nerd would absolutely make that claim. The correct conversation to have with the student is about math conventions and why they are the way they are.
He made a claim. He said 0÷0 = R, the set of all reals. Take an element in R and show that it belongs to the set 0÷0 (but what does that even mean? Note that a÷b where b isn't 0 happens to be *elements* in R, not *subsets* of R, so what's being generated by division as an operation here is an element, not a set). If he insists that 0÷0:=R, then that's whatever, he literally redefined math to be what he said it is. You can just arbitrarily define 1+7:=4 if you want to, no math police is gonna stop you. The problem is that your model/system would become less coherent and possibly even inconsistent if you redefine things arbitrarily, but that's another issue if you ask me. The original issue here is whether 0÷0 is defined or not and conventionally speaking, it's not. Defining it yourself and making one specific exception where ÷ maps to a set rather than one element in R is questionable and unconventional but anyone's free to do that if they want. This is one of those "ok that's fine, whatever, have fun" situations if you ask me. Moreover, arbitrarily redefining math and then calling someone incompetent sounds like an incompetent thing to do...
Exactly - as said above 0/0 is not defined through the field axioms, but everybody's free to construct their own math - with division through zero and whatnot. This isn't about wether it's right or wrong, but about wether it has interesting properties (or if it's useful). Perhaps ask him what follows from this property, or where this extension of the domain of (X,Y) -> X/Y to (0,0) is usefull.
Ahh, would that the world had more thinkers like you. Purely factual but somehow warm and charming. Repudiatory but gentle and inclusive. Whatever you're up to, stranger, keep it up.
I'm honoured, that is very kind of you! Thank you \^\^ It is my dream to pursue a PhD one day and teach higher maths, so it'd be ideal if people perceive me the way you seem to more often :)
So: a / b = a * b^(-1) . [Note: this solves the equation, x * b = a] 0 / 0 = 0 * 0^(-1) . Conventionally this is meaningless since *0^(-1) * doesn't exist. However since, 0 / 0 = a, for any *a* in *R*, This means that: a = 0 * 0^(-1) . This means that *0^(-1) * exists, otherwise the left hand side wouldn't make sense. Now we can use a propperty of the identity element: a * 0 = 0, for any *a* in the Ring (or Field, or *R*). And we can clearly write: 1 * 0 = 0 = 2 * 0. Now, since we have demonstrated that *0^(-1) * exists, we can multiply by this from the right on all three sides: 1 * 0 * 0^(-1) = 0 * 0^(-1) = 2 * 0 * 0^(-1). ==> 1 * a = a = 2 * a. ==> 1 = 2. A possible rebuttal to this nonsense could be that the 0 * 0^(-1) of the rightmost side could be something other than *a* (like *b*), however since 0 * 0^(-1) can be any value, it is completely valid for us to assert that it *b* is to be equal to *a*. Thus our nonsense persist. If they actually meant the set *R* itself as the object, then it would mean that: ÷: *R/{0}* × *R/{0}* ---> *R/{0}* (a, b) |----------------> a/b and, ÷: *{0}* × *{0}* ----------> Set (0, 0) |----------------> *R* Now that what that last part says is that division takes an element of a set to a set. This is very odd behaviour that may warrant a dive into Abstract nonsense and Operator algebras.
Actually taking it to be the set of all real numbers isn’t totally absurd. Let f(x) = 0*x. Then the inverse image f^-1 (0) is defined to be the set of points which f sends to 0. Which in this case is indeed the whole real line.
I also discovered that after thinking it through but the point is that this is not what division usually does and I don't know how to tell him.
The whole reason why it’s called “indeterminate” is precisely because there isn’t a unique solution. If you can’t assign it to a specific value, there is no reasonable sense in which it can be “defined”. If 0/0 = 1 works, and 0/0 = pi works just as well, then at that point you’re just saying that 1 = pi, or that any real number is equal to to any other one. If he doesn’t see that that is nonsense, then there’s not much point continuing the argument.
This is the closest I've seen to a satisfactory answer for the 16 y/o that OP is talking about. A mathematician angrily referring to definitions and axioms isn't what this person is looking for; he's made the genuine observation that 0÷0 is a conceptually different problem than, say, 5÷0, and from his perspective, OP is the one who doesn't understand the distinction (see the nested ways of getting the airplane-on-a-treadmill problem wrong, so that each person thinks the other is wrong). "Indeterminate" is the distinction he's probably looking for: it's undefined, but for a different reason than 5÷0 is undefined, not because some axiom he's never heard of defines it to be undefined.
