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Math Help - injective, surjective or bijective (no2)

  1. #1
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    injective, surjective or bijective (no2)

    Are the following functions injective, surjective or bijective

    1 f: R -->R s.t f(x) = 2x +1
    2 f: R -->R s.t f(x) = x^2
    3 f: [0, oo) --> R s.t f(x) = x^2
    4 f: [0, oo) -->[0, oo) s.t f(x) = x^2
    5 f: R -->R s.t f(x) = e^x
    6 f: R -->R s.t f(x) = log(x)

    Whether a relationship is injective, surjective or bijective is really messing with my head and the standard definitons aren't helping me much. I'll try to explain what I think for each of these. These definitions help a little

    Injective means that every member of "A" has a unique matching member in "B". You won't get two "A"s pointing to one "B", but you could have a "B" without a matching "A"

    Surjective means that every "B" has at least one matching "A" (maybe more than one). Does this mean for every B being in the codomain?


    1. Injective because every x has a unique corresponding f(x). Each x only maps to one f(x).
    Surjective because for all of the codomain there is at least one x. Is this a correct way of thinking of it?
    This means it is bijective.

    2. Not injective because f(-2) = f(2). Not every x has a unique f(x).
    Not surjective because the codomain is R and the function is not defined in R-.

    3. Injective. By changing the domain to [0, oo) this removes the problem of each x (apart from zero) having two corresponding f(x).
    Not surjective. Still not defined in R-.

    4. Injective. Same as 3.
    Surjective. By limiting the codomain to [0, oo) this sets the codomain equal to the range.

    5. Injective. Each x maps to a unique f(x)
    Not surjective. Parts of the codomain are undefined. Would this be surjective if we made the codomain R+

    6. Not injective. R- of the domain is undefined for f(x).
    Surjective. The whole codomain has been mapped onto by x.

    Any comments on how I approached this? Mistakes? Thanks!
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  2. #2
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    1 f: R -->R s.t f(x) = 2x +1

    surjective. better to prove more explicitly:

    let y \in \mathbb{R}. We need to show \exists x \in \mathbb{R} s.t. f(x)=y. Note that y= 2x +1 \Leftrightarrow y-1= 2x \Leftrightarrow \frac{y-1}{2}=x. so let x:=\frac{y-1}{2} so that f(x) is onto.
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  3. #3
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    Quote Originally Posted by GaloisTheory1 View Post
    1 f: R -->R s.t f(x) = 2x +1

    surjective. better to prove more explicitly:

    let y \in \mathbb{R}. We need to show \exists x \in \mathbb{R} s.t. f(x)=y. Note that y= 2x +1 \Leftrightarrow y-1= 2x \Leftrightarrow \frac{y-1}{2}=x. so let x:=\frac{y-1}{2} so that f(x) is onto.

    So for the second one I'd say someting like y = x^2 but x ≠ (y)^0.5 for all the codomain?
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