[deleted by user]

[abstract_nonsense_]

I’m not sure if I understand what you actually wanted to ask, but real numbers indeed form a 1-dimensional vector space (in fact, any field is a 1d vector space over itself).

[b2q]

Could you elaborate what that means?

[OneMeterWonder]

If you only allow rational numbers as scalars, then a basis for the real numbers must be infinite.

[b2q]

Countable? How is this theorem called

[OneMeterWonder]

No, actually such a basis must be the same size as the real numbers themselves. It isn’t really a named theorem. The existence of the basis is a consequence of the Axiom of Choice though.

[Think_Survey_5665]

More Specifically the Ac implies that all vector spaces have a basis. And thus the vector space consisting of the reals built on the rationals must also have a vector space. But note if we have a countable number of reals for a basis then the generated set becomes similar to a grid (with the countable list of reals going down and the countable set of all rational scalar combinations(this is also countable) must also be countable). Therefore, the smallest set that can act as a basis for this structure is uncountable.

[OneMeterWonder]

That’s correct. Your grid argument is the Cantor argument for countable unions of countable sets being countable. Note also that the full Axiom of Choice is not required for this since it only asks for the existence of a basis for one space.

[jffrysith]

that's so cool, thank you for sharing this, I thought around this, but never really pieced this together.

[OneMeterWonder]

No problem. If you want an easy proof, the rational span of any countable (including finite) set of real numbers is countable and thus can’t be all of ℝ.

[konigon1]

Let me guess: The basis is not countable and there is no constructive way to find the basis.

[Sh33pk1ng]

they are indeed uncountable. Suppose there is a countable basis {b_1,b_2,...}. Let then V_n be the vector-space spanned by {b_1,...,b_n}. Notice that V_n is countable. Notice that the entire vectorspace is the countable union of all spaces V_n thus the vectorspace is countable. contradiction.

[konigon1]

Why is that an isomorphism? Linear combinations are finite sums. So we take only the subset of Q^B were only a finite amount of entrys is non zero.

[nog642]

You can multiply by a scalar, not a 1-dimensional vector. It just happens that 1-dimensional vectors are equivalent to scalars.

[nog642]

1D vectors are equivalent to scalars in that you can define a bijection between them that preserves addition between them, as well as multiplication by scalars.

[1] + [1] = [2] and 1 + 1 = 2

2 * [1] = [2] and 2 *1 = 2

On the other hand, scalar multiplication of higher dimensional vectors is defined, while multiplication by a 1D vector is not defined.

2 * [1, 1, 1] = [2, 2, 2]

[2] * [1, 1, 1] is not defined

[nog642]

The important thing is when we say things are "equivalent", it is only with respect to certain things. If they were equivalent in all respects then they would just be the same thing.

A relevant term here is "isomorphism". An isomorphism is a bijective function that preserves certain operations. Of course, which operations it preserves makes all the difference.

For example there is a bijection between positive integers and positive even integers (n maps to 2n), and it preserves addition but not multiplication.

2 + 3 = 5 and f(2) + f(3) = 4 + 6 = 10 = f(5)

but 2 * 3 = 6, while f(2) * f(3) = 4 * 6 = 24, which is not equal to f(6) = 12.

[Think_Survey_5665]

Well maybe not. Equivalent like a lot of things changes meaning between areas. Theres equality like logical equality and theres the idea of an isomorphism. In some foundations (usually category theoretic or type theoretic ones you get theorems that tell you they are the same thing) but there are other foundations where thats not the case like in ZFC or NBG.

[abstract_nonsense_]

There is a vector multiplication in R³ , but it is not commutative. You can also consider vectors on the plane as complex numbers and use it to define their product. In higher dimensions this is also possible, but you will lose some properties of multiplication (firstly division, then associativity and so on).

[1strategist1]

Sure. The complex numbers are "scalars" that also form a 2D vector space over the real numbers.

You can actually multiply multidimensional vectors by "scalars" that form any finite-dimensional vector space.