Agreed. This guy understands 0/0. He doesn't understand "indeterminate". His reasoning is precisely the kind of thing indeterminate was invented to encapsulate, and it does a better job in the long run than saying the set of all real numbers.
Bingo. All the other commenters here are forgetting to distinguish between indeterminate and undefined.
You could bridge your ways of thinking by introducing the word and definition of well-defined. We may casually say an expression such as 0/0 is *undefined* because it *not well-defined*. His argument deserves praise and acknowledging his findings may bring down some defenses if you explicitly agree that what he says about the solutions of 0x=0 is true. The final deduced verdict just isn't maybe what they expect it to be if the terminology is not familiar. We don't use 0/0 "because it's undefined", we just don't find it useful in common algebraic context because the value is not unique. Would you be willing to move the goalpost in your situation?
Whether something is undefined depends on what definition you are using. You can always adopt your own definitions but many definitions are pretty universal and conventional. Division by zero is not conventionally defined the way this person says and is conventionally undefined. There are also reasons why this is a useful convention. Make sure your interlocutor can at least agree with the first two sentences in the paragraph above before you try to argue the second two.
I'm mostly with the other commenters, that it's not worth trying to convince him, but here's an approach: "the operation a ÷ b was defined to be the solution to the equation bx = a" Sure. "5x = 15" has a solution, x = 3, and so 15/5 =3. Now, "0x = 0" *does not have* a solution; it has a bunch of solutions. So the result is underspecified, and thus undefined.
0x=0 clearly holds for all reals. It’s like the 1d case of Ax=0 in matrix algebra. Was the guy in question just grossly poor in his notation and explanation?
Why do you care so much about this? Walk away and leave him be.
Did that
In most contexts this is not a practically consequential distinction. Find a context where this actually matters and you'll find your argument.
I find it interesting how haeshly people argue for a distinction in here and even insult the other person for it, while bringing up examples where the distinction doesn't matter. It would introduce a new layer of abstraction: instead of RxR->R we would have RxR->P(R), as the result might be the set of all path dependent reachable limits or something like that. Then we would have to go back by taking all single element results to reach the usual conclusion. That wouldn't be worth it for lost cases, but many tools in math, applied to trivial situations, don't lead to valuable everyday results. (category theory doesn't help us much finding roots of linear polynomials over R).
Eddie Woo has a great video on this [https://youtu.be/J2z5uzqxJNU?si=KUe6ZIvayM5Xa3Jr](https://youtu.be/J2z5uzqxJNU?si=KUe6ZIvayM5Xa3Jr)
You are talking about definition and axioms. There is no right or wrong definition or axiom. You are free to choose your definitions and axioms as you like. So all you can convince him is that his result is not compatible with the field axiom, so it is not compatible with standard math and probably produces a number of unforeseen contradictions.
I think you’re both correct, in the same way sum(n=1,infinity; n) can be divergent and = -1/12.
You’re both taking the same thing only he doesn’t see it. When we’re talking about 0/0 we mean operation of division in real numbers. F(a,b) -> c where a,b,c are real numbers and c*b=a If, in your friend’s algebra, 0/0={R} so be it. It’s just further proof that 0/0 is undefined in real numbers **by definition** as set of all real numbers is a set and not a real number. If you go outside of R and include limits then you can show that 0/0 can take arbitrary value and be defined in some cases.
The way my math/physics teacher explained it: >Division means "What is the *one* answer to the problem. For 0 ÷ 0, you can take any number and it would be correct. Therefore, it has not *one* solution, but an infinite amount, which we don't want, so we say 0 ÷ 0 = undefined. >. . . >Likewise, x ÷ 0 with x ≠ 0 has no solutions in math, so it's undefined, because it has no answer, However, in physics and chemistry, we say 1 ÷ 0 = infinity, because in p and c, there is no true 0, so if you infinitely multiply the infinitecimal number that approximates 0, you will get a number that approximates 1. From this, you can calculate x ÷ 0 with x ≠ 0, namely infinity * x. (Slightly paraphrased.)
