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Thread: Bijective

  1. #1
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    Talking Bijective

    Hi, does anyone how to solve the following problems:

    In each of the following cases, determine if the given function is bijective. If the function is bijective, find its inverse.

    (a) The function f : R -> R defined by f(x) = 2x-3
    (b) The function f : Z -> Z defined by f(x) = 2x-3


    I don't really know what's the difference between (a) and (b).
    I've done part (a), anything wrong?

    Ans: (a)
    Let yER, then if xER, then
    y is the image of x under f <=> f(x)=y
    <=> 2x-3=y
    <=> x=(y+3)/2
    so each yER is the image under f of a unique image of xER.
    Therefore, f is bijective and has an inverse (y+3)/2

    Thanks very much!!!!
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  2. #2
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    Quote Originally Posted by suedenation
    Hi, does anyone how to solve the following problems:

    In each of the following cases, determine if the given function is bijective. If the function is bijective, find its inverse.

    (a) The function f : R -> R defined by f(x) = 2x-3
    (b) The function f : Z -> Z defined by f(x) = 2x-3


    I don't really know what's the difference between (a) and (b).
    I've done part (a), anything wrong?

    Ans: (a)
    Let yER, then if xER, then
    y is the image of x under f <=> f(x)=y
    <=> 2x-3=y
    <=> x=(y+3)/2
    so each yER is the image under f of a unique image of xER.
    Therefore, f is bijective and has an inverse (y+3)/2

    Thanks very much!!!!
    a)Prove it is bijective. First it is a function (TRUE). Next, it is one-to-one (injective) because if, $\displaystyle f(a)=f(b)$ then, $\displaystyle 2a-3=2b-3$ thus, $\displaystyle a=b$ (TRUE). Next, we need to show it maps $\displaystyle \mathbb{R}$ onto $\displaystyle \mathbb{R}$. Which means $\displaystyle \forall,x\in\mathbb{R}\exists,y\in\mathbb{R}$ such as, $\displaystyle x=2y-3$. Which, is true because if you take $\displaystyle y=\frac{x+3}{2}$. Thus, $\displaystyle f:\mathbb{R}\to \mathbb{R}$ defined as $\displaystyle f(x)=2x-3$ is a bijective map. By, the property by bijective maps we replace the ordered pair $\displaystyle (x,y)$ for $\displaystyle (y,x)$ Thus, in $\displaystyle y=2x-3$ we have, $\displaystyle x=2y-3$ thus, $\displaystyle y=\frac{x+3}{2}$ is the inverse function.

    b)Problem with this map is that it is not surjective. Meaning $\displaystyle \forall, x\in\mathbb{Z}, \exists, y\in \mathbb{Z}$ such as, $\displaystyle x=2y-3$. Now neccesarily true because $\displaystyle \frac{x+3}{2}$ can be a non-integer thus, $\displaystyle \frac{x+3}{2}\not \in \mathbb{Z}$

    $\displaystyle \cal{Q}.\cal{E}.\cal{D}$
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