Closed under addition and closed under scalar

• negation
In summary: or closed under both addition and scalar multiplication?closed under both addition and scalar multiplication
negation
Could someone explain to me about what closed under addition and closed under scalar multiplication means? I have a patchy idea of what it is but how does it relates to A = {(x,y) | x^2 + y^2 <= 1}?
What does A stands for? What does the language implies?

Edit: My interpretation: Let's suppose there exists a field k with R^n where n = 2) and A is a subset of the field k.
An element is closed under addition iff an element, uA, and, vA such that
u^2+v^2 = <=1.
If u^2+v^2 <=1, then, u and v is a subset of A.

Last edited:
negation said:
Could someone explain to me about what closed under addition and closed under scalar multiplication means? I have a patchy idea of what it is but how does it relates to A = {(x,y) | x^2 + y^2 <= 1}?
What does A stands for? What does the language implies?

In other words, you are considering ##R^2## as a vector space over the reals. Addition means ##(u,v)+(x,y) = (u+x,v+y)## and scalar multiplication means ##t(x,y) = (tx,ty)##, the usual vector operations in ##R^2##.

Edit: My interpretation: Let's suppose there exists a field k with R^n where n = 2) and A is a subset of the field k.
An element is closed under addition iff an element, uA, and, vA such that
u^2+v^2 = <=1.
If u^2+v^2 <=1, then, u and v is a subset of A.

You have a very confused interpretation. You don't talk about whether an element is closed under addition, you talk about whether the subset ##A## is closed under addition. That is, if ##(a,b)## and ##(c,d)## are in ##A##, is ##(a,b)+(c,d)## always in ##A##? If so, ##A## is closed under addition. Similarly, if ##(x,y)\in A## is ##t(x,y)\in A## or not.

LCKurtz said:
In other words, you are considering ##R^2## as a vector space over the reals. Addition means ##(u,v)+(x,y) = (u+x,v+y)## and scalar multiplication means ##t(x,y) = (tx,ty)##, the usual vector operations in ##R^2##.

You have a very confused interpretation. You don't talk about whether an element is closed under addition, you talk about whether the subset ##A## is closed under addition. That is, if ##(a,b)## and ##(c,d)## are in ##A##, is ##(a,b)+(c,d)## always in ##A##? If so, ##A## is closed under addition. Similarly, if ##(x,y)\in A## is ##t(x,y)\in A## or not.

How does x^2 + y^2 relates the the closed addition and scalar multiplication?

negation said:
How does ##x^2 + y^2\color{red}{\le 1}## relates the the closed addition and scalar multiplication?

That's how you tell if ##(x,y)\in A##. So if you add two elements in ##A## is the result in ##A##?

LCKurtz said:
That's how you tell if ##(x,y)\in A##. So if you add two elements in ##A## is the result in ##A##?

So x^2 + y^2 <=1 is a condition. If vector u and vector v both of which are elements of S where S is a set of vectors and if u+v <= 1 then it is closed under addition.
And if it fulfils 2 other necessary conditions- namely, closed under scalar multiplication and where 0 vector is an element of S then S is a subspace.

negation said:
So x^2 + y^2 <=1 is a condition.
Yes, this is the condition by which you can determine whether a vector <x, y> is a member of set A (the name of the set from post #1). S is not mentioned in post #1.
negation said:
If vector u and vector v both of which are elements of S where S is a set of vectors and if u+v <= 1 then it is closed under addition.
No, that isn't it. If u and v are members of A, and if u + v is also in A, then A is closed under vector addition.
The condition, as you call it, is how you can tell whether a given vector is in set A.
negation said:
And if it fulfils 2 other necessary conditions- namely, closed under scalar multiplication and where 0 vector is an element of S then S is a subspace.
Just to be clear, assuming that u is in A, and t is a scalar, if tu is also in A, then A is closed under scalar multiplication.

Mark44 said:
Yes, this is the condition by which you can determine whether a vector <x, y> is a member of set A (the name of the set from post #1). S is not mentioned in post #1.
No, that isn't it. If u and v are members of A, and if u + v is also in A, then A is closed under vector addition.
The condition, as you call it, is how you can tell whether a given vector is in set A.

Just to be clear, assuming that u is in A, and t is a scalar, if tu is also in A, then A is closed under scalar multiplication.

Is it the same thing to say A closed under addition, closed under scalar multiplication and has zero vector and A is the subspace of a vector space?

negation said:
Is it the same thing to say A closed under addition, closed under scalar multiplication and has zero vector and A is the subspace of a vector space?
Not quite. You need some ifs and a then.

For a set A, a subset of a vector space, if
1. A is closed under vector addition, and
2. A is closed under scalar multiplication, and
3. The zero vector is an element of A,

Then A is a subspace of the vector space. (Whatever vector space we're working with.)

1. What does it mean for a set to be closed under addition?

When a set is closed under addition, it means that if you add any two elements from the set together, the result will also be an element of the set. In other words, the set contains all possible sums of its elements.

2. How can you tell if a set is closed under addition?

To determine if a set is closed under addition, you can pick any two elements from the set, add them together, and see if the result is also an element of the set. If it is, then the set is closed under addition.

3. What does it mean for a set to be closed under scalar multiplication?

A set is closed under scalar multiplication if multiplying any element in the set by a scalar (a real number) results in an element that is still part of the set. In other words, the set contains all possible scalar multiples of its elements.

4. How do you check if a set is closed under scalar multiplication?

To check if a set is closed under scalar multiplication, you can choose any element from the set and multiply it by any real number. If the result is also an element of the set, then the set is closed under scalar multiplication.

5. Why is it important for a set to be closed under addition and scalar multiplication?

Closure under addition and scalar multiplication is important because it ensures that operations on elements of the set will always result in elements that are still part of the set. This allows for consistency and predictability in mathematical operations and makes it possible to prove theorems and solve equations using the properties of the set.

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