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Thread: Help please

  1. #1
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    Help please

    Discuss whether or not the given function is injective, surjective, and/or bijective.

    f: N into Q, where f(n) = 1/n

    g: X subset of R^3 into [1,infinity), where g(x,y,z) = e^sqaure root of x^2+y^2+z^2-4

    h: X subset of R^3 into R^3, where h(x) = 2x/2 norm of x
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by jburks100 View Post
    Discuss whether or not the given function is injective, surjective, and/or bijective.

    $\displaystyle f: \mathbb{N}\longrightarrow\mathbb{Q}$, where $\displaystyle f(n) = 1/n$

    $\displaystyle g: X\subseteq\mathbb{R}^3 \longrightarrow [1,\infty)$, where $\displaystyle g(x,y,z) = e^{\sqrt{x^2+y^2+z^2-4}}$

    $\displaystyle h: X\subseteq\mathbb{R}^3\longrightarrow\mathbb{R}^3$, where $\displaystyle h(x) = \frac{2x}{||x||_2}$
    $\displaystyle f$ is certainly injective, because it has a well-defined inverse. It is not surjective because not every point in $\displaystyle \mathbb{Q}$ is hit. It's not bijective because both aren't satisfied.

    $\displaystyle g$ is not injective because the points $\displaystyle (2,3,4)$ and $\displaystyle (2,4,3)$ (for example) yield the same value of $\displaystyle g$. It is, however, surjective, as every point in $\displaystyle [1,\infty)$ is hit. It's not bijective because both aren't satisfied.

    $\displaystyle h$ isn't injective, because, for example, $\displaystyle h(1,2,3)$ and $\displaystyle h(2,4,6)$ both yield the same value. It isn't surjective either, because the only points in $\displaystyle \mathbb{R}^3$ that are hit are vectors of length $\displaystyle 2$, which certainly don't compose all of 3-space.

    PS: Make sure that the way I interpreted your functions is correct.
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