Can someone explain to me why subtraction and division is both commutative? The reason I believe that they are commutative are as follows: $$ 3-2 = 3+(-2) $$ $$ 3+(-2) = (-2)+(3) $$ I've read a website that says this doesn't prove the above because I'm still "commutating the addition". But why is this a problem? What exactly differentiates this from commutating with a negative sign? Can you both explain to me intuitively why it makes a difference, and prove why this is true for all numbers?
The same with division: $$ 3*(1/6)=3/6 $$ $$ 3*(1/6) = (1/6)*3 $$
Can you explain to me intuitively why it makes a difference, and prove why this is true for all numbers?
Oh, and one final thing. I'm still a beginner. Can you explain it to me as simply as possible, using as few advanced math terms as possible?
A binary operation (like addition or subtraction, etc.) on a set $S$ can be thought of as a function from $S \times S$ to $S$. For example, subtraction on $\mathbb{R}$ is a function which takes each pair of real numbers $(x,y)$ to the real number $x - y$.
A binary operation $f$ is commutative if $f(x,y) = f(y,x)$ for all $x$ and $y$ in the set. Subtraction is not commutative because, taking your example, $f(3,2) = 3-2 =1$ and $f(2,3) = 2-3 = -1$.
When you write $3-2 = -2 + 3$, you are actually verifying the commutativity of addition, because what you’re actually showing is that $3+(-2) = (-2) + 3$, or in other words, $g(3,-2) = g(-2,3)$ where $g$ is the function that represents the binary operation that is addition.
The same goes for multiplication. If $h$ is the function representing division, then $h(x,y) = x/y$. And, taking your example, $h(3,6) = 3/6$ whereas $h(6,3) = 6/3$, which are not equal, so division is not commutative. What you are verifying is that multiplication is commutative, just as you did for subtraction.