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  1. #1
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    Equinumerous

    Show that the following pairs of sets S and T are equinumerous by finding a specific bijection between the two sets in each pair.

    a. S= [0,1] and T = [1,3]

    b. S= [0,1] and T = [0, infinity]
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  2. #2
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    Quote Originally Posted by GoldendoodleMom View Post
    Show that the following pairs of sets S and T are equinumerous by finding a specific bijection between the two sets in each pair.

    a. S= [0,1] and T = [1,3]
    Think geometrically. Length of $\displaystyle [0,1]$ is $\displaystyle 1$ and length of $\displaystyle [1,3]$ is $\displaystyle 2$. Thus, strech $\displaystyle [0,1]$ to $\displaystyle [0,2]$. Now shift this interval by one to the right this gives $\displaystyle [1,3]$.

    Now the streching can be desribed by function $\displaystyle f :[0,1] \to [0,2]$ by $\displaystyle f(x) = 2x$. And the shifting can be described by $\displaystyle g:[0,2]\to [1,3]$ by $\displaystyle g(x) = x+1$. Thus, $\displaystyle g\circ f: [0,1] \to [1,3]$ is what you are looking for. Finally $\displaystyle g(f(x)) = 2x+1$ is the mapping you want.
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  3. #3
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    Quote Originally Posted by GoldendoodleMom View Post
    Show that the following pairs of sets S and T are equinumerous by finding a specific bijection between the two sets in each pair.

    b. S= [0,1] and T = [0, infinity]
    Consider $\displaystyle f:[0,1) \to [0, \infty)$ defined by $\displaystyle f(x)=\frac{x}{1-x}$ for $\displaystyle x \in [0,1)$

    RonL
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    Quote Originally Posted by CaptainBlack View Post
    Consider $\displaystyle f:[0,1) \to [0, \infty)$ defined by $\displaystyle f(x)=\frac{x}{1-x}$ for $\displaystyle x \in [0,1)$

    RonL
    Also an alternative is to use $\displaystyle \tan$ and/or $\displaystyle \tan^{-1}$ (inverse of $\displaystyle \tan$), these really help too much.
    $\displaystyle f:[0,1)\to[0,\infty)$
    .........$\displaystyle t\to f(t)=\tan\bigg(\frac{\pi}{2}t\bigg)$
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