Determining injectivity and surjectivity Are these functions injective or surjective? Also, how should I go about proving this?
The function maps $ℕ×ℕ$ to $ℤ$.


*

*$f(a,b) = 4a+5b$

*$f(m,n) = m^2-n$

*$f(p,q) = 5^p·3^q$


Thanks!
 A: It's surjective
For injectivity you check that if $$f(a,b)=f(a_1,b_1) \implies (a,b)=(a_1,b_1)\\4a+5b=4a_1+5b_1\\4(a-a_1)+5(b-b_1)=0\\5(b_1-b)=4(a-a_1)\\a-a_1=5\\b_1-b=4$$
So it's not injective,for surjective you check if every number from $\mathbb{Z}$ is contained in the function.For example lets take
$$f(a,b)=3\\4a+5b=3$$
Logically if either $a$ or $b$ are $\geq1$ left side is bigger than the right side,so it's not surjective
$$f(m,n)=m^2-n\\f(m,n)=f(m_1,n_1)\\m^2-n=m_1^2-n_1\\m_1=\pm m\\n=n_1$$
Which means it's not injective but it's surjective(try to prove it)
3rd one is not surjective but is injective
A: *

*Neither.

*Surjective, but not injective since $f(n,n^2) = 0$ for all $n$

*Injective but not surjective since $-1$ has no preimage.
A: Note 
$$
4\cdot 30+5\cdot4=5\cdot(24+4)=4\cdot 5+5\cdot 24\\
(30,4)\mapsto 120 \\ 
(5,24)\mapsto 120
$$

So $$f(a,b) = 4a+5b$$ is not injective.
It is also not onto as $$-1\ne f(a,b)$$ for any $(a,b)\in \mathbb{N}\times \mathbb{N}$
Now,
$$
f(m,n)=m^2-n
$$
is surjective as $$f(n+1,n^2+n+1)\mapsto (n+1)^2-(n^2+n+1)=n$$
$$
f(n^2,(n^2+n))\mapsto n^2-(n^2+n)=-n
$$
It is not injective as:
$$
\forall n, f(n,n^2)\mapsto 0 
$$
Now,
$$
f(p,q)=5^p 3^q
$$
is not surjective as 
$$
\forall (p,q)\in \mathbb{N}\times \mathbb{N}, \\ f(p,q)\ne 0 
$$
It is injective though as:
$$
5^p3^q=5^r3^s\longrightarrow p=r \land q=s \longrightarrow (p,q)=(r,s)
$$
Because every natural number admits only one prime factorization.
