Is it safe to assume in Math that we will always have n - 1 operators for n operands?

Question

In math (or maybe more specifically basic arithmetic), is it safe to assume the for the number of operators involved in an operation will always be 1 less than the number of operands? Also would this apply to logical operations? (AND OR expression evaluation)

Answer 1

No. Consider $\sqrt5$, which has one operand ($5$) and one unary operator ($\sqrt{{\vphantom 5}\cdot{}}$). I wouldn't rule out operators which take more than two operands either, like the multinomial coefficients.

For logical expressions, you have the NOT operation, which gives you a similar counterexample. And in programming, the ternary operator A ? B : C is quite common (the operator has two separate symbols, but it really is only one operator).

If we want to have fun with this, we could consider a symbol like $5$ itself a null-ary operator, taking no operands and giving you the actual number you want back. But I see no reason to do that seriously.

Answer 2

An $k$-ary operator replaces $k$ operands by a single one, hence the operators in a balanced expression must be such that the sum of their "arities" minus one equals $n-1$.

If there are just $m$ binary operators, $m(2-1)=n-1$.

Answer 3

For binary operators, the given statement is true for simple arithematic. Though, this may not hold true when we deal with matrices. Two matrices may have elements greater than $1$ and still be represented by a single operand. For instance, $\begin{bmatrix}a&0\\c&d\end{bmatrix} $+$ \begin{bmatrix}e&0\\f&g\end{bmatrix}$

Also, for unary operators,e.g. negation, the no. of operators may be more than 1, e.g., $\lnot( A \cup B)$ has $2$ operators and $2$ operands.