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Math Help - 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 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".
    ----------------------
    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}
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