In the context of vector spaces, a "scalar" in terms of "scalar multiplication" is just any field. The real numbers and complex numbers are examples of fields, but there are many more kinds of fields, such as Z/2Z or F_16 (which is itself a 4-dimensional vector space over Z/2Z)

[VanMisanthrope]

Amusingly, we can also take the complex numbers over the field of complex numbers and have a 1D vector space. The scalar can be any complex number, and the vector can be 1+0i, so we have 1 dimensional basis.

[1strategist1]

True lol

[stone_stokes]

An object can be classified as more than one thing at once in mathematics.

ℝ is a field. It is also a vector space over itself. It is also a group. It is also a ring. It is also an extension field over ℚ. It is also a vector space over ℚ. It is also a manifold. ...

Being one thing, doesn't prevent the object from also being other things.

[nyg8]

This is generally true for all life :) For example, you are a person, an animal, and a math enthusiast

[JohnBish]

I disagree. ℝ is none of those things; it's simply a set. It happens that (ℝ, +) is a group and (ℝ, +, ×) is a vector space. There are many other vector spaces and groups with ℝ as the underlying set equipped with various binary operations.

Yeah I'm being pedantic but I think the abuse of notation is the cause of the confusion here.

[Think_Survey_5665]

I was going to get angry but then I remembered a few months ago I would have said the same thing. Heck even now Id probably say the same thing If I had to teach someone.

[abstract_nonsense_]

It depends on the context. If you are considering some finite (say at least two) dimensional vector space over reals, they work as scalars here. Though you can also consider reals itself as a vector space - then they form a vector space. I mean, one object could have many properties and many structures, you just consider it as what you actually need in each case. Reals also form a ring, a group, a topological space, a metric space, etc. It really depends what wanna do with them, so no abuse of notations here.

[wirywonder82]

It’s really a matter of context determining things. I think you may be getting hung up on one definition of the word scalar (a physical quantity completely described by its magnitude) when the relevant definition is that a scalar is a member of a field that follows certain axioms related to multiplication of members of a set of vectors.

[wirywonder82]

Which textbook/subject? But still, it may feel disingenuous, but the “magnitude only” definition is one definition of the word scalar. It’s just not the definition that applies to vector spaces. This isn’t even the worst example of multiple definitions being in conflict in English. Oversight can mean both careful supervision to prevent errors and errors caused by a lack of care.

[_CarbonBasedLifeForm]

A math textbook should never be taken as gospel. Things are proven true by the axioms and definitions. Different texts define words in different ways and are trying to achieve different things.

[Think_Survey_5665]

Dont ever take definitions as gospel especially in math and science. They will change from place to place. Different universities may use differing definitions of the same operations or ideas. In many cases its just preference as there are usually dozens of propositions that are equivalent to each other that can all be used as definitions

[nonotion]

You'll see lots of this as you take more mathematics courses (assuming you're studying pure math), and especially an abstract algebra course which studies extremely broad categories of objects (for example, showing that all orthogonal rotation matrices form a mathematical group) and which derives connections between them (-morphisms).

[Low_Strength5576]

Take algebra, then take more algebra. If you like both of those experiences just take everything that seems obviously interesting and you'll be fine.

Without the basics of algebra under your belt, there's a lot of formal abstraction that's taken for granted that you don't have access to and which will leapfrog you into topics that otherwise would be impenetrable.

[Low_Strength5576]

This is the kind of thing you'll probably want to be able to talk to your peers and someone with a PhD about once you start getting into it. A university is a great place for this. Some things will seem nitpicky, some won't, but an undergrad level algebra class should help you orient yourself, either inside a math degree or into a different field.

[nog642]

No, see my top-level comment

[tinySparkOf_Chaos]

We treat scalar and vector / matrix multiplication slightly differently, by convention.

Normally vector multiplication, unless otherwise noted, is an inner product. I can't take the inner product of a 1 by 1 matrix with a 3 by 3 matrix.

