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Thread: Subspaces

  1. #1
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    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?
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  2. #2
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    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 $\displaystyle V$ 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".
    ----------------------
    Quote 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.
    Quote 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 $\displaystyle \mathbb{Q}$.
    Quote 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.
    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 $\displaystyle \mathbb{R}^3$. 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 $\displaystyle \mathbb{Q}$
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