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Function domain and range

  1. Jul 14, 2009 #1
    Say we have a function, [itex]f(x)=x^3[/itex]

    one would say "[itex]f[/itex] is a function from [itex]\mathbb{R}[/itex] to [itex]\mathbb{R}[/itex]" or [itex]f: \mathbb{R}\to\mathbb{R} [/itex]

    Then say we have a vector function, [itex]\vec{g}(t)=<t^2+1,t> [/itex].

    How would one use the above notation? Would it be [itex]\vec{g}: \mathbb{R}\to\mathbb{R}^2 [/itex]?

    And could one say that [itex] \mathbb{R}^2 [/itex] is the same as the vector space [itex] \mathbb{R}^2 [/itex]?

    What is the difference between a set of vectors and a set of points?
  2. jcsd
  3. Jul 14, 2009 #2
  4. Jul 15, 2009 #3
    So . . .

    Is this right then?

  5. Jul 15, 2009 #4
    (from page 2)
    So if a vector is an ordered pair of points, does this mean that a collection of vectors is a collection of points?
  6. Jul 15, 2009 #5


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    No, that means a collection of vectors is a collection of ordered pairs of points!

    If you mean to think of the "pairs of points" as simply a collection of points, you lose the "ordering" which is an important part of vectors. For example, if P, Q, and R are points and your collection of vectors is {(P,Q), (P,R)}, that is very different from the collection of points {P, Q, R}. Note, for example that the collection of vectors {(Q,P)(Q,R)} contains different vectors than the first example but would "reduce" to the same collection of points, {P, Q, R}.
  7. Jul 15, 2009 #6
    Is my post from #3 correct????

    Ok, so in linear algebra I recall reading something along the lines of
    Theorem 1:
    Suppose , [itex] \vec{u} [/itex] and [itex] \vec{v} [/itex] are vectors in [itex] \mathbb{R}^n [/itex]. We say that [itex] \vec{u} [/itex] and [itex] \vec{v} [/itex] are orthogonal if [itex] \vec{v}\cdot \vec{u}=0 [/itex].

    So... say we are "in" [itex] \mathbb{R}^2 [/itex]. Is is true that [itex] \mathbb{R}^2 =\mathbb{R}[/itex] x [itex] \mathbb{R}=\{ (x,y)|x\in \mathbb{R}[/itex] and [itex]y \in \mathbb{R} \}[/itex]?

    Then say we have a point (2,3), is it true that (2,3) [itex]\in \mathbb{R}^2[/itex]?

    Then is it also true that the vector [itex]\vec{u}=<2,3> \in \mathbb{R}^2[/itex]?
  8. Jul 15, 2009 #7
    A follow up to this...


    wiki says :

    "For any non-negative integer n, the space of all n-tuples of real numbers forms an n-dimensional vector space over R, which is denoted R^n and sometimes called real coordinate space. An element of R^n is written [itex]\vec{x}=(x_1,x_2, \ldots x_n) [/itex] . . ."

    Does this mean that since (2,3) is a point and does not have the "ordering" that vectors have, the point (2,3) is NOT an element of R^2, but the vector [itex]\vec{u}=<2,3>[/itex] IS an element of R^2? (According to wiki an element of R^n needs to be a vector)

    ISSUE #1 IN MY HEAD: if [itex]\vec{g}: \mathbb{R}\to\mathbb{R}^2 [/itex] (see post #1)
    and (x,y) is some arbitrary point, (and thus from my understanding as of now this point (x,y) is not an element of R^2)

    then how would one write the function definition of [itex]h(x,y)=x^2+y^2 [/itex]?
    I couldn't write [itex]h:\mathbb{R}^2 \to \mathbb{R} [/itex] since (x,y) isn't a vector? Or do you make it a vector, then plug it into [itex]h [/itex]?
    Last edited: Jul 15, 2009
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