Is the Set of Positive Ordered Pairs Closed Under Scalar Multiplication?

I agree that my last sentence was a bit confusing. What I meant was that if x> 0, then -x< 0. That is, it is not true that x> 0 and -x> 0.
  • #1
aznkid310
109
1
[SOLVED] Closed real vector spaces

Homework Statement


Determine whether the given set V is closed under the operations (+) and (.):

V is the set of all ordered pairs of real numbers (x,y) where x>0 and y>0:

(x,y)(+)(x',y') = (x+x',y+y')

and

c(.)(x,y) = (cx,cy), where c is a scalar, (.) = multiplication


Homework Equations


To show if they are closed or not, i know that i must satisfy a set of conditions such as:

u(+)v = v(+)u
u(+)0 = u
c(.)(u+v) = c(.)u(+)c(.)(v)
et...

I also know that (x,y)(+)(x',y') = (x+x',y+y') is closed but c(.)(x,y) = (cx,cy) is not. So how do i show this? Just use arbitrary numbers?


The Attempt at a Solution



I tried plugging in x = y = c = 1 for simplicity, but if i do that, it shows that both are closed
 
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  • #2
Why not try c=(-1)? There aren't any restrictions on the values of c, except that it's a scalar.
 
  • #3
if i do that, c(.)(x,y) = (-1,-1) = -2, which is the same as (cx,cy) = (-1,-1) = -2. But the answers are suppose to be different
 
  • #4
aznkid310 said:

Homework Statement


Determine whether the given set V is closed under the operations (+) and (.):

V is the set of all ordered pairs of real numbers (x,y) where x>0 and y>0:

(x,y)(+)(x',y') = (x+x',y+y')

and

c(.)(x,y) = (cx,cy), where c is a scalar, (.) = multiplication


Homework Equations


To show if they are closed or not, i know that i must satisfy a set of conditions such as:

u(+)v = v(+)u
u(+)0 = u
c(.)(u+v) = c(.)u(+)c(.)(v)
et...

I also know that (x,y)(+)(x',y') = (x+x',y+y') is closed but c(.)(x,y) = (cx,cy) is not. So how do i show this? Just use arbitrary numbers?


The Attempt at a Solution



I tried plugging in x = y = c = 1 for simplicity, but if i do that, it shows that both are closed

aznkid310 said:
if i do that, c(.)(x,y) = (-1,-1) = -2, which is the same as (cx,cy) = (-1,-1) = -2. But the answers are suppose to be different

In what sense is "(-1, -1)= -2"? These are operations on pairs of real numbers. I see nothing there that says you are to add the two components.

You said that "c(.)(x,y) = (cx,cy)". Why do you then say "they are supposed to be different"?

And exactly why do you believe that this set is not closed under the defined scalar multiplication?
 
  • #5
No, you do not "just use arbitrary numbers". You have to show that they are true no matter what numbers you use.

To show that the set is "closed under addition", you need to show that (x, y)(+)(a,b) is also in the set: in other words, you need to show that (x+ a, y+ b) is an "ordered pair of real numbers".

To show that the set is "closed under scalar multiplication", you need to show that c(.)(x, y)= (cx, cy) is an "ordered pair of real numbers".
 
  • #6
aznkid310 said:
if i do that, c(.)(x,y) = (-1,-1) = -2, which is the same as (cx,cy) = (-1,-1) = -2. But the answers are suppose to be different

The point is the (-1).(x,y) for x>0 and y>0 is (-x,-y). -x and -y are negative. (-x,-y) is not in V.
 
  • #7
Dick said:
The point is the (-1).(x,y) for x>0 and y>0 is (-x,-y). -x and -y are negative. (-x,-y) is not in V.

So to show that c(.)(x,y) = (cx,cy) is in v, then they must satisfy the fact that x and y must be greater than zero no matter what c is?

Btw, I know that c(.)(x,y) = (cx,cy) is not in V b/c that's what the answer is in the back of the book
 
  • #8
aznkid310 said:
So to show that c(.)(x,y) = (cx,cy) is in v, then they must satisfy the fact that x and y must be greater than zero no matter what c is?
Yep. I will confess that I didn't notice the "x> 0, y> 0" at first!

Btw, I know that c(.)(x,y) = (cx,cy) is not in V b/c that's what the answer is in the back of the book
That's not a very good answer! It would be better to say "I know that c(.)(x,y)= (cx, cy) is not in V, for all c, because if c= -1 (as Dick suggested) -1(.)(x, y)= (-x, -y) and since x> 0, -x< 0".
 

Related to Is the Set of Positive Ordered Pairs Closed Under Scalar Multiplication?

1. What is a closed real vector space?

A closed real vector space is a mathematical concept that refers to a set of vectors that is closed under addition and scalar multiplication. This means that when two vectors are added together or multiplied by a scalar, the resulting vector is also within the same set.

2. How is a closed real vector space different from an open real vector space?

A closed real vector space is different from an open real vector space in that it includes all boundary points, while an open real vector space only includes interior points. In other words, a closed real vector space contains all possible combinations of vectors, while an open real vector space only contains a subset of those combinations.

3. What are some examples of closed real vector spaces?

Examples of closed real vector spaces include the set of all real numbers (R), the set of all n-dimensional vectors (R^n), and the set of all matrices with real entries (MxN). These spaces are closed because any combination of vectors or matrices within the set will still result in a vector or matrix within the same set.

4. How are closed real vector spaces used in scientific research?

Closed real vector spaces are used in scientific research to model and analyze physical systems. They are particularly useful in physics and engineering, where they can represent properties such as forces, velocities, and accelerations. By studying closed real vector spaces, scientists can better understand and make predictions about these systems.

5. What is the importance of closed real vector spaces in linear algebra?

Closed real vector spaces are an important concept in linear algebra because they are the basis for many other mathematical concepts, such as subspaces, linear transformations, and eigenvalues. They also provide a framework for solving systems of linear equations and understanding the properties of vectors and matrices.

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