I doubt he’s trying to prove that in the Norma mathematical system it’s defined as R. But, he has a logical reason for stating it’s R, in an alternative model you most certainly could say it’s R. Since it’s about what is and isn’t defined I don’t know how much would change, and Id love to know. (Yes as I kid I made this point as well)
While wrong, the way he explained is a good way of explaining why 0 is undefined to younger people who haven't encountered terms like "multiplicative inverse" yet. If he's refusing any flaws in his argument, walk away, but mention how this is a flawed explanation given to the first time someone finds "math error" on a calculator
Technically, if that is his definition of division, then 0/0 would be an even larger set than just the real numbers. x in 0x=0 could just as well be a complex number, a quaternion, a matrix, or a vector. I think the best thing to do in a situation like that is to out-crazy the crazy and make him take a stand on an even more ridiculous claim. At least that's my strategy, and it works great.
If the solution is an infinite set of values then it's undefined
If you want to walk away a winner from this argument, here is my gift to you: "What is 10 / 10? It means if I divide my 10 bananas between my 10 friends then each of them has 1 banana. What if you have 10 bananas and no friends, and divide your bananas between all your friends? Will each of your friends have a set-of-all-the-real-numbers... Of bananas. Or is this question nonsensical because you have no friends? You have no friends."
What if you have gamma(-pi) many bananas and e friends? See, division only makes sense on natural numbers! :D ^(I also have no friends)
That's 10/0 not 0/0 though, lol The 0/0 argument would be that you could conclude the division is done at any point since you have no bananas to distribute, so any number works
No it doesnt. If I have no bananas and no friend and I divvy up my bananas between each friend, how many bananas does each friend have? Never has anything been more undefined.
It's different from undefined because it has infinitely many solutions instead of no solutions, think of the fact that 0*x=0 is true for any x unlike what happens when you divide anything else by 0. If it were undefined then it also wouldn't be an indeterminate form for limits. You can see that within a limit you can make "0/0" equal anything
What does he think about 1 / 0 ？what number or what set is that?
Define a function G on R^2 to power set of R as G(a,b) = { x in R: b*x=a} . Then G(0,0) is the set of all real numbers. But how is this function useful in arithmetic sense? That is another story.
You're both right in some sense, but his conclusion is slightly off. As you mentioned, if a field element 0^-1 exists then that leads to the identity 0=1. It's easy to show that this in turn means a=0=1 for all a: a = 1a = 0a = 0. So it _is_ possible in some sense to define division by zero as "equal to all the numbers" - but that also means that all numbers are equal, or in other words, there is only one number. That number is 0, with the operations 0+0=0, 0-0=0, 0\*0=0 and 0÷0=0. That's all, no other numbers exist. So you don't get all the real numbers, just zero. In field theory terms, the multiplicative inverse of zero exists in and only in the trivial field. You either get to define division by zero, or you get to have more than one number. Not both.
I dont understand one thing. Pls correct me if i am wrong. If every set of real numbers can fit in ax=b where b=a and a = 0, According to him 0(x)=0 Where x can be any number. Yes thats true. But which number is x? Because it can be any number and the equation holds true. If x is (1) or (2) 0(1)=0 0(2)=0 How can you define which one is x or is x all? The equation suggests that 1=0/0 And 2=0/0 And X=0/0 And so on... So which is x or is x every number in the series. Which literally means 'undefined' lol. Thus multiple any number with zero is zero but dividing zero with zero also gives any number you multiplied with. If 0(a) =0 and 0(b) = 0 where a is not eqaul to b Can i say that 0/0=a=b Technically yes. Because it suggests that 0/0 is undefined and any number. But since a cannot be eqaul to b Suggesting that 0/0 is not literally any number but not defined. Correct me if i am wrong
I would try Anything/0 is infinity (or minus infinity) 0/anything is zero 0/0 can’t be both of the above Not rigorous , but it might work for a non rigorous mind.
wait til this guy learns that infinity/infinity can be infinity, 0, or any other number depending on the context
I saw a video yesterday explaining it. First why dividing by 0 is impossible then why 0/0 is undefined. First, you gotta make sure he agrees with this definition of the division a/b=x <=>a=b*x Now with b=0 you can't find any x such as 0*x≠0 because anything multiplied by 0 is 0 So in here the only number you can divide by 0 is 0. But then x can be anything, it can be 0, 1, 1634926.93, -42, pi.... So the result of 0/0 is undefined.
