Question

i do not understand the closure property of the axioms

Answer 1

which axioms? My understanding is basically that if a set of numbers satisfy the closure property under a certain operation then when you apply that operation you will always get an element of that set as a result. For example, whenever you add real numbers you always get real numbers as a result, therefore the set of real numbers is closed under addition.

Answer 2

my question is what closure property mean? literally.

Answer 3

In mathematics, a set is said to be closed under some operation if performance of that operation on members of the set always produces a unique member of the set. For example, the real numbers are closed under subtraction, but the natural numbers are not: 3 and 8 are both natural numbers, but the result of 3 − 8 is not.

Answer 4

i understand this too simple, yet i need few examples to comprehend.

Answer 5

It's just what Barkante and I said, the closure property involves a set of numbers (or even arbitrary elements) and an operator. The set and operator (typical operators are subtraction, addition, multiplication, etc, but you can make up any operator you want), satisfy closure if you can take all the elements of the set and apply the operator and get an element of the set as a result. As an example, if the operator if the square root function and the set if all real numbers then closure is not satisfied because if you take the square root of a negative real number you get a imaginary number, which isn't in the set of real numbers. So the set of real numbers with the square root function as the operator does not satisfy closure.

Answer 6

Vectors are a good example. They are closed under the cross product operation because when you cross two vectors the result is a vector. However when you take the dot product, the result is a scalar value (its just a number, not a vector). So not closed under dot product.

Answer 7

Thanks xavier and Jprahman, How do i determine whether the following set forms a subspace of R2? {(x1, x2)T | x1 + x2 = 0}
The properties are, meaning all I need to prove is that:
1. 0 in v
2. X in v
3. x + y in v

Answer 8

While it's not a proof I can say that in order for x1+x2 = 0 to be true then we know that x1 and x2 must be additive inverses, beyond that I don't know a lot in the way of advanced linear algebra and the definitions of space and dimensionality.

Answer 9

all i need to do in here is prove those 3 properties, such as:
1. 0 vector in v
2. x vector in v and then
3. x=y vector in v

Answer 10

I looked some of it up. I'm confused as to what the T is. The last one seems to go like: let u=(x1, -x1) and v=(x2,-x2) be elements in v. Then u+v=(x1+x2,-x1-x2) and since (x1+x2)+(-x1-x2)=0, x+y is in v.

Answer 11

One thing I have thought of is that you can visualize the subset of R2 in question as the set of all points on line with slope -1 running through the origin. Maybe that will lead to a insight?

Answer 12

Perhaps to prove 0 in v you could say that x1=0,x2=0, x1+x2=0, therefore the 0 vector (0,0) is in v.

Answer 13

Ah i see. I'm still unsure about the T, though. Another axiom states that it needs to work when multiplied by a constant. That might be the second condition with the capital X. Let u=(x1,x2) be in v then u=(x1,-x1) Also, cu=(cx1,-cx1) and since cx1-cx1=0, cu is in v.

Answer 14

Yeah, I don't understand the purpose of the T either, huntershoaib may have to explain that further.
http://en.wikipedia.org/wiki/Euclidean_subspace actually describes in greater detail what properties a Euclidean subspace must satisfy.

What about: Let u=(x1, x2), x1+x2=0 (definition of the set), cu=(cx1, cx2), cx1 + cx2 = ? -> c(x1 + x2) = ? -> c(0) = 0 -> cx1 + cx2 = 0, thus for all values of scalar c and vector u in v we find that cu is a member of v.

Answer 15

I'll have to read about that tomorrow (2am). I did something similar but I don't know too much about the subject to know if its allowed. Since the set is defined by x1+x2=0 i let x2=-x1 and worked with that. So cx1 + cx2 = cx1-cx1=0. Your method seems a more polished somehow >_>. Anyways, seems like an interesting topic.