# Proof that a pair is injective and surjective

Sorry in advance for my poor maths skills. I've picked up maths in a computing course after over a decade of being away from the subject. I have a question where I need to show if the below function is injective and / or surjective.

$f: \mathbb{Z}{\times}\mathbb{Z} \to \mathbb{Z}$ $f(a,b)=a + b + 3$

If I use a non-pair I know how to solve this, I would say $f(a) = f(b)$ and solve this and see if it works for integer numbers. But with a pair as the input I'm a bit confused as to how to show whether this is injective / surjective or not.

for example:

$f(a_0, b_0) = f(a_1,b_1) \\ a_0 + b_0 + 3 = a_1 + b_1 + 3 \\ a_0 + b_0 = a_1 + b_1$

If I then say $a_0 = 0, b_0 = 1, a_1 = 1, b_1 = 0$

$0 + 1 = 1 + 0$

So the equation works even though the inputs were different. Is this proof that this function is not injective?

You indeed proved that $f$ isn’t injective. It is surjective as any element $c \in \mathbb Z$ is the image of $(c, -3)$.

Not Injective:

Rephrasing:

$f(a,b) = a+b + 3;$

$f(c,d) = c +d +3 ;$

$f(a,b) =f(c,d) \iff$

$a+b = c+d.$

Example:

$a+b=7$, with: $a=3, b=4.$

$c+d=7$ , with $c=5, d=2$.

Not injective:

Two different points $(a,b)$ and $(c,d)$ are mapped onto the same image point.

Surjective :

Let $b=0$, and consider any $z \in \mathbb{Z}.$

Then : $f(a,0) =a +3 =z.$

Set $a = z-3$ to obtain

$f(a,0) =z.$

Yes it is. You found two distincts elements $$(0,1) \text{ and } (1,0)$$ with the same image by $f$, so this function is not injective.