How do I determine if these relations are functions, surjectives or injectives?
So here is Question 1:
$\mathbb{f = \{(a, b) ∈ N^2 × N | a ∈ N^2 , a = (a_1, a_2), b, a_1, a_2 ∈ N, b = a_1a_2\}}$
And Question 2:
$\mathbb{f = \{(x, y) ∈ S^2| y = x^2\},\text{ where }S = \{x ∈ R | x ≥ 0\}}$
I am having a lot of trouble trying to figure these two problems out. The problems themselves are very overbearing and I have no idea where to start.
First Relation:
$$\mathbb{f = \{(a, b) ∈ N^2 × N | a ∈ N^2 , a = (a_1, a_2), b, a_1, a_2 ∈ N, b = a_1a_2\}}$$
Function: Yes
This one is a function, Because $b=a_1.a_2$ is always unique. It's means that always multiplying two integer numbers result is an unique integer number. So it's a function.
Surjective: Yes
Since all numbers in $\mathbb{N}$ is covered by multiplying at least a pair of two integer numbers, so it's a surjective.
Injectives:No
Assume that $b=16$ so we may have these possible $a_1$ and $a_2$. $$a_1=1 \; a_2=16$$ $$a_1=2 \; a_2=8$$ $$a_1=4 \; a_2=4$$ $$a_1=8 \; a_2=2$$ $$a_1=16 \; a_2=1$$
For all of these we have $b=16$ so this function is not Injectives, since result of multiplying all these pairs will be same and $16$
Second Relation
$\mathbb{f = \{(x, y) ∈ S^2| y = x^2\},\text{ where }S = \{x ∈ R | x ≥ 0\}}$
Function: Yes
Since result of $x^2=x.x$ is always unique. It's means that for each $x$ we always have a unique result for $y=x^2$.
Surjective: Yes
Since it's covers all number in $S = \{x ∈ R | x ≥ 0\}$. Assume there is a real positive number $Y$, since $$X=+\sqrt{Y} \;or\; X=-\sqrt{Y}$$ We must have a positive value for $x$ so $$x=X=+\sqrt{Y}$$. So this relation is Surjective
Injectives:Yes
Since for each $y$ there is only one $x$ so this relation is injectives.