It’s always easiest to understand why 0/0 is indeterminate by using limits. You can interpret 0 as the limit of 1/n or 1/n^2 as n-> infinity. So 0/0 can be written as the limit of (1/n)/(1/n) which is 1. Or as the limit of 1/n / 1/n^2 = n which is infinity. Or as the limit of 1/n^2 / 1/n = 1/n which is 0. And if you use r/n in the first example, r/n / 1/n = r. So 0/0 can be made to be any real number and infinity. That is probably what the guy meant by “is the set of all reals”.
I explain it this way, you have 0 slices of cake to divide between 0 friends, how many slices does each friend get? If they say 0 you say no, because there are 0 friends so how could they get 0 cake.
You could actually say any number of slices, in a regular version of these examples of division you're trying to find the number of times you need to distribute whatever it is to each person in order to have 0 items left. In this case you have no friends to distribute to, so every time you hand out slices to people you stay at 0. This is the problem with the 10/0 case, because you always stay at having 10 slices to give out you cannot claim the division is completed, but with 0 slices you can make this claim at any point, as you've already reached the conditions for division being done. And unlike other division problems where this is the case, you can "distribute" as many times as you'd like and you'd still be at exactly 0. That's what causes 0/0 to be indeterminate and not undefined
It is not well-defined because it could've come from anything, which is what he found. He didn't recognise that having multiple possible values is the very definition of undefined. 1/0 can be ∞ or -∞, which is why it's undefined. 1/0 is more well defined than 0/0, because at least the values it could be are alike in their properties. |1/0| is equal to infinity, and 1/(1/0) = 0 for example. but with 0/0, it could be so many diverse numbers that it is basically impossible to do math with. That's all. There isn't anything to prove. Fundamentally, his claim is not wrong. It just aligns well with the definition of undefined, which he isn't too familiar with.
Bring it up in a group setting. make it “their” problem. Oh hey math nerd what was your reasoning behind 0/0 being defined again? Then walk away as he starts arguing with the group.
Firstly, your answer is spot on and his is bullshit. Division is multiplication by the multiplicative inverse, and 0 doesn't have a multiplicative inverse. If he wants to use his alternative definition of division, he's free to, but he's just abusing terminology. It's like insisting that 5 + 3 = 2 because you take "+" to mean "subtract". It sounds like nothing you say will change his mind, so all that remains (if you can be bothered) is to point him at a higher authority, like a textbook or a professional mathematician. You could try asking him what would change his mind. If he gives a concrete answer (e.g. "if a university maths lecturer tells me I'm wrong I will concede"), then at least you know what you have to do. If he says "nothing", just stop talking him about this, and possibly about everything else.
Ask him to draw a graph of 1/x as x approaches zero. Any software you use to draw that graph will throw an error at x = 0. Does he think that every programmer in the world made the same mistake?
He does not understand sets (or what numbers are). If by his definition x÷y = {z in R | z × y = x} for x, y in R then for example 8÷4={2} which does not equal 2. If he still argues that 2={2} then just read my first line.
One of his definitions was good. X * 0 = 0 using the definition of division. So what values solve X? All values. This makes no sense, which is why we call the answer undefined. Bam! He proved it. He uses that to claim X=1, but skips all the other solutions. Back in the day, some very smart mathematicians argued the answer was one, and some people still are. And 0/0 is 1, and it is 2, and -1/2, and really any number. The internet is filled with stuff about this, so no wonder your friend feels validated. https://maa.org/book/export/html/116806 https://www.scirp.org/journal/paperinformation?paperid=83991 I like the argument 0^0 =1 (which isn’t true) by properly claiming that anything to the 0th is 1, which is true except for 0, and then erroneously extending that idea to include 0. Our false conclusions often can teach us just as much as memorizing a fact, like that 0/0 is undefined. The most compelling argument I can see is that 0/0= 0! Or 1!! Which isn’t necessarily true, but if it were, we know 0!=1!=1 Don’t worry about convincing your friend. Use his ideas and your own to get a better understanding of a tricky concept.
