Determining if function is surjective Question:
Let $\\f: \ \mathbb{R} \to \mathbb{R} \times \mathbb{R}$ via $ f(x) = (x+2, x-3)$. Is $f$ injective? Is $f$ surjective?
I was able to prove that $f$ is injective. However, I am not quite sure if $f$ is surjective. If it is surjective could someone please tell me how to prove that. If not, could someone provide a counter example.
Thanks 
 A: Hint:
Let $y=f(x)=( y_1 , y_2 ),$ notice that $y_1-y_2=5$
Now consider point $(6,0)$, does it have a preimage?
A: Well, you need to show that for any $(a,b)\in\mathbb{R}^2$ there exists $x\in\mathbb{R}$ such that $f(x)=(a,b)$. Since $f(x)=(x+2, x-3)$ then $a=x+2$ and $b=x-3$. In particular $x=a-2$ and $x=b+3$ and therefore $a-2=b+3$ and finally $a=b+5$. So that's the constraint on $(a,b)$ in order for it to be in the image of $f$. Finding a counterexample is easy now, e.g. $(a,b)=(0,0)$.
A: I think you need to understand the intuition behind surjectivity :
A fonction is surjective if it at least fully "fills" the "end space" (if someone knows the real term feel free to comment) here $\mathbb{R} \times \mathbb{R}$. 
But since it take $x$ values in $\mathbb{R}$, it won't be able to "fill" $\mathbb{R} \times \mathbb{R}$.
Here your fonction $f(x) = (x+2, x-3)$ will map $x$ to a point on the line of equation $y=x-5$ in $\mathbb{R} \times \mathbb{R}$, so everything outside of that line would not be reached, hence it is not surjective.
Another thing you should keep in mind, for a function to be bijective (injective and surjective) the starting space and ending space need to be the same dimension, so if $f:\mathbb{R} \mapsto \mathbb{R} \times \mathbb{R}$ is injective it can't be surjective since $\dim(\mathbb{R}) \ne \dim(\mathbb{R} \times \mathbb{R})$.
I hope those quick tips helped you get a better understanding of those notions.
