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I just want to make sure I understand this correctly. The problem:

Prove whether or not the set of all pairs of real numbers of the form $(0,y)$ with standard operations on $\mathbb R^2$ is a vector space?

Here are the axioms that define a vector space:

  1. $\mathbf{u}+\mathbf{v}$ is in $V$. Closure under addition.
  2. $\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}$. Commutative property.
  3. $\mathbf{u}+(\mathbf{v}+\mathbf{w}) = (\mathbf{u}+\mathbf{v})+\mathbf{w}$. Associative property.
  4. $V$ has a zero vector $\mathbf{0}$ such that for every $\mathbf{u}\in V$, $\mathbf{u}+\mathbf{0}=\mathbf{u}$. Additive identity.
  5. For every $\mathbf{u}\in V$, there is a vector in $V$ denoted by $-\mathbf{u}$ such that $\mathbf{u}+(-\mathbf{u}) = \mathbf{0}$. Additive inverse.
  6. $c\mathbf{u}$ is in $V$. Closure under scalar multiplication.
  7. $c(\mathbf{u}+\mathbf{v}) = c\mathbf{u}+c\mathbf{v}$. Distributive property.
  8. $(c+d)\mathbf{u}=c\mathbf{u}+d\mathbf{u}$. Distributive property.
  9. $c(d\mathbf{u})= (cd)\mathbf{u}$. Associative property.
  10. $1(\mathbf{u}) =\mathbf{u}$. Scalar identity.

Forgive me, but I will simply state the conclusions I reached for each axiom without a FULL proof.

Four definitions to be fulfilled:

  1. Vectors are of the form $(0,y)$
  2. Scalar multiplication: standard operation of $\mathbb R^2$
  3. Vector addition: standard operation of $\mathbb R^2$
  4. Scalars: $c \in \mathbb{R}$

The ten axioms:

  1. True, a real number plus a real number produces a real number. Closure under addition is fulfilled.
  2. True, again it is a property of real numbers.
  3. True, properties of real numbers once again.
  4. True, the zero vector is simple $(0,0)$.
  5. True, $(\overrightarrow{-u})$ is $(0,-y)$
  6. True, property of real numbers. Closure under scalar multiplication is fulfilled.
  7. True, property of real numbers.
  8. True, property of real numbers.
  9. True, property of real numbers.
  10. True, property of real numbers.

Now, there are a few things to be said. One, all I have to do is prove the two closure statements and have implied the rest are true due to properties of $\mathbb R^2$. Also, I would have to assume $y \in \mathbb{R}$, otherwise it fails axiom 6 (i.e. $y=0$ fails the stated form $(0,y)$).

Basically, is the above work correct in proving that the stated vector space exists?

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Seems fine. For first, instead of what you wrote informally, write $(0,a)+(0,b)=(0,a+b)$, so the sum of two elements of $V$ is in $V$. –  André Nicolas Oct 2 '12 at 21:35
    
In your opinion, is it implied that $y \epsilon \mathbb{R}$? And thanks for your help. –  Rick Oct 2 '12 at 21:37
1  
In the definition, it says all pairs of real numbers of the form $(0,y)$. I guess it is not explicitly stated, but yes, it is at least implicit that $y$ ranges over the reals. –  André Nicolas Oct 2 '12 at 21:39
    
I figured it would be, the book I have doesn't define their sets very well - especially when it comes to polynomials (all?, can $a=0$? etc.). –  Rick Oct 2 '12 at 21:41

1 Answer 1

As André Nicolas remarks (slightly edited):

Seems fine. For the first axiom, instead of what you wrote informally, write $(0,a)+(0,b)=(0,a+b)$, so the sum of two elements of $V$ is in $V$.

(OP: Is it implied that $y \in \Bbb R$?)

In the definition, it says "all pairs of real numbers of the form $(0,y)$". I guess it is not explicitly stated, but yes, it is at least implicit that $y$ ranges over the reals.

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