>A vector space just generalizes a field.
<No? Vectors are a whole separate set of objects you introduce. It's not just a generalization.
>But if you make the vector space the same field as the scalars it's the same.
<But vectors are different from scalars.
>But it's the same object.
<But they're different types. We write vectors in bold.
>Math doesn't have types.

>The next simplest example is the field F itself. Vector addition is just field addition, and scalar multiplication is just field multiplication. This property can be used to prove that a field is a vector space. Any non-zero element of F serves as a basis so F is a 1-dimensional vector space over itself.
...
Surely it causes problems if you can't assume vectors and scalars are distinct?

This is everything I hate about mathematics in one example. Are mathematicians not capable even of using tagged unions? Imagine if the vector space contained the objects used to construct the scalar field. It's insane and will make mathematics implode under special cases eventually. It's like math isn't made to find mathematical truths directly but so some "wiggers" (jewish) can go "UM ACKCHUALLY" and use intentionally hidden structure to make some important seeming abstractions in a way that somehow avoids helping anyone else understand. It's utterly disgusting.
Post other examples of this stuff itt, if you want.

who are you quoting

You can always get your own maths, with nominative types and row polymorphism.

...

Ackhually, your sentences gave me an idea: what if we complement numbers with strings and concatenation as the group op?

String concat is not commutative, and string mult is still to be defined. So no string vectors for you.

>Math doesn't have types.
Balderdash. Let's do this with types.

A field is
- a type
- two distinguished elements of the type (the additive and multiplicative identities)
- four operations involving the type (addition, additive inverse, multiplication, multiplicative inverse)
- a bunch of properties all of the above must satisfy

A vector space is
- a field, called the scalars
- a type, called the vectors
- a distinguished element of the vector type (the additive identity)
- an addition operation for two vectors
- an additive inverse operation for the vectors
- a multiplication operation between a vector and a scalar
- a bunch of properties all of the above must satisfy

There's no reason why the type of scalars can't be the same as the type of vectors, and in fact it's sometimes useful to make them the same.

I think the problem you're envisioning has more to do with operator overloading than making the scalar and vector types the same. If you set the scalar and vector types to the same thing, and you multiply a scalar and a vector, do you use the scalar-scalar multiplication or the scalar-vector multiplication? Sure, you could solve this by making a multiplication operation that takes a tagged union of the scalar and vector types, but that just overcomplicates things. The right way to solve the problem is to distinguish between the two "multiplication" operations in cases where there's any possibility of confusing them.

Of course it so happens that in the most obvious and useful way of making the scalar and vector types the same, you're also choosing the exact same operation to serve as scalar-scalar multiplication and scalar-vector multiplication. Exercise for everyone: Is it possible to make a vector space where the scalar and vector types are the same yet scalar-scalar multiplication and scalar-vector multiplication of the same objects has different outcomes?

P89321
Another thread by the stupid person who can't understand anything.

Summary:
>correct description above OP's pay grade
<reeee vectors are records i can't into math

A vector is DEFINED as a member of a vector space. The most frequently-used vector spaces are Cartesian products of fields, e.g., R^n, which produce the familiar kind of vector notated as a tuple. But vector spaces are by no means limited to Cartesian products.

>seething over 1-dimensional vector space being just the field that coefficients are drawn from
You are literally stupid. This is what makes mathematics great: the degenerate case is just the simpler thing. Mathematics' defining feature is its internal consistency. Where any other discipline would introduce an infinite number of nonsensical exceptions, mathematics keeps its shit straight. And here you are, a mathlet, whining about it. Go back to /g/.

>Imagine if the vector space contained the objects used to construct the scalar field.
No, you dumb *****, that's not what is being said. What the writer did was give an EXAMPLE of a 1-dimensional vector space: a field. He is not saying multi-dimensional vector spaces have lesser-dimensional values floating around in them, breaking their consistency. That is a hallucination on your part caused by your low intelligence.

>muh wiggers
You are an embarrassment to wigger haters everywhere. Don't ever use this word again. Your innumeracy is a you problem.

P89401
Respectfully, you can't teach your way out of this problem. I know this from experience. OP is not asking a question, he is attacking the truth. He is not interested in learning. The only thing you can do here is disengage, or attack back.

Something precious was lost during the Renaissance, when ancient Greek writings were recovered, but the recourse against a bad-faith interlocutor was lost. The ancient Greeks would resolve impasses by fighting, because that's all you can do anymore.

This is why debate is fruitless now, and the advantage goes to sophists who spew lies faster than faucets spew water: the final recourse that enforces good faith has been removed.

The best we can do on this board is fighting words, which are nowhere near as effective as actual fighting. But, at the very least, it can signal to others that they shouldn't waste their valuable time and energy on a disrespectful idiot.

P89410
> But vector spaces are by no means limited to Cartesian products.
Wait, what?

