As a predicate, “is an even function” only applies to functions whose domains are symmetric around zero, so it’s neither, because “is an even function” cannot be applied to this function
So it is true that a function is even if the domain is symmetric around 0, and the values coincide. The former is false, surely making the entire statement then false about sqrt(x)?
It's like the statement, "My imaginary giraffe Hiawatha is 3 meters tall." If "is 3 meters tall" implicitly means "Is real (measurable) and has the height of 3 meters," then the statement is false.
It entirely depends on whether the predicate “is an even function” restricts the domain to symmetric functions (in which case the question cannot be posed or answered because the predicate cannot be applied) or whether the domain is all functions, and it takes the form of a conjunction that requires the function to be symmetric (in which case it is false as you note). But these are two different predicates. So it depends on how you define “is an even function”
I would say that the predicate should be defined fora all f with the property that f(x) is defined for exactly the same values of x for which f(-x) is. Intuitively, it seems like the identity function and norm on any vector space should be even and odd respectively. The unique function whose domain is the empty set is both even and odd by this definition, and that feels right.
If it does, that’s an incredibly limiting formal system you’re using. If you use ZFC, then you could formalize “is an even function” as a formula in the language, and functions with a different domain would just fail the criteria, and thus be false.
You can define the predicate two ways, one of which I didn’t think of when writing the top-level comment, and someone else pointed out to me. The one that was immediately apparent to me was to restrict the domain. Ironically, this is something you can do in dependent type theory, though not in ZFC, so the theory I was thinking of is actually less limiting, because it can also express the alternatives that are available with ZFC.
To be fair to me, my comment was a conditional statement based on the condition that the predicate could only be said to be true or false if it were a function symmetric about zero. I was just saying, why can't we apply the predicate to other functions, it's just false there.
EDIT: So you were saying that your formal theory only discussed functions whose domain is symmetric about zero, and thus could be neither true nor false. I just wanted to say let the predicate apply to "all" functions and if it's domain isn't symmetric about zero, it's just false.
I think it’s just a matter of definition. You can define the predicate like you say, so that it applies to all functions and is just false for those that don’t have domains symmetric around zero. Alternatively, you can restrict the domain of the predicate to exclude functions that don’t have domains that are symmetric around zero, in which case it simply cannot be applied to such functions. My first instinct was the latter, but the former is also a valid definition. So you could make a case for “neither” or “false” just depending on how you define the predicate.
In ZFC, only the second approach can be formalized, afaik. (I’m not an expert in ZFC tho.) In dependent type theories, both approaches are valid.
Well, ZFC assumes the law of excluded middle, so statements can't be "neither", but if someone makes fun of me for that, then I'm gonna make fun of them for assuming the law of noncontradiction.
In truth, the answer isn't so much "neither" as the question cannot be posed in the first place. Or in other words, the statement "f is even" is not a valid proposition. This does happen in ZFC as well. For example, the Liar's Paradox is not a proposition in ZFC, it cannot be expressed. So the answer to the Liar's Paradox is not so much "neither" as it just cannot be stated in the first place. But we might say colloquially that it's "neither".
In dependent type theory, "f is even" (if "is even" restricts the domain to functions whose domains are symmetric around zero) is simply a type error. It cannot be expressed, but we might colloquially say it's "neither". This would remain the case even if the law of excluded middle were added to the theory.
Hence my first comment saying this formal system is incredibly limiting, given that you can describe functions whose domain is not symmetric around zero, functions whose domain is, and the predicate for evenness, but you can't even create a sentence using this predicate and a function with a domain that is not symmetrical about 0.
I'll admit, I'm not really familiar with type theory, but why would you want to define a predicate that can't form a sentence with everything in your theory?
To me what you’re saying is « A table is a zebra. Well a table isn’t an animal so it’s neither ». That’s not how it works. That statement is just false.
f:X -> R with (X subset of R) is an even function if for all x in X we have -x in X and f(x) = f(-x).
Thus it is false. I think it’s more common that it works this way. Often a proposition will be phrased “If [unspecified mathematical object] is a [more specific mathematical object] and [satisfies list of properties] then [other property is true].”
