Thread: injective, surjective or bijective (no2)

1. 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!

2. 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.

3. Originally Posted by GaloisTheory1
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?