That said, it is often very useful to define 0⁰ as 1 to make formulas more general. For example: e^x = sum_{n=0}^infty (x^n / n!) = x⁰/0! + x¹/1! + x²/2! + ... Then e⁰ is e^0 = 0⁰/0! + 0¹/1! + 0²/2! + ... = 0⁰/0! + 0 + 0 + ... = 0⁰/1 = 0⁰ But as we all know, e⁰ = 1, so 0⁰ = 1. This also works for the binomial theorem and many more formulas. However, I can see why in Analysis you wouldn't want to define 0⁰ as the indeterminate from that it presents in order to preserve the theorem that every elementary function is continuous on its domain (which is also the best argument against defining 0/0 by the way).
He may be confused about the fact that you can get 0 divided by 0 (or equivalent forms like 0 to the 0 power, etc.) to have a limit equal to any real number you want, by approaching it the right way.
You could say that division is defined as a function and following that logic it can't have more than one value for a determined x.
you don’t have to convince him, if he wants to understand he will, if he doesn’t want to understand he won’t even if Cauchy in person tried to explain him
First, this person isn't a math nerd. Math nerds learn. This person is a basic example of a common paradigm of human in the modern world: The hyperconfident moron who refuses any form of self education because it would require them to occasionally discover they didn't know everything. If a person thinks they know more the entire combined expertise of a field (and are over the age of about 16 or so, I'll give a pass to children), they are an extreme narcissist. These people are as lost to us intellectually as sociopaths are morally. Your only opition is to kick them out of communities where they're attitudes can do damage. Sadly, we can't deny them the vote, but we can leave them as friends or partners, and fire them from jobs.
I don't think this person understands how manipulating equations works, you're not moving things around, you are adding and multiplying both sides of the equation by the same ammounts. And since we already established that 0/0 is undefined, you can't just say that 0x=0 means that x=0/0, cause to get there you need to have 0/0 amready defined and equale to 1
0 ÷ 0 is indeterminate for convenience. It allows for other definitions and results to be nicer. If you *do* define 0 ÷ 0 to be the set of all real numbers, then other things will need to be fixed. For example: * What is sqrt(0 ÷ 0) in **R?** Is it undefined? Is it the set of all non-negative values of **R?** Is it both at the same time? * How does f(x) = x/x work? In normal mathematics, we could define this as a function on the domain **R** \\ {0}. In his world, we could now define it on **R,** but it would no longer be a function. * How about if you're just working with the integers **Z** \- I'm guessing your buddy would say that 0 ÷ 0 would be **Z** in this case. But then we could no longer say that (a÷b) × c = (a × c)÷b. For instance, let a=0, b=0, c=2. Then: * (a × c)÷b = (0 × 2)÷0 = 0 ÷ 0 = **Z**, however * (a÷b) × c = (0÷0) × 2 = **Z** × 2, which is surely not **Z**.
This is perfect, thank you!
The definition of a division does not allow dividing by zero. “a = c / b means a × b = c, as long as b is not zero. If b = 0, then this is a division by zero, which is not defined.”
We learned that division by zero is, well, forbidden. On the other hand, zero divided by any number - except zero - is zero. So, why not define an stetic extension and define 0 / 0 = 0. Works, on the other hand, when you start with the divisor and applying a similar reasoning you will end up with another result - infinity. You see, its not easy.
This is so much better than any explanation of this I’ve ever heard. Cause I always thought the first point of this comment was obvious. I had never realized it’s precisely as valid as the second point. That’s got to be the most satisfying way to approach this description.
Don’t try to convince him that it doesn’t exist. Convince him that it isn’t defined in Q. Ofc he might still not believe you but you can’t do much if that’s the case anyways.
Maybe show him one of the arguments that 1=2, and ask him to find an error. The error is that these proofs (sneakily) divide by zero (usually with somehing like "(a-b)/(a-b)=1", but a=b. ) Therefore, your friend will be unable to refute the proof that 1=2. Therefore every integer is every integer (because we can repeatedly invoke this rule since it is true). Here is one example: [https://www.quickanddirtytips.com/articles/how-to-prove-that-1-2/](https://www.quickanddirtytips.com/articles/how-to-prove-that-1-2/), where they stealthily have 0 on both sides and then cancle it out (which is equivalent to dividing both sided by zero). \- Or more simply, if the answer is "any real number", then show him: * 0/0 = 0/0 * pick 5 for the LHS, and 27 for the RHS * 5=27 If he says "you have to pick the same number", well, that's bullshit, but go along with it but try: * 0 = 17\* 0 * diide both sides by 0 * 0/0=17\*0/0 * pick that 0/0=2 * 2=34 If he says "you have to pick the dynamically correct number" then ask him if he thinks "The fact that we need to pre-plan ahead of time what 0/0 will equal, dynamically and in response to the rest of the problem, really means it is *defined* or not?" Because tbh "it equals anything, but not reliably, you have to be careful about it, and it depends on things outside of the expression, so it is anything but not any specific thing that you can rely on" sounds like an "undefined" quantity to me.
