# What are Mathematical Operations?

1. Dec 19, 2015

### Drakkith

Staff Emeritus
What exactly is a mathematical operation? Wiki states that an operation in math is "a calculation from zero or more input values (called "operands") to an output value."

That sounds oddly like a function, but I'm assuming that not all operations are functions? What is the relationship between functions and operations? I notice that, according to the article again, operations share several properties with functions, such as having domains. It looks like a function is just a special type of operation?

Thanks all.

2. Dec 19, 2015

### HallsofIvy

The difficulty is that "operation" is so general. An 'operation' way of changing one mathematical object into another. It might be a "function" in the strict sense or it might not- "function" is a subset of "operation".

3. Dec 19, 2015

### micromass

It's the converse. An operation is a special type of function. For example, the addition operation on $\mathbb{R}$ is a function
$$+:\mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R}$$
which takes a pair of real numbers $(x,y)$ and sends it to a real number $+(x,y)$ (usually denoted as $x+y$).

Something not really being an operation on $\mathbb{R}$ is division. This would be a function
$$/:\mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R}$$
but this is not defined in $(1,0)$ for example. So it's not a function and thus not an operation.
It is an operation on $\mathbb{R}\setminus\{0\}$ however.

4. Dec 19, 2015

### Samy_A

I always thought that operations were functions on a subset of a product space. But I guess it is a question of definition.

But then I don't see the difference between a function and a unary operation ...

5. Dec 19, 2015

### Drakkith

Staff Emeritus
Seems like there's some contradiction here between Micromass and HallsofIvy. Any thoughts?

6. Dec 20, 2015

### Svein

In my world, a function is something that takes one ore more parameters and return one value. An operation is more general.

Examples:
• Rotating a vector is an operation (it could be expressed as a vector function, but then you would have to define that rigorously).
• Finding the inverse image of a function is an operation, but usually not a function

7. Dec 20, 2015

### micromass

All of these can be described as functions though.

Drakkith, as you can see, there is a lot of disagreement on the word "operation". In my post, I took the word "operation" as in abstract algebra. Evidently, not everybody takes this view. So you can probably expect different books to have different kinds of definitions for operation. So I think it might be better not to think much on the word operation, since it is not really standardized.

8. Dec 20, 2015

### Drakkith

Staff Emeritus
I see. I didn't realize it wasn't a standardized term.

9. Dec 20, 2015

### chiro

If you want a general idea then it is best served by looking at an algebra - which starts from groups and gets more general.

If you want a specific idea then it is a function of the structure being used.

Although operators don't have to have the sorts of constraints that the field of algebra imposes it is a good idea to do so.

Note that algebra's usually assume that every object has the same representation which makes things a lot easier and you also get consistency and sensible results when these constraints are followed (like in groups and higher level structures).

As an example - you can't divide by zero with real numbers but you can use any rotation matrix in that algebra (for that dimension) and everything has an inverse which in terms of consistency means that any combination of those structures given the operator in the group can always be "un-done".

10. Dec 20, 2015

### Drakkith

Staff Emeritus
Whew, this seems like it's going to be a little above my head. I don't even know what a 'group' in algebra is.

11. Dec 21, 2015

### AgentSmith

Informally, a group G is an algebraic structure that has an associative operation, an identity element, and an inverse for every member of the group.
More formally, (G,o) such that:
1: The operation o is associative.
2: There exists an identity element e belonging to G for o.
3: Every x belonging to G has an inverse x^-1 in G.
4. If o is commutative, then G is an Abelian group. Addition, subtraction, multiplication, are examples of operations.

Sorry, I don't know how to do this with all the correct symbols.

12. Dec 21, 2015

### MrAnchovy

Don't worry about it, I think the key point is that the term 'operation' does have a specific meaning in abstract algebra but there is no useful general meaning. In what context have you come across the term?

13. Dec 21, 2015

### MrAnchovy

You missed arguably the most important property - closure. And your no. 4 is not part of the definition of a group.

Drakkith I find that examples are the best way of introducing algebraic concepts:
1. Closure - integers under division are not a group because they are not closed: $\frac12$ is not an integer
2. Associativity - integers under subtraction are not a group because subtraction is not associative: $(1 - 1) - 1 \ne 1 - (1 - 1)$
3. Identity - positive integers under addition are not a group because the identity element for addition (0) is not a positive integer
4. Inverse - integers under multiplication are not a group because the multiplicative inverse of z is $frac 1x$ which is not in general an integer
However integers under addition are a group because:
1. The sum of any two integers is also an integer so they are closed
2. Addition is associative: (x + y) + z always equals x + (y + z)
3. The identity element for addition, 0, is in the integers: x + 0 = x
4. Every integer k has an additive inverse -k in the integers: (x + k) + (-k) = x

14. Dec 21, 2015

### Drakkith

Staff Emeritus
I haven't really. I tutor Calculus and below at my college so I use the term when tutoring. I just got curious one day about what an operation is.

Doesn't this conflict with number 3 in your first list:

15. Dec 21, 2015

### micromass

Notice the word positive in the second quote.

16. Dec 21, 2015

### chiro

Basically you add constraints in algebra that may not have been there if you looked at things that are two general.

One problem that most people understand by high school (possibly even primary school) is that you can't divide by zero and that you can't make normal arithmetic consistent if you allow it. If you do this under a normal one dimensional number system (or even the complex numbers) then everything breaks down completely.

But if you have the sort of thing that a group has then things retain that consistency and you also get a situation where regardless of what operator (a binary one for a group) exists, you can always do algebra with it and the structures within the group.

It's kind of a baseline so that you don't have to worry about inconsistencies and problems and you have to add constraints to do that.

Also note that mathematicians try and take really abstract things and find properties or attributes that tell us something fundamental about that class of representations.

One of the most important attributes of a common structure (a square matrix) is its determinant. This tells us not only the hyper-volume but also whether it has an inverse.

With group theory, the same incentive and aim exists but it exists for algebra in an even more abstract sense.

Just be aware that one of the most important aspects in all of mathematics - regardless of field or specialization is consistency. If something is found not to be consistent then that is a big problem and I would say that algebra did in some way facilitate solving the problem that zero created in the one dimensional and complex numbers when division and its inverse multiplication were involved.

17. Dec 21, 2015

### Drakkith

Staff Emeritus
I did notice it. Does that somehow change things?

18. Dec 21, 2015

### pwsnafu

The set of integers contains zero, the set of positive integers does not. Zero is the only candidate for the identity under addition, hence the latter cannot form a group.

19. Dec 21, 2015

### Drakkith

Staff Emeritus
Oh I see. If Identity is a property of a group, then the positive integers by themselves are not a group since 0 is not a positive integer.

20. Dec 22, 2015

### WWGD

But there are other obstacles to the positive integers becoming a group (under addition) . If you choose the nonnegative integers, you will not have the (additive) inverses needed to make a group. There is an interesting idea , I think, associated with this: the group generated by EDIT a subset S of the group, as Micromass pointed out, which is the smallest (cardinality-wise) group that contains the set S. In this case, the group generated by the nonnegative integers is the set of all integers under addition.

Last edited: Dec 22, 2015