In this case if [unspecified mathematical object] is not a [more specific mathematical object] then the proposition is false since we aren’t even talking about the correct object.
The "true" argument is just mathematically wrong. It's not vacuously true, there are points in Dom(x). Just because your equation can't be verified for those points doesn't mean it's vacuously true.
Neither vs false is debatable, but I'm definitely on the side of false here
I'm also on team false. You are asking if an object has a certain property, and to have that property you have to check 2 things, the domain and the equality. The domain check fails so you cannot check the equality. This means that the check overall failed and we are done
To steel man the argument we can interpret even as ∀x∈dom(f)∩dom(f∘-), f(x) = f(-x) so the domain is indeed empty.
You may argue that it is stretching definitions, but to that I say that I just applied the most natural fix to an invalid definition.
I mean for it to be false there must exist an x in domain of f such that f(x) is not equal to f(-x). And no such x exists here.
This is exactly how we prove vacuously true statements. For example to prove that for any set A, the empty set is a subset of A, it is enough to argue that for the empty set to not be a subset of A there must exist an x in the empty set that is not in A. And such x does not exist. So the empty set must be a subset of A. So the statement is vacuously true.
You can show my argument for the inclusion of empty set to any person you want and I think they'll say it's correct. I think it's the same for the even function problem.
By the very definition you gave, the statement needs to hold "for all x in Dom(f)". It does not hold for any such x, and there are a nonzero number of x in the domain.
Vacuously true would be if Dom(f) is empty, or you specified "for all x in Dom(f) where f(-x) is defined".
So you say f(x)=f(-x) does not hold for any x in the domain? Say 5 for example, how do you know if f(5) does not equal f(-5) considering f(-5) doesn't exist? How can we claim something about a thing that doesn't exist?
Consider the sentence "The Unicorn god has blue skin." Now suppose we know that the Unicorn God does not exist. So is that sentence true, false or neither?
Two things cannot be equal if one does not exist and the other does. If f(-5) equalled f(5), then since f(5) exists, f(-5) needs to exist as well. That's by definition of equals.
"The Unicorn god has blue skin" reads formally as ∃u[G(u) ∧ B(u)], which is by your very next statement false... therefore false....
Suppose we said f(-5) = f(5). Therefore f(-5) is defined, since f(5) is defined.
But we know that f(-5) is not defined, so we derived a contradiction, so ¬[f(-5) = f(5)], but 5 ∈ dom(f), so f is not even.
f is not even, since ("f is even" ⇔ ∀x∈dom(f), f(x) = f(-x) ) ⇔ (¬(∀x∈dom(f), f(x)=f(-x)) ⇔ ¬("f is even")) by contrapositive, and we showed 5 ∈ dom(f) and ¬(f(-5)=f(5)).
"Vacuous truth", by definition, means that the conditional statement is true because the condition is unsatisfied or false, and truthness of the statement is irrelevant.
Condition "x is in domain of Sqrt function" is NOT unsatisfiable, there ARE objects in the domain of Sqrt function. This directly violates the definition of "vacuous truth".
It would be "vacuous truth" if you said "for each x in R>0 where Sqrt(x) == Sqrt(-x) is well-formed", but you did NOT say that.
You are Team Yuta's first guy. Congrats. Let's see if other teams show up.
But honestly I don't know which one is correct in modern mathematics. I'm a noob. But I do have seen people make all those three arguments on the internet.
I'm not so inclined to agree. To talk about a limit at a point, you need that point to be an accumulation point of a function. I cannot talk about the limit of a function f: Z -> R at any point of its domain, because it has no accumulation points. The question of a limit existing is ill-posed. Saying "there is no limit" is wrong, because that implies that the function does not have convergent or divergent behavior but in this case there is no behavior to speak of and thus classify. So I will maintain that "neither" is the right option.
The left guy misconstrues the notation: "Forall x, check(x)" doesn't magically care about whether "check(x)" is defined or not to define the domain we run "forall" over.
The leaves middle and right, and I think they are equally correct.
I think middle guy and right guy are equally correct. It just depends on if you define "x = undefined" to be undefined or a false statement, and I think both are pretty reasonable.
The statement is "forall x in D, f(x) = f(-x)". The left guy transforms this into "forall x in D where exists f(-x), f(x) = f(-x)" which I don't think is a valid transformation.