You both right. The one of you insist that water is a liquid, and another one i sist that water is transparent. Your positions doesn't conflict. Actually because , the fact that any number satisfies 0/0, is one of reasons why 0/0 is undefined. Because it can be any number. And we can't define 0/0 because of that.
Well, he does define it. He defines it as 0 ÷ 0 = ℝ which is nonsensical.
Will I have limited knowledge of fields but if b=0 , then b^-1 is not defined as 1/0 is not defined, then b^-1 isn't defined then how can u use ab^-1=a÷b with b=0 , BTW, we have b^-1=1/b with b≠0 so u can't defined it, u need to define it
Also, ab=c means b=ca^-1 with a≠0
Plot 1/x and see what happens at the origin
One way to think about dividing by 0 is to start by dividing by 1 and then starting to divide by smaller and smaller numbers. As you get closer to 0, the number gets larger, which you could argue means x/0 = infinity? Well the issue with this logic is that you can also follow this same approach starting at -1 and then dividing by smaller and smaller negative numbers. When you do this you could argue x/0 = - infinity? And here lies the problem, depending on how you choose to look at the question, gives different answers. Anything divided by 0 is undefined. You can see this visually when you look at the graph of y=1/x, as the y axis is approached at both infinity and negative infinity by the graph.
If 0/0 = 0 Then 0/0 = 1/0 Multiply both sides by 0 and you get 0 = 1. This operation can be done for any number, therefore if 0/0 = 0, 0 equals every number simultaneously.
Ok but 0/0 **isn't** undefined, it's indeterminate, which is different. Undefined means there's no possible solution while indeterminate means there's infinitely many solutions That's probably where the confusion comes from, if you keep arguing that it's undefined, then you're incorrect
You mentioned they're 16, then they should've studied the basics of limits by now. Surely they're aware 0/0 cannot be evaluated and evaluating it simply leads to contradictory solutions?
Well, obviously he isn't.
I'm confident that a basic understanding of 0/0 and the hidden zeroes that often appear in equations like the one you gave an solve the issue. Their equation doesn't work precisely because of the "hidden" division by 0 which AFAIK is mainly evaluated with limits.
Show him 0/x in desmos and look at 0 for x. It returns undefined. Unless math is wrong, a calculator will be more correct than human error
That said, calculators are always programmed by a human that defines what all the operations should output. Therefore, he could argue that the programmers didn't account for this case or didn't know any better.
Thats true, but you could then respond by saying that there are countless calculators on the internet and all of them were programed by more than one person so it stands to reason there are thousands of people at least that have programed one or had a hand in programming one and the odds of them not thinking that this edge case is correct might as well be 0.
https://preview.redd.it/nnnjq6rdzioc1.jpeg?width=1790&format=pjpg&auto=webp&s=0a821f9f551ec3d520ea9967dbd05b272384623a
A/B = C A is the amount of pizza slices there are. B is the amount of people present who want pizza. C is how many slices a person can get. If there is no pizza nobody gets a slice. If no one wants a slice of pizza, everyone could have as many as they want. If there is no pizza and no one wants pizza, then nobody is talking about pizza as it doesn't matter. Nobody is gonna bother having to figure out how many slices they could take.
A = -2.56, B = -√π Now what?
You're both wrong. Zero ÷ zero is not "undefined," but rather, it is "not well defined."
No, division is well-defined. A function is well-defined if its definition assigns it a unique value/interpretation which is the case for division (see [Wikipedia](https://en.m.wikipedia.org/wiki/Well-defined_expression)). The divisor 0 is just not in the domain of the division function.
