# Subspaces

• Apr 7th 2006, 05:28 AM
TexasGirl
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?
• Apr 7th 2006, 06:49 AM
ThePerfectHacker
Quote:

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 $V$ is a "vector-space" over field $F$. When there exists a binary operation:
$*:F\times V\to V$ such as,
$a*\alpha \in V$
$1*\alpha=\alpha$
$a*(b*\alpha)=(ab)*\alpha$
$(a+b)*a=a*\alpha+b*\alpha$
$a*(\alpha+\beta)=a*\alpha+a*\beta$.
Definition:If subset of a vector space is a vector space over the same field. It is called a "subspace".
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Quote:

Originally Posted by TexasGirl
(a) R2 is a subspace of R3 over R

No, because we need that $\mathbb{R}^2$ be a subgroup of $\mathbb{R}^3$ but that is not true because $\mathbb{R}^2$ are pairs, while $\mathbb{R}^3$ are triples. Thus, these sets have totally differenet elements.
Quote:

Originally Posted by TexasGirl
(b) {(0,0,0)} is a subspace of R3 over Q

Yes, because the binary operation we defined on $\mathbb{R}^3$ over $\mathbb{Q}$ stasfies conditions 2,3,4,5. It only is questionable whether or not this binary operation is closed for $\{0,0,0\}$ over $\mathbb{Q}$, we can see that it is closed. And finally $\{0,0,0\}$ is your trivial subgroup of $\mathbb{R}^3$ thus, $\{0,0,0\}$ is a subspace of $\mathbb{R}^3$ over $\mathbb{Q}$.
Quote:

Originally Posted by TexaxGirl
(c) Q3 is a subspace of R3 over R

No, because what happens if $x\in\mathbb{R}$ is irrational. Then, the binary operation is not closed because rationals multiplied by irrationals become irrationals.
Quote:

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 $\mathbb{R}^3$. Checking, whether it is closed. Notice $\{x,y,0\},x\in\mathbb{R},y\in\mathbb{Q}$ when multiplied by a rational numbers becomes,
$\{rx,ry,0\}$ notice that $rx$ is still real and $ry$ is still rational. Thus,
$\mathbb{R}\times\mathbb{Q}\times\{0\}$ is a subspace of $\mathbb{R}^3$ over $\mathbb{Q}$