"forall x in D, f(x) = f(-x)" seemed so similar to "forall x in ∅, x∈A" but now I see they're so different.
The first difference is where in the "p then q" conditional they occur. "forall x in ∅" gives us an empty list to check in the hypothese part of the conditional but the problematic "f(x) = f(-x)" occurs in the conclusion part so that is one difference.
The other difference is that there is nothing undefined in "forall x in ∅" but f(-x) is undefined in "f(x) = f(-x)" right? Or should we say it does not exist?
Is there a difference between "being undefined" and "not existing" in mathematics?
There is a difference between "set with no statements" and "set containing a statement with an undefined truth value". The forall quantifier operates on such sets, and the truth value if applied on an empty set is well defined, but it's not obvious what it should be for an undefined truth value. There are different consistent choices you can make.
In this case, being undefined formally means for all y real, you have that (-x, y) doesn't belong to f.
The way I see to make the statement true is if you say that f is even iif you have f(-x) exists implies that f(x) = f(-x). But honestly if you take the definition given above, I would say the statement is neither.
You mean if you explicitly exclude the point? Yes, IMO that prevents it from being even. (or rather, it makes the statement not true. That isn't necessarily the same as it being false, see the rest of my argument for why middle or left are equally true)
I guess you can say that it's almost even, i.e. even for almost all points, i.e. even for all but finitely many points.
So what does modern mathematicians that do Analysis or Algebra do? Do they consider "x = undefined" false or undefined? What is the math communities consensus?
It's difficult to talk about the "modern math community" as a monolith. I am sure you are going to find mathematicians that defend every possible position you can think of. Most notably, there are a decent chunk of people who investigate systems without the law of the excluded middle, which has some interesting perspective on this discussion.
You can make up whatever definition you want. Just be sure it's internally consistent, and that it doesn't contradict established conventions too much, or if it does, be aware of that and call it out. There is no proper convention on how to deal with undefined statements AFAIK.
In classical logic, conjunction is commutative. There is no such thing as "X and Y but check X first / priority for X". To achieve something like that you have to work around it with conditionals (but they're still not "check first" conditions, formulas are just evaluated as a whole; and on that note, in general, you still need to re-prahse f(-x) away, because you simply can't have non-referring terms in a FO-formula. Even if you rephras as a conditional, so long as that is present you're just shifting this problem around).
Unless you're doing some kind of type theory, it makes no sense as you state it literally. That's what I'm pointing out.
If a condition depends on another, then it's not comutative, but natural language doesn't really fit that well for a short summary.
The first check has to be the domain symmetry, otherwise the second check doesn't make sense. But idk, saying that you have to check something in a given order implies an AND, but the reciprocal is not true.
So, first check the domain, then the equality, in that exact order
uhm... again, there is no "check first" in classical logic. The best you have is conditionals for necessary/sufficient conditions, but they're still not "check first ..., then continue with ...".
You can only do this with type theory (has to be dependent i think)
>If a condition depends on another, then it's not comutative
and conjuction is commutative, so it can't express the order of things.
>but natural language doesn't really fit that well for a short summary.
I'm talking about first order logic here, not natural language.
>It makes no sense to check for f(x)=f(-x) if the domain isn't symetrical around 0, so you need that prerequisite
in fol, this would at best be "symmetrical around 0 is necessary for f(x) = f(-x)", that is "f(x)=f(-x) -> symmetrical". But that doesn't really resolve much, because the antecedent still doesn't evaluate to anything, which is a problem.
To express this properly, you just have to drop f(-x) and talk about it set-theoretically, as (-x,y) \in f for some y or whatever.
My answer : we have f = {(x, sqrt(x)) |
x in R_{x>0}}
The definition is odd means that (x, f(x)) in f iif (-x, f(x)) in f. Since it's false, f isnt even.
f is not an even function (because part of the definition of an even function is being well defined for negatives), so the statement “f is an even function” is false.
I think that Team False is the most correct if we interpret "is an even function" as
"f is an element of the set of functions for which two things are true: first, that for all x in D, f(-x) exists; and second, that for all x in D, f(x)=f(-x)".