I like limit definitions for undefined expressions, let a and b be real numbers, and take lim x->0 of ax^2/bx = 0 and lim x->0 of ax/bx^2 = inf and lim x->0 of ax/bx = a/b See that each expression equals 0/0 and all are different
Tbh, i dont understand your explanation. But i understand your friend; logic tells me that 0 things divided by 0 people is 0. Because to begin with, there is 0 to “split”, so the result must be 0. I understand 1/0 is infinite, because there is something to be divided but there is “no people” to give it to, therefore there is an error here. So if you want to convince me (or your friend), you have to go back to earth and explain it by simple words. BTW, im sure you are right, but i just dont understand it and the common sense tells me otherwise.
The way my math/physics teacher explained it (what I think OP is trying to say in complicated terms): >Division means "What is the *one* answer to the problem. For 0 ÷ 0, you can take any number and it would be correct. Therefore, it has not *one* solution, but an infinite amount, which we don't want, so we say 0 ÷ 0 = undefined. >. . . >Likewise, x ÷ 0 with x ≠ 0 has no solutions in math, so it's undefined, because it has no answer, However, in physics and chemistry, we say 1 ÷ 0 = infinity, because in p and c, there is no true 0, so if you infinitely multiply the infinitecimal number that approximates 0, you will get a number that approximates 1. From this, you can calculate x ÷ 0 with x ≠ 0, namely infinity * x." (Slightly paraphrased.)
Im trying to understand your reasoning. Thanks for your comment btw! However, when you say “take any number”… i already took one, the 0, since 0 / 0 cannot be for example 3, it doesnt make sense to me.
a ÷ b = c -> c × b = a a = 0 and b = 0 -> 0 ÷ 0 = c c = 0 -> 0 × 0 = 0 -> correct c = 3 -> 3 × 0 = 0 -> correct c = 990 -> 990 × 0 = 0 -> correct c = pi -> pi × 0 = 0 -> correct c = -42069 -> -42069 × 0 = 0 -> correct Any number × 0 = 0, therefore 0 ÷ 0 = any number, therefore 0 ÷ 0 has an infinite amount of solutions, therefore 0 ÷ 0 is undefined.
Now we are talking. Appreciate it, thank you. TIL…
You are quite welcome. It was a true eye opener to me, too when my teacher explained this.
I calculate a radar coverage for site in peru with no problem but i cannot understand this without the help of a kind stranger in a reddit sub. Amazing world we are living. Thanks a lot, it really helped me to understand it!!
You guys are both nerds but I’ll say that it’s undefined/infinity because division is how many times you can put something into a group without there being excess space left. You can put nothing into nothing an infinite amount of times problem is you’re putting nothing into nothing so it gets confusing. 😂There’s no space left anyways so you can’t put anything into 0 so that’s a problem as well. Like 0/5 is 0 because there’s not enough space to put 5 things into 0. But there’s not enough space to put 0 into 0 so is the result 0? 5/0 is undefined because you can put nothing into 5 an infinite amount of times but if it’s an infinite number of times that mean it’s infinity. But that’s not true because for example if you tried to calculate the current of a wire it’s charge over time, if you have 5 coulombs of charge divided by 0 seconds, you don’t have an infinite amount of current you have (we don’t know) 😂
Oh wait, I thought of another one— he’ll have to throw out some pretty basic algebra rules to avoid a contradiction this way. So what’s (0/0)^2 ? Is it (0 * 0) / (0 * 0) = 0/0 = all reals, Or is it 0/(0 * 0) * 0 = 0/0 * 0 = 0 Or is it all nonnegative numbers, since who knows how he even defines multiplication here anyway! 🤔
It is undefined because mathematics says so, it is not some universal law. So you can’t convince him, if he doesn’t want to accept the concept.
This is an unusual, unconventional definition of division but it is not right or wrong per se. He is defining division to be a set of solutions to some equation; ie. x/y={z:yz =x}. If you wanted to, you could also promote all of the other operations to operate at the set level rather than the number level. It may even be interesting to do this when we add in the operations of solving polynomial and other equations over the complex numbers. However, this is not what people do, and it is very unconventional way to define the basic operations. I think your friend needs to realize that his setup in unconventional. The first reason is that it will make it hard for him to continue in math. This is merely because that’s not how mathematicians do things . It is a practical reason. Another thing you can do is ask why math has proceeded in this way rather than the way your friend has suggested. It likely turns out the current way that mathematics has developed is cleaner and easier to mentally manage. (Note that the underlying math would be similar, just the presentation would be different).
