To divide $60$ by $12$ using subtraction:
$$\begin{align*}
&60-12=48\qquad\text{count }1\\
&48-12=36\qquad\text{count }2\\
&36-12=24\qquad\text{count }3\\
&24-12=12\qquad\text{count }4\\
&12-12=0\qquad\;\text{ count }5\;.
\end{align*}$$
Thus, $60\div 12=5$.
You can even handle remainders:
$$\begin{align*}
&64-12=52\qquad\text{count }1\\
&52-12=40\qquad\text{count }2\\
&40-12=28\qquad\text{count }3\\
&28-12=16\qquad\text{count }4\\
&16-12=4\qquad\;\text{ count }5\;.
\end{align*}$$
$4<12$, so $64\div 12$ is $5$ with a remainder of $4$.
Let's clarify:
$$9 - 4 + 3 \color{red}{\ne} 9 - (4 + 3) \tag{1}$$
It appears that you are confusing what is means to group together, or associate, the operations.
- Yes, addition and subtraction are commutative: The operations can be performed in any order.
- Yes, addition and subtraction are associative:
The terms can be grouped in any order before conducting the operations.
BUT, the mistake in the statement $(1)$ above is that the terms haven't been grouped correctly. The correct way to associate the latter two terms is:
$$\begin{align*}9 - 4 + 3 &= 9 + (-4 + 3) \tag{2}\\
&=9-(4-3)
\end{align*}$$
In the original statement $(1)$ at the top of this post, what you have done is introduced a second minus sign.
$$\color{red}{9 - (4 + 3) = 9 - 4 - 3} \tag{3}$$
To conclude, there is no ambiguity to what either $(2)$ or $(3)$ means. But they mean completely different things. The parenthesis, used for grouping in this example, must adhere to the multiplicative property of distribution. If we stick parenthesis into a math statement at will, then we run the risk of completely altering the results. To group items properly, we must make sure that our result conveys the same message - the same order of operations.
Best Answer
I claim it is impossible. Let $a = b = \sqrt{2}$. Then any number you can make with your operations is a rational multiple of $\sqrt{2}$, and in particular you cannot make $ab = 2$.
edit: this answer was given when it was unclear whether the use of arbitrary real constants was allowed. I assumed they were not, but the question has since been edited to indicate they are. I'm keeping this answer anyway because people seem to like it.