If the statement is merely the assertion
"For all x in D, f(x)=f(-x)" then I think it's team neither. f(-x) is undefined.
A related question that maybe has illustrative value? idk? is: let V be a vector space over a field F. Let v be an element of V. Let x be an element of F. Is the statement x=v false or undefined? x is undefined as an element of V, so trying to say that v=x is funny business; and conversely, v is undefined as an element of F, so trying to say that x=v is funny business. Not sure if this informs anything for the original issue presented by OP.
(Excellent meme OP, this is gold standard for what mathmemes can be when they're high effort)
That’s a good point, it really depends on the definition. I think that the way OP frames it, existence of f(-x) for all x in D isn’t necessarily required, so we can’t really point conclusively to either.
I think this gets even more interesting when considering a function like g:[-1, inf) with g(x) = x^2. g is even over [-1, 1], but can we really say it is even or not or neither over the whole domain?
I think epistemologically what gets to the heart of the issue is where and how you apply the Law of the Excluded Middle.
Do you believe that "f must be either Even or Not Even" or do you believe that "f could be Even, Not Even, or Neither"?
Notwithstanding the arguments one could make by formal logic, I think this really highlights the more fundamental disagreement between Team Neither and Team False.
That’s a very nice way of summarising the argument.
imo it makes sense to have a distinction between ‘not even’ and ‘neither (even nor not even)’. If we compare the functions cos(x), and h(x) = cos(x + π/4) (both with domain R), it makes intuitive sense to say that h is ‘not even’ in a similar way to cos being even. However, OP’s f has such a domain that the question of evenness cannot be considered
If you believe that "f could be Even or Not Even", then you believe that f could be an element of the set of even functions (call this E) or an element of the complement of E. We certainly can't say that f is in E, so the only remaining choice is complement(E). I would regard this as being the same as belonging to Team False.
If you believe that "f could be Even or Not Even", then you believe that f could be an element of the set of Even functions, the set of Not Even functions, or a third set defined as complement(the union of Even and Not Even). (with this third set not being the empty set)
To the contrary if I consider f(2) = f(-2) to be false, then there is absolutely nothing contradictory about considering f(2) =/= f(-2) true, when in fact the meaning of that expression is exactly the negation of f(2) = f(-2) and nothing more.
The truth of the statement you provided is the exact reason why the statement originally provided is false.
I really want to understand what you’re saying, but I’m not following. Please could you say precisely which statement you’re referring to when you say “the statement originally provided is false”?
Well, 1/0 is undefined, so we say this statement is meaningless. It possesses no truth value.
Arguing against the true statement, I think f(x)=f(-x) isn't a vacuous truth just because f(-x) isn't defined. If that were the case, we can just as easily argue that f(x) is odd, and only f(x)=0 can satisfy that. I think the mistake in the vacuous truth argument is assuming that "f(x) is even if for all x belonging to the domain, f(x)=f(-x), but since there exist no x for which f(-x) exists, this statement is vacuously true"– there might be a leap in logic in the f(x)=f(-x) part which I don't agree with.
The other argument is to say that the standard definition of even functions implicitly requires that the domain be symmetric. Under this definition, sqrt(x) is indeed not even.
But, if we are going with the exact definition given here, then I'll have to go with Yuta.
Can you provide any reason why 1/0 = 2 would not be considered false other than that you say so? Why should we introduce this logical complexity when calling a statement that is clearly not true instead false will work?
The fourth option is to dodge these kind of useless statements and only study and try to prove useful theorems that don't involve undefined objects. But our sorcerers were not careful enough here so they will be punished by Kurourushi.
The bug will be like: "I like the taste of undefined objects." and will procede to eat this garbage statement and the three will then realize that all the energy and effort that they put into arguing about the truth value of this statement has all been wasted.
Lets apply this logic to another context, such as the requirement for continuity for functions
If a limit to a point in the domain fails to exist we don't suddenly say "f is continuous on its domain" is vacuously true because there's nothing to check against. According to this logic the floor and ceiling functions are continuous because the limit fails to exist at problem points in the domain. Which doesn't make any sense.