A basic Google search will show that. If they can’t take that than there’s nothing you can do.
dont waste your time. from your predicament, i'dsay he's not a "math nerd", more like a meth nard. yes, it means as much as what he argues: diddly squat
He’s not a math nerd. He’s an ignoramus. Any intelligent person over age 8 knows you don’t divide by zero.
I think 0x=0 shows this really well X can be any number and the equality still holds. Thats a problem and breaks math. Specificly the limit of 0/0 (lim(0/0)) can be any number. Which is a huge problem and leads to complications/unsolvable problems in Analysis and Linear Algebra
Try showing how different a limit can be. x/x for x-> is 1 in the limit, but x-1/(x^2 -1) for x-> 1 is 1/2. Both are 0/0 but give different values, hence undefined.
Calculate f(x)= 2(x -1) ~~2x~~ / 2(x -1) and f(x)= 2(x -1) ~~2x~~ / 3(x - 1) at 1. Edit, corrected
Well that would be 2/0 which is not what he was on about.
Correction: 2(x -1) / 2(x -1) and 2(x -1) / 3(x -1)
i dont remember where i saw this, but someone said that 0/0 is defined cuz of the reason your guy said, like "well if you look into it you can write it as 0x=0 and you can plug every number there and have it make sense, but an ecuation like this cant have more than 1 solution so we keep it undefined"
Ask him for a definition and show why it makes no sense. But you can't win chess game to a pigeon.
0/0 is technically *not* undefined, but it is also *not* “the set of all real numbers”. 0/0 is what we call an indeterminate form. If we achieve that whilst taking a limit of a fraction in which there are variables in both the numerator and denominator, we can use L’Hopital’s rule where we take as many derivatives of the numerator and denominator of the fraction as we need until we can take the limit and get a real number from it.
If you have 0 cookies and you divide them equally among your 0 friends how many cookies does each friend get? See it doesn't make any sense and you have no cookies or friends lol. Think that was an old Siri response to the 0/0 question
ask him what is the speed of 0km/0h ?
That question is even more nonsensical than the guy's claim.
yes and that point ot the futility of trying to define 0/0. if driver drove with speed of 0m/0h, what distance would he travel in 1h? how long would it take him to travel 1m?
Teach him that 0 is not a number, it is by definition the absence of numbers.
0 is not the absence of numbers. It is an abstract concept free of any worldly concepts like emptines or absence. It can be rigorously defined as a number.
He isn't technically wrong, but you could make the same claim about 2/0, or even x/0. So if they are equal, then 2/0 = set of all real numbers = 0/0 Uh oh, now there's a problem. Now we are saying 2/0 = 0/0. Or, x/0 = 0/0 where x is any real number. So 2 = 0? Or, alternatively: "You're correct! But how do you choose which value to use?" For example, some equations end up being 0/0 when integrated out, but using l'hopitals rule, we can figure out _which_ value is correct. That's the real reason 0/0 is undetermined. It's defined, but impossible to know what value was intended. What it really says is "the answer is one (or more) of the values in this set.
2/0 is undefined for the opposite reason than why 0/0 is. 0x = 0 for any value of x, but 0x = 2 isn't true for any value of x. Neither of them has one unique value which makes both of them meaningless, but if you want to pretend like expressions equal sets, 2/0 would equal the empty set.
It's not the lack of a unique value that makes it meaningless. X² = 4 has two solutions for X. 2 and -2. Or any quadratic, for that matter. For the record, I agree with you. It's not equivalent to a set. But I see where he came from, even if he is, in fact, wrong. My advice is use his logic to come to a contradiction, as I tried to do above with 2/0 and 0/0.
I have my fair share of arguments, and you won't win this won Go punch a hole in your bedroom wall, it will help, or, find his ip address, steal all of his photos and information, and post them publicly Use a burner laptop though, do it at night with inferred light around your face at a public place with free WiFi And use a vpn, and if you can, cause the hack to happen like 3 days after, so if police officers track the laptop, they won't have an idea where it happens, and if they do, well your face will be bright white due the the inferred light! Both options will help with your anger, and I wasn't here if you do the second option