P89401
>I think the problem you're envisioning has more to do with operator overloading than making the scalar and vector types the same. If you set the scalar and vector types to the same thing, and you multiply a scalar and a vector, do you use the scalar-scalar multiplication or the scalar-vector multiplication? Sure, you could solve this by making a multiplication operation that takes a tagged union of the scalar and vector types, but that just overcomplicates things.
Except that the notation distinguishes vectors and scalars, not vector and scalar operations.
And in either case it's insane to call it a generalization of a field.

>Is it possible to make a vector space where the scalar and vector types are the same yet scalar-scalar multiplication and scalar-vector multiplication of the same objects has different outcomes?
With F=V=R, scalars as normal, but the objects of V shifted by +1 while all operations act as though on the unshifted objects (and the zero vector is 1).

P89410
>He is not saying multi-dimensional vector spaces have lesser-dimensional values floating around in them, breaking their consistency. That is a hallucination on your part caused by your low intelligence.
You idiot, that's not what I meant.
Imagine if the intersection of F and V was non-empty accidentally because V was a set of objects containing the ones that happened to be used to construct F. This doesn't happen just with tuples of elements of F.

>What the writer did was give an EXAMPLE of a 1-dimensional vector space: a field.
This only works if there is a single overloaded operation for multiplication, and there is no additional information to distinguish vectors and scalars. That would make my example above unrepresentable.

>This is what makes mathematics great: the degenerate case is just the simpler thing. Mathematics' defining feature is its internal consistency.
That you instinctively want to ignore specifics to make important sounding generalizations shows you have little interest in internal consistency.

>Go back to /g/.
I will.

P89411
Do you know this "truth"? Or do you just worship mathematicians who can obfuscate things so only their intuitions are useful.

P89321
>No? Vectors are a whole separate set of objects you introduce. It's not just a generalization.
A field is just is a vector space over itself where scalars and vector happen to be the same kinds of mathematical entity.
What's the problem exactly ?

P89410
>Where any other discipline would introduce an infinite number of nonsensical exceptions, mathematics keeps its shit straight.
If anything, mathematics tries to keep its shit straight. Just look at the proof of abc conjecture by Mochizuki as a counter-example.
I hope proof assistants will make research easier in the future once it get recognized as the most practical method of proving theorems/lemmas.

P89425
Truth doesn't exist. In the realms of mathematical logic, consistency and completeness are the key properties that matter when it comes to building foundations.

OP believes that a set can't be a field and a vector space at the same time because he views a field as its own object that encapsulates the underlying set and the field operations instead of the set itself being a field under the given operations. You can decide either to say that a set F is a field under operations + and *, or that the tuple (F, +, *) is itself the field (or whatever equivalent encoding). OP chooses the latter convention. OP believes it's wrong to say that [tex: \mathbb{R}] is both a field and a vector space because in his view the underlying field is ([tex: \mathbb{R}], +, *) and the underlying vector space is ([tex: \mathbb{R}], +, [tex: \mathbb{R}], +, *), which are not equal. OP just refuses to accept the other convention. see log.txt for long pointless conversation

P89893
([tex: \mathbb{R}], +, *, [tex: \mathbb{R}], +, *) rather, where ops are vector addition, scalar multiplication, field addition, field multiplication respectively

and yes op every finite nonempty set S is a group because you can relabel its elements {0,1,...,|S|-1} and let the group operation be +. same for any set with the cardinality of the integers, reals.

P89420
Yes, exactly as you read. The most useful vector spaces by far are the result of Cartesian products (R^n, etc.), but an extension field qualifies as a vector space. My abstract algebra textbook devotes a brief chapter to it while discussing extension fields.

The complex numbers are an example of this: i=sqrt(-1) is algebraic over the reals (footnote 1). And, in our interactions with it, we openly exploit the fact that is is a vector space: we model it as a plane with basis {1, i}! And it is simultaneously a field, being a ring with unity, commutative addition, and multiplicative inverses for all nonzero elements.

Students are usually eased into dealing with complex numbers by being to told to pretend it's R^2 but multiplying by i causes a 90 degree CCW rotation. But we don't construct C by Cartesian product, we construct it by algebraic extension of the reals with i.

Footnote 1: There exists an irreducible polynomial with real coefficients whose root is i: f(x) = x^2 + 1.

op just refuses to accept the convention that "field" refers to the set itself (implying the existence of operations) instead of the tuple containg the set and its operations. this means when he hears someone say "a field is a vector space" he doesn't (or didn't) understand this is a statement about sets. now he understands but still insists he is right, saying this use of language is wrong, bad design and "obfuscation". op should calm down and try to understand instead of fighting.

P90057
>op should calm down and try to understand instead of fighting.
This is called being unteachable. A person who has an anger response to new information can't learn.

Vector space? You mean module over a field?

Isn't a vector that can be a scalar called tensor?

P90602
no

P90596
lolz

Where is this log from? I want to bully mathlets (and maybe learn something) too.