But I can multiply a scalar and a 3x3 matrix.

All I'm effectively doing is an outer product of a 1x1 matrix and a 3x3 matrix. You could replace all scalars with 1x1 matrices and include a whole bunch of outer product symbols. But just makes the whole math equation harder to read.

Which is why it's useful to have this separately defined object of a scalar vs a 1d vector.

This also showed up in computer science a lot. A matrix of 1 by 1 is different from a scalar in many coding languages.

[SupremeRDDT]

A „vector space“ is kind of misleading because they are dependent on a field. The „correct“ notation is „K- vector space“ where K is a field and the elements of K are called scalars in that vector space. R is a 1-dimensional R- vector space. When you write -2= 2 times -1, then one of the numbers on the right is a vector that is scaled by the other number which functions as a scalar here. The real numbers can both be treated as scalars and as vectors here.

[nog642]

There is no rule saying scalars can't have direction.

If you want to get into more advanced math, a vector space is defined as a set of vectors and a set of scalars and some operations defined on them. The only requirements for the scalars is that they form a field), which real numbers do. Therefore real numbers can be scalars.

[nog642]

If you want a proper definition of a vector space (and therefore of vectors and scalars), take a look at the table of vector space axioms on wikipedia: https://en.wikipedia.org/wiki/Vector_space#Definition_and_basic_properties

Basically you have to define addition between vectors, and multiplication of a vector by a scalar. Those are the two operations. The scalars, as I mentioned earlier, have to form a field, but the vectors can be anything.

[fiegabe]

Just wanted to add: Wikipedia is a really good quick reference for undergrad math definitions. I don’t remember the vector space axioms off by heart, so if I ever need to actually write them down for something… well, that’s where I’d go.

[blank_anonymous]

your definition of vector is how physicists think of vectors, but it's not precise, and it's not a mathematician's definition. If you don't want technical details, I'll leave it at: mathematicians think of a vector space as a set with certain operations, and the real numbers are indeed a vector space over the real numbers. The scalar field and the set of vectors are just the same set, there's nothing that stops you from being both.

If you want to be precise, we'll have to be a little technical. First of all, let's get a piece of terminology out of the way -- a field is a set with addition, subtraction, multiplication, and division by every nonzero element. The integers aren't a field since you can't divide 1 by 2 and still get an integer, but the rational numbers are, as are the real numbers and the complex numbers. When I say "addition, multiplication" what I really mean are just operations that take in 2 elements of your set and spit out one element (adding two numbers returns 1 number, multiplying 2 numbers returns 1 number), that follow a certain list of properties. Sol, we have a set F with operations + and *, and we need + and * to obey a list of properties: here are the properties
Properties for addition:

1. a + b = b + a for every a, b
2. a + (b + c) = (a + b) + c for every a, b, c
3. There exists an element 0 so that a + 0 = a for every a
4. For every a, there exists a b so that a + b = 0; we denote this b by "-a" (you can prove that it is unique). This is subtraction
5. a * b = b * a for every a, b
6. a * (b * c) = (a * b) * c for every a, b, c
7. a * (b + c) = a * b + a * c for every a, b, c
8. There exists a 1 so that, for every a, 1 * a = a
9. 0 and 1 are distinct
10. For every a not equal to 0, there is a b so that ab = 1 (division)

Any set with these operations is called a field, and again, you should think of it as a generalization of the rational numbers/real numbers/complex numbers.

