I know how to determine if a function is surjective but it takes a lot of time and it's not always correct. So, are there any tricks?
For Example:
f : $\Bbb N$ x $\Bbb N$ → $\Bbb N$, $f(a,b) = \left(\dfrac{ab(a+b)}{2}\right) $ a surjection?
I don't think it's surjective because we can never get 2.
Not only $2$, but any prime number greater than $3$ cannot be reached. This is because we would want $$ab(a+b) = 2p$$This means $a \vert 2p$. Hence, $a=1$ or $a=2$ or $a=p$. If $a=1$, we would need $$b(b+1) = 2p$$ Not possible for any prime $p>3$. If $a=2$, we need $$b(b+2) = p$$which is clearly not possible and lastly if $a=p$, we need $b(b+p) = 2$, which is again clearly impossible.
Although $f$ is not a surjection (as others have answered) if treated as a function on the natural numbers, here are some tricks to show that $f$ is a surjection if treated as a function on the reals, $f: \mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}.$
Consider fixing one of the input coordinates and see what the resulting function does.
For example, let $a=2$, then $f(2, b)= 2b+b^2$. This is a polynomial which has a minimum at $f'(b)=0$, i.e., $0= 2+2b \rightarrow b=-1$. At $b=-1$, $f(2,-1)= -1$. This polynomial has the range $(-1, \infty)$. So we have just shown that the original function $f$ hits $(-1, \infty)$.
Now let $a=-2$, then $f(-2,b)= 2b-b^2$. You can see that this polynomial peeks somewhere above $0$ and its range includes $(-\infty, 0)$.
So, putting the two cases together, we can see that, yes, $f$ is surjective.
Similar tricks of restricting inputs can be used for other functions. Hope this helps.
Suppose $f: A \to B $ is a function. $f$ is said to be surjective if given a point $b \in B$, then we can always find a point $a \in A $ such that $f(a) = b $. So, In other words, your task is to find an element in $A$.
Another way to show surjectivity is to show that $Im f = B $. In fact, your function is not surjective since $ Im f \subset \mathbb{N} $
are there any tricks?
Is monotonicity a trick? The numerator of the rhs is monotonically increasing in both a and b.