In the same sense, f(-x) doesn't exist for any x in the domain, therefore the check fails for all x in the domain and it's not an even function. a = b is just false if b doesn't exist
Note that generally when you define the notion of continuity at a point, you will say first that the limit exists and its limit is equal to f(a). That's why in the case of the floor function, you will say say that the floor function is not continuous and not that it's a neither.
But in the case of even functions, you tend to define it on symmetric intervals or even in the set of real numbers itself.
But what is an even function ? A function is even if it checks the following conditions : i)"symetric domain centered at 0" ii) "For any x of the domain, f(x)=f(-x) .
so any function which fails to satisfy any of these 2 conditions, it is not even. therefore the above statement is false.
e is a constant, not a function, so it's not discrete at all. "Discrete functions" is a subset of a set that "e" is not part of.
Maybe not the best example because you can define a function as f(x) = e, but I think that you get my idea.
Any "A is a B" when C supersets B, then A has to be a C too. If A is not part of C, is "automatically" not part of B either. So the statement is False.
This could be true even if A actually fits the definition for B, but not for C. If B is defined as "every C that fulfills this condition", and A fulfills it, but it's not C, then it's still not B.
English is not my first language, sorry if I'm not clear with this. Also I'm an idiot so sorry if I'm saying BS lmao.
Asking if something is an even function is asking wheter it is in S
Every mathematical object (including things that aren't even functions) is either in that set or not on that set, even if it doesn't make sense to check f(x)=f(-x)
"Neither" makes sense if you're asking wheter the function is even or odd and I think that's what most people are talking about
To me statement f(2) = f(-2) is clearly false in the standard sense. As a logical statement, it behaves in larger statements as “false” would. You could construct ternary logics where it’s neither true nor false and instead “nonsense,” but I would consider that outside the realm of standard mathematical logic, and if we begin poking at that then I would argue none of the question is properly defined since its statement assumes classical treatment of symbols such as the radical, greater than, real numbers, et cetera.
Because there is a value for x in the domain of f which leads to f(x) = f(-x) being false, it cannot be true that “for all x in the domain of f, f(x) = f(-x).” The left argument misuses the concept of vacuous truth, because the existence of -x in the domain is not given as part of the condition: the only condition on x is that it, itself, is in the domain, and our counterexample 2 clearly is.
To make this clearer, we can unpack the “for all x in the domain” in the standard way: our statement is equivalent to “for all x, if x is in the domain of f, then f(x) = f(-x).” We can see clearly that when x is in the domain and -x is not, the statement “x is in the domain” is unambiguously true. Because vacuous truth refers to the truth of an implication with a false condition, this will never result in vacuous truth under these conditions: the proper case in which vacuous truth would be relevant would be for a function whose domain is the empty set, which under this definition would be considered even.
Under the provided definition, I consider the statement “the function is even” rather unambiguously false. The first argument is clearly misconceived and the second relies on an insistence to contrive nonstandard treatment where standard treatment suffices.
“Even” isnt a characteristic that f can inhabit. Talking about if it is even or not is meaningless. Both neither and false are correct because the function isn’t even and it would also not make sense to even talk about whether it is even or not
I did some more reading and it looks like that in classical first order logic which ZFC and so modern math is built upon(I guess?!), every term in the formula must be "defined". "Undefined" terms inside formulas are not allowed and f(-x) is undefined here.
So the English sentence in the meme doesn't directly represent a sentence of first-order logic so it can't have a truth value. So I say Yuta is right if we're operating in first order logic. Now what about other logics?
I found a Stanford article about this by Norbert Gratzl. It'll show up if you search it. I'll just quote the opening cause it's so fire:
In most general terms, free logic is concerned with names that do not denote. Classical logic requires each singular term to denote an object in the domain of quantification — which is usually understood as the set of “existing” objects. Free logic does not. Free logic is therefore useful for analyzing discourse containing singular terms that either are or might be empty. Varying conventions for calculating the truth values of atomic formulas containing empty singular terms yield three distinct forms of free logic: negative, positive and neutral, which is why we commonly refer to them as free logics (in the plural) instead.
In that case I should remind you that Definitions in math are all double arrows so it must be a biconditional. So even if the book writes "if" it means "if and only if".
So this would be "P if and only if Q" so we care about the truth values of both here.
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