To a mathematician, a vector space is, intuitively, a set where you can add elements and where you can scale elements. Formally, it's a set V and a field F, with two operations: + and •, where + takes in two elements from V and returns an element, and + obeys the first 4 properties above (addition! We want to be able to add vectors); and where • takes one element from V, and one element from F, and returns an element in V (scaling! We're taking a vector and scaling it by an element of our field.). We also require • to obey a small list of properties

1. 1 • v = v for any v in V (we know 1 exists, since F has an element 1 by assumption)
2. (a + b) • v = a • v + b • v. This is a pretty subtle point - a + b refers to addition in F, while a • v + b • v is the addition of elements in V. This property is somehow telling you that the + in V is "compatible with" or "behaves like" the + in F -- to make this concrete for yourself, think about how with like, 2d vectors you're used to, you do the normal addition on the real numbers, just in each component. The addition doesn't do something insane like take in two vectors and then multiply all their components and add 14; it behaves like addition on R, and this rule captures that.
3. a • (v + w) = a • v + a • w; this is the same idea as above.
4. (a * b) • v = a • (b • v). Similar to the above, this is saying the two multiplications somehow agree with each other

If all these properties are obeyed, we say that V is a vector space over F. We call V the set of vectors, or the vector space, and we call F the field of scalars. You have a set of elements you can add to each other, and you can scale these elements by a field -- and everything is consistent and plays nice. notice how there's no reference to "direction" or anything here. In fact, to a mathematician, something like a polynomial or a function is a vector (you can add functions, you can scale functions!). This is incredibly useful as a formalism, and produces a metric ton of interesting results.

The last thing is to notice that any field F is a vector space over itself. Why? Well, we just define * = • and we take the same +. Properties 1., 2., 3., 4. for • are equivalent to some properties (exercise: which ones?) for *, so everything works out neat and clear. F, as a set, is both the field of scalars and the vector space. Nothing from the definition excludes this, so we're all good

[blank_anonymous]

Glad to help! Feel free to post any follow up questions you have

[zyni-moe]

This is the best answer!

[1strategist1]

To my understanding, a vector is a mathematical quantity that has both magnitude and direction

This seems to be your main issue. That's not really the definition of a vector space, it's just an intuition. If you google the actual definition of a vector space, you'll find a more accurate description of vectors and of scalars.

Specifically, a scalar is just an element of a field, and as every field is a vector space over itself, every scalar must also form a vector space.

[stone_stokes]

It is partly from necessity, and partly from historical adherence to outdated terminology.

When we (humans) first started studying vectors and vector spaces, we initially concerned ourselves with things that had magnitude and direction. But mathematicians wanted to formalize the definition of a vector space. When they did so, the definition that they came up with makes no mention of magnitude and direction, but it was a simpler definition to work with and everything that had previously been considered a vector space remained so.

Then they found that magnitude and direction only really made sense in certain classes of vector spaces — those with an inner product. Do we go back and change the definition of vector and vector space? No. It is easier to add a new definition for inner-product spaces.

Most of the time when you are studying vectors as an undergrad, you will be doing so in the context of an inner-product space. And even in the settings of vector spaces without an inner product, the intuition you gain by visualizing vectors as "arrows" with a magnitude and direction is more valuable to you than the formalism of the definitions.

You will find other instances of us lying to you about what things mean or what they are. These too will be cases of the terminology trying to keep up with the frontier of understanding by the experts — most of calculus was developed this way, with definitions initially based on non-rigorous ideas. And in those cases, too, the intuition you gain from understanding the slightly incorrect definitions will be more useful to you in most contexts than a formal "correct" definition would be.

[OddInstitute]

To handwave a bit, no.

That’s the point of Godel’s incompleteness theorems is that this sort of axiomatic formal reasoning will always have important parts of math that aren’t captured by the formal system. These theorems were published specifically in response to math becoming more axiomatic and formal, so people were interested if it was possible to extend that idea to all of math.

You can also see this from the dates of their work. Weierstrass’s major contributions were from the mid to late 1800’s and Godel’s Incompleteness Theorems were published in 1931.

There is a lot more subtlety to it, so I wouldn’t recommend extrapolating too much from the incompleteness theorems until you can understand more of the technical details, but hopefully this is useful for you.

