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

[aznkid310]

[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

 

[Dick]

Why not try c=(-1)? There aren't any restrictions on the values of c, except that it's a scalar.

 

[aznkid310]

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

 

[HallsofIvy]

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

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?

 

[HallsofIvy]

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".

 

[Dick]

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.

 

[aznkid310]

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

 

[HallsofIvy]

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".