1. ## Subspaces

Discuss whetehr or not:

(a) R2 is a subspace of R3 over R
(b) {(0,0,0)} is a subspace of R3 over Q
(c) Q3 is a subspace of R3 over R
(d) RxQx{0} is a subspace of R3 over Q

I understand the requirements of a subset; however, I do not understand what it means for a set to be over say R. What does this mean and how does it play into deciding if one set is a subspace of another?

2. Originally Posted by TexasGirl

I understand the requirements of a subset; however, I do not understand what it means for a set to be over say R. What does this mean and how does it play into deciding if one set is a subspace of another?
Let me begin with a definition.
Definition:An abelian group $\displaystyle V$ is a "vector-space" over field $\displaystyle F$. When there exists a binary operation:
$\displaystyle *:F\times V\to V$ such as,
$\displaystyle a*\alpha \in V$
$\displaystyle 1*\alpha=\alpha$
$\displaystyle a*(b*\alpha)=(ab)*\alpha$
$\displaystyle (a+b)*a=a*\alpha+b*\alpha$
$\displaystyle a*(\alpha+\beta)=a*\alpha+a*\beta$.
Definition:If subset of a vector space is a vector space over the same field. It is called a "subspace".
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Originally Posted by TexasGirl
(a) R2 is a subspace of R3 over R
No, because we need that $\displaystyle \mathbb{R}^2$ be a subgroup of $\displaystyle \mathbb{R}^3$ but that is not true because $\displaystyle \mathbb{R}^2$ are pairs, while $\displaystyle \mathbb{R}^3$ are triples. Thus, these sets have totally differenet elements.
Originally Posted by TexasGirl
(b) {(0,0,0)} is a subspace of R3 over Q
Yes, because the binary operation we defined on $\displaystyle \mathbb{R}^3$ over $\displaystyle \mathbb{Q}$ stasfies conditions 2,3,4,5. It only is questionable whether or not this binary operation is closed for $\displaystyle \{0,0,0\}$ over $\displaystyle \mathbb{Q}$, we can see that it is closed. And finally $\displaystyle \{0,0,0\}$ is your trivial subgroup of $\displaystyle \mathbb{R}^3$ thus, $\displaystyle \{0,0,0\}$ is a subspace of $\displaystyle \mathbb{R}^3$ over $\displaystyle \mathbb{Q}$.
Originally Posted by TexaxGirl
(c) Q3 is a subspace of R3 over R
No, because what happens if $\displaystyle x\in\mathbb{R}$ is irrational. Then, the binary operation is not closed because rationals multiplied by irrationals become irrationals.
Originally Posted by TexasGirl
(d) RxQx{0} is a subspace of R3 over Q
Yes, again properties 2,3,4,5. Are true because they depend on the same binary operations that are on $\displaystyle \mathbb{R}^3$. Checking, whether it is closed. Notice $\displaystyle \{x,y,0\},x\in\mathbb{R},y\in\mathbb{Q}$ when multiplied by a rational numbers becomes,
$\displaystyle \{rx,ry,0\}$ notice that $\displaystyle rx$ is still real and $\displaystyle ry$ is still rational. Thus,
$\displaystyle \mathbb{R}\times\mathbb{Q}\times\{0\}$ is a subspace of $\displaystyle \mathbb{R}^3$ over $\displaystyle \mathbb{Q}$