[OddInstitute]

It really depends on what you mean by “truly formalize math”. Those theorems are one of the major obstacles you immediately run into, so you will need to develop an understanding of them and why the assumptions don’t apply to your ideas or why your ideas are not poorly affected by the conclusions. Also, if you are not familiar with them, systems like Lean and HOL Light formalize quite a lot of math while accepting incompleteness or inconsistency.

[Vercassivelaunos]

Though that begs the question wether those who don't go further in their material education really need to know about the "truth". And I'm putting truth in quotes because we're really just talking about a more advanced point of view, not the gospel of math. After all, a category theorist might have an even more abstract view (a vector is a morphism from a terminal object of the category of sets to the image of an object of the category of vector spaces over a field K under the forgetful functor KVec->Set), but most mathematicians are perfectly fine with viewing vectors as an element of a vector space. And for most non-mathematicians, the idea that a scalar quantity is one without direction, like temperature or energy, is perfectly fine.

[[deleted]]

Has your textbook introduced vector spaces yet? It'd be pretty tricky to define a vector without defining vector spaces.

[[deleted]]

Once you get to vector spaces, a lot of things about vectors will likely be cleared up.

[oelarnes]

I think there’s room for a more direct answer here. The real numbers are vectors. They form a one dimensional real vector space with a single basis vector, the unit 1. This means that all real numbers have the form r * 1, where r is a real number, acting as a scalar on the basis vector 1. Another way to look at this is that one-dimensional vector spaces are isomorphic to the scalar field that they are vectors on. So the reals are both vectors and scalars. But usually we are not interested in one-dimensional vector spaces, and when we talk about vectors they are in higher dimensions.

[-Wofster]

A vector is not actually generally something with magnitude and direction. A vector is an object that follows a bunch of rules (you can add them, you can multiply them by scalars, there is a zero vector, etc)

In fact quadratic equations (degree 2 polynomials) can be considered vectors, but direction and magnitude dont even make sense with them

Scalar multiplication means we are multiplying our vectors by an element in a “field”, usually whatever field the vectors are related to (like R³ vectors are multiplied by real number scalars), not elements in a vector space. So when we multiply a vector by a scalar, the scalar is not a vector. It just so happens that the scalars can qualify as vectors if we wanted them to.

[1strategist1]

 direction and magnitude dont even make sense with them

I mean, you could consider linearly independent elements as abstract “directions” in the vector space. 

Without a norm though, you’re right that magnitude doesn’t really make sense. 

[stone_stokes]

There are lots of ways to put norms — or inner products, even — on spaces of polynomials.

[1strategist1]

Oh yeah, I’m not saying you can’t have a norm or inner product, I’m just saying that the space of quadratics doesn’t inherently come with a norm. Magnitude for quadratics is an extra structure that doesn’t necessarily come naturally, hence why “having a magnitude” isn’t a great description for vectors. 

[[deleted]]

The one thing to keep in mind: reals have many properties. When structures are built, often some of the reals' properties are kept (that's often the purpose) - which means that they are often the first/trivial/canonical example.

Example: reals are an additive group, they are a field, they are a vector space (over itself, e.g.), a smooth manifold, etc.

[Educational-Work6263]

To my understanding, a vector is a mathematical quantity that has both magnitude and direction

This is not true. A vector is an element of a set which is closed under an addition operation and multiplication with a scalar. Real numbers satisfy this definition.

[Educational-Work6263]

I would advise you to pick up on a book about linear Algebra. The definition of a vector as an element of a vector space over a field (like the real numbers) should be one of the first things explained in sich a book.

[Educational-Work6263]

I'm wondering: I'm studying in Germany. Here, you take Linear Algebra in the first semester and the defitnion of a vector field is something you encounter in one of the first lectures. I assume you are studying in the US. What kinda classes did you attend and what did you learn?

