Is subtraction an operation on integers?

Question

I have a question regarding operations. I use the following definition: An operation * on A is a rule which assigns to each ordered pair (a,b) of elements of A exactly one element a*b in A.

This implies that a*b must be UNIQUELY identified. I don't exactly understand what it means.

For example: Is subtraction an operation on integers?

If a and b are integers then for example:

for (3,2): 3-2=1

for (4,3): 4-3=1 is the same as for (3,2)

Does it mean that subtraction is not an operation on integers?

Thanks,

Leszek

Answer 1

An operator is a function. This means that $4 - 3$ (or think of it as $-(4, 3)$ if you want to look at it as a function) maps to a single point. That is, $4 - 3 = 1$, and there isn't a value different from $1$ such that $4-3$ is equal to that value. So for example, you don't map $f(x) = 3$ and $f(x) = 5$, for the same $x$, right? The same thing applies to operators.

Answer 2

Subtraction is indeed a binary operation on the integers.

For every ordered pair $(m, n)$, $m - n = a\in \mathbb Z$, and there is no other integer $b \neq a$ such that $m - n = b$.

So indeed, for every pair of integers in $\mathbb Z$, there is one and only one such integer $a$ such that $m - n = a$.

There is no problem if there are two other integers $m', n'$ such that $m - n = m' - n' = a$. What matters is that if $m' - n' = a$, then for every integer $b \neq a$, $m'-n' \neq b$.

Answer 3

It might clear up your confusion to see an example of a function that is not an operation. Suppose we define a function $f$ on rational numbers in this way: $$ f\left(\frac{a}{b}\right)=a+b. $$ This looks perfectly reasonable at first glance. The problem is that our "function" can have different outputs given the same input. For instance, $2/1=4/2$ but $f(2/1)=3$ while $f(4/2)=6$. The technical term is that $f$ is not well-defined.

Answer 4

Subtraction is a binary operation on integers: given two integers one can subtract the second from the first. Obviously it is not commutative $a-b\neq b-a$. But what is problematic is that subtraction is not associative: $(a-b)-c$ is not the same as $a-(b-c)$. So we have to be careful and use brackets.