Order of variables $f(a,b)$ vs $f(b,a)$ I wanted to ask a question about the order of variables in a function. 
Coming from a tutorial on mathematics for medicine, we discussed that if you had a function $f_{1}(a,b)$ and then swapped round the order $f_{1}(b,a)$ there appears to be a property that:
$$f_{1}(a,b) = -f_{1}(b,a)$$
which is especially relevant in situations involving imaginary numbers, and I was struggling to work out why. 
I read a page on Wikipedia about Binary relations to try and make understanding of this, but was left hard to understand why this works. 
I pondered why this really doesn't matter for when the function is a product i.e. $ab$ and this makes sense since you are only multiplying the values so the order is irrelevant e.g. $3 \times 2 \times 1$ is identical to the result of $1 \times 2 \times 3$
yet I couldn't work out why the order of variables matters, especially when related to the inner product. 
I did think briefly about cartesian equations such that $f(x,y)$ is not the same as $f(y,x)$ since you are starting on the $y$ axis and ending on the $x$ axis compared to the other way round for $f(x,y)$ but that didn't give me a definitive reason for the need of a negative sign in front of the function.
Why does there have to be a change in sign when I change the order of variables?
 A: You can think of a function of two variables as assigning a number $f(x,y)$ to each point $(x,y)$ in the plane. The graph of $f$ will be a surface - the height above $(x,y)$ is $f(x,y)$. 
Swapping the coordinates of a point reflects that point over the line $y=x$. To see that, experiment with, say $(1,2)$ and $(3,3)$.
Sometimes the value of a function at the reflected point is related to the value where you started. In your particular example, it's the negative. For example, the function
$$
f(x,y) = x^2 - y^2
$$
has that property. here's a numerical check:
$$
f(2,1) = 4-1 = 3
$$
while
$$
f(1,2) = 1 - 4 = -3.
$$
The graph is above the plane when $x > y$ and below when $x < y$. The value when $x=y$ is $x^2 - y^2 = 0$.
Most functions of two variables will not behave this way. There may be some particular function in your medical application that does.
A: 
Coming from a tutorial on mathematics for medicine, we discussed that if you had a function f1(a,b) and then swapped round the order f1(b,a)
there appears to be a property that:
f1(a,b)=−f1(b,a)

I think the tutoral was taling about one particulare function.  This is not at all true in general.
Suppose $f(a,b) = \cos a + b^2$.
Obviously it is not true that $f(a,b) = \cos a + b^2 = -\cos b - a^2 = f(b,a)$.  Why on earth would that be true?
I suspect that they were talking about a specific function.  For example: (I'm making this up off the top of my head) lesseee....oh!  This would be the simplest!  $f(a,b) = a-b$.  Then $f(b,a) = b-a = -(a-b) = -f(a,b)$.
There are probably less trivial examples.  $f(a,b) =a^3 - 3a^2b + 3ab^2 - b^3$ for example.  $f(b,a) = b^3 - 3b^2a + 3ba^3 -a^3 =-(a^3 - 3a^2b + 3ab^2 - b^3)=-f(a,b)$