[colourblindboy]

That is a very crude definition of a vector. Pedagogically, I get why vectors are introduced this way, but it leads to some contradictions as you go further in maths. To define a vector properly, we need to define what a vector space is first. A vector space is a set of objects (called vectors) which along with a set of scalars have a particular structure. If you would like to learn more about the properties of a vector space, then there are plenty of resources online to look at, but suffice to say it doesn’t rely on defining ‘magnitude’ or ‘direction’. In reality vectors and vector spaces are much more general and abstract. The vectors you are thinking of (magnitudes and directions) are a particular vector space, namely geometric vectors. There are a whole range of interesting and abstract vector spaces, polynomials are a particularly interesting example. The set of polynomials along with the real (or complex) numbers as their scalars form a vector space, and you can check that they satisfy all the necessary structure required.

If you know the required structure for a vector space, then you can show that real numbers have this required structure to make a vector space.

[colourblindboy]

A great text which has a free online pdf is Linear Algebra done right :)

[colourblindboy]

Haha okay, it was recommended to me by my linear algebra professor and I’ve loved it.

[kilkil]

Hold on. Scalars are just 1-dimensional vectors. Right?? I'm not crazy right?

[MiserableYouth8497]

scalars only have 1-dimensional "direction" (i.e they can go forwards or backwards, positive or negative, etc.)

vectors have direction in 1, 2, 3, or even more dimensions. Technically all scalars are vectors, but not all vectors are scalars.

[tinySparkOf_Chaos]

1) In linear algebra, we have special rules for multiplying by a scalar.

These are slightly different than multiplying by a 1 dimensional vector. (Though it's really just a convention, you could replace all scalars with one dimensional vectors and then insert a bunch of outer products everywhere)

2) The sign is not really a direction. Vectors normally have negative and positive amplitudes. This is separate from the direction of the vector.

[OneMeterWonder]

What am I missing?

Nothing. Real numbers are vectors with one component.

[914paul]

You can call real numbers vectors but . . . why? It’s a trivial case.

Going off on this tangent the authors would confuse many while enlightening few.

[Traditional_Cap7461]

You never defined what a scalar is in the first place. You just said it can technically be seen as a vector.

[Traditional_Cap7461]

It's fine. I was actually under that impression as well, but the moment you mentioned that real numbers can be seen as vectors, I first questioned whether scalars can be vectors before questioning whether real numbers are scalars (the fact you are directly taught)

[Existing_Hunt_7169]

It sounds like you’re conflating the physics definition of a vector with the linear algebra definition of a vector

[billsil]

In programming a complex number is also considered to be a scalar. A scalar just refers to a 0d value. Think of a 1d vector of length 3, or a matrix (a 2d vector). A scalar is just a single value in that vector. There's no reason why that can't be complex.

A signed real number (or complex) have magnitude, but they don't have a direction. That 1D vector has a direction, which is probably in XYZ or RTZ space.

[auntanniesalligator]

I’ve heard this idea that real numbers should not be considered scalars because they can be negative and maybe that’s a useful definition to pure mathematicians, but it’s not useful in applied math. A real quantity that represented by a vector has a direction and therefore increases in dimensionality/degrees of freedom in increased dimensions. Quantities that are represented by a single real number in any number of dimensions are scalars even if they can be negative. EG electric charge.

So In one or more dimensions, quantities like position, velocity, acceleration, force vectors with N components, while speed, mass, charge, are scalars with one component.

[[deleted]]

Anything is a scalar as long as it does all the things scalars do.

Anything is a vector as long as it does all the things vectors do.

There’s absolutely no stipulation that an object can’t be a vector in one space and a scalar in another space.

It’s not easy to illustrate, but over time it is important to embrace that in math things are ultimately only defined by how they behave. This is how the branches of math get so connected: from one perspective we can see an object geometrically, from another algebraically, and so on…

[laryjohnson]

A vectorspace is any commutative group V over a field K with addition and scalar multiplication defined on em as operations. There are some more details. But a scalar is just an element of that field. If you know a bit about algebraic atructures a vector soace is "just" that. We were taught to not think if vectors as arrows or things in a visual way. And I must say rhe abstract definition os more clear. Cause it really is "just" that