Binary Operations: Addition & Subtraction

[bentley4]

I am unsure what is meant by binary operation.
Is the binary operation of addition : addition and subtraction or is it just addition?

 

[tiny-tim]

hi bentley4!

"binary" just means that it has two inputs …

a + b is a binary operation …

and a - b is a different binary operation! ​

 

[bentley4]

Thnx for your answer Tiny-tim.
But on the wiki page of ring:

The most familiar example of a ring is the set of all integers, Z.

And the definition of ring acc. to wiki is: "An algebraic structure consisting of a set together with two binary operations usually called addition and multiplication, where the set is an abelian group under addition (called the additive group of the ring) and a monoid under multiplication such that multiplication distributes over addition."

So if we have the binary operation of just addition, we can do addition and subtraction since int. numbers can also be negative. But we also use devision in integer calculation next to multiplication so this is a third distinct binary operation. Yet it says a ring exists only of 2 binary operations, I don't understand.

Also, how can we prove the associativity of a ring if we only have 2 inputs? Or for what would associativity ever be useful if we can only use 2 inputs?
So when I calculate in Z I can only restrict myself to 2 inputs?? I don't understand.

 

[tiny-tim]

hi bentley4!

And the definition of ring acc. to wiki is: "An algebraic structure consisting of a set together with two binary operations usually called addition and multiplication, where the set is an abelian group under addition (called the additive group of the ring) and a monoid under multiplication such that multiplication distributes over addition."

So if we have the binary operation of just addition, we can do addition and subtraction since int. numbers can also be negative. But we also use devision in integer calculation next to multiplication so this is a third distinct binary operation. Yet it says a ring exists only of 2 binary operations, I don't understand.

there are infinitely many binary operations on a ring (a + 2b, a + 3b, etc) …

but two of them are defined as being the addition and multiplication in that particular ring

Also, how can we prove the associativity of a ring if we only have 2 inputs? Or for what would associativity ever be useful if we can only use 2 inputs?
So when I calculate in Z I can only restrict myself to 2 inputs??

a(b + c) and ab + ac are ternary operations (3 inputs) …

we can have operations with as many inputs as we want

 

[bentley4]

hi bentley4!

Hey tiny-tim You seem a really kind person.

there are infinitely many binary operations on a ring (a + 2b, a + 3b, etc) …

Aha, ok.

but two of them are defined as being the addition and multiplication in that particular ring

Don't we need a proof to make this kind of extrapolation?(or it is just unprovable and we assume it is right?)
Ok, but why isn't division defined for a binary operation in the ring of Z? We need to define it if we want to use devision right? Or does the ring of Z not completely encompass Z as we normally use(such as in informatics)?

a(b + c) and ab + ac are ternary operations (3 inputs) …

So a unitary operation is just a number then?

we can have operations with as many inputs as we want

Where is this defined that you can do that, is it axiomised?

 

[Fredrik]

So a unitary operation is just a number then?

No.

A unary operation on a set X is a function from X into X. (The word "unitary" means something else entirely).
A binary operation on a set X is a function from X×X into X.
...
An n-ary operation on a set X is a function from X×...×X (that's n copies of X) into X.

Where is this defined that you can do that, is it axiomised?

The existence of Cartesian products is guaranteed by the axioms of set theory, and given any two sets X and Y, the set of functions from X into Y is also guaranteed to exist by the axioms of set theory. This implies that the existence of n-ary operations for any n is also guaranteed to exist.

Don't we need a proof to make this kind of extrapolation?(or it is just unprovable and we assume it is right?)

You seem to be asking for a proof of the fact that the term "addition" is used only for one binary operation on the set. Obviously, that isn't something that needs to be proved. If you meant to ask something else, you will have to clarify.

Ok, but why isn't division defined for a binary operation in the ring of Z?

Here you seem to be asking why 1/n isn't always an integer.

 

[tiny-tim]

hey bentley4!

Don't we need a proof to make this kind of extrapolation?(or it is just unprovable and we assume it is right?)
Ok, but why isn't division defined for a binary operation in the ring of Z? We need to define it if we want to use devision right? Or does the ring of Z not completely encompass Z as we normally use(such as in informatics)?

Z is a set

(Z,+) is a group

(Z,+,x) is a ring

(Z,|| ||) is a metric space

all these Zs are the same numbers, and we can put whatever structure we want on them

if we only define the + operation, for example, then we can easily make multiplication by a whole number (just by adding that number of times), but not multiplication by √2 (unless we have also defined a distance, || ||)

division is the inverse of multiplication …*in some rings, that works, and division exists, but in other rings it doesn't

So a unitary operation is just a number then?

no, it can for example, be a vector … t -> r(t) is a unary operation

Where is this defined that you can do that, is it axiomised?

that question doesn't make sense …

we can define (on a set) anything we like ​

 

[bentley4]

Hi Fredrik, thnx for your response again.

No.
A unary operation on a set X is a function from X into X. (The word "unitary" means something else entirely).
A binary operation on a set X is a function from X×X into X.
...
An n-ary operation on a set X is a function from X×...×X (that's n copies of X) into X.

I still don't quite understand a unary operation.
Suppose we have a set L={2,3}
Then a binary operation on L is a function of LxL={(2,2),(2,3),(3,2),(3,3)} into L={2,3}.
What do you mean with into?
From what I understand a binary operation would be (3,3)+(into?)(2,3)
This would equal (5,6), but this is not an element of any of these two sets.(only true in every case for an infinite set right?)
A unary operation would be L={2,3} into L={2,3}.
So 2+2 for example? I don't get it

The existence of Cartesian products is guaranteed by the axioms of set theory, and given any two sets X and Y, the set of functions from X into Y is also guaranteed to exist by the axioms of set theory. This implies that the existence of n-ary operations for any n is also guaranteed to exist.

Ah. I think I understand now. I just meant to say in just R^1 for example, we can do a multiplication. For this multiplication, no cartesian product needs to be done.
Multiplication can be derived from addition. such as 4*2= 4+4. But is addition defined? The fact that we say that 3 > 2 is due to language not set in the ZFC. I guess this is just accepted to be trivial. Just want to cover my grounds so I don't have to think of it again.

Here you seem to be asking why 1/n isn't always an integer.

So, for division, Z is not closed and isn't a ring(for division)?

 

[bentley4]

Hi Fredrik, thnx for your response again.
Ah. I think I understand now. I just meant to say in just R^1 for example, we can do a multiplication. For this multiplication, no cartesian product needs to be done.
Multiplication can be derived from addition. such as 4*2= 4+4. But is addition defined? The fact that we say that 3 > 2 is due to language not set in the ZFC. I guess this is just accepted to be trivial. Just want to cover my grounds so I don't have to think of it again.

This is answered https://www.physicsforums.com/showthread.php?p=3301695&posted=1#post3301695" by stephen Tashi.

 

[Fredrik]

I still don't quite understand a unary operation.
Suppose we have a set L={2,3}
Then a binary operation on L is a function of LxL={(2,2),(2,3),(3,2),(3,3)} into L={2,3}.
What do you mean with into?

It seems that the concept you're having difficulties with is "function". Loosely speaking, a function from X into Y is an assignment of a member of Y to each member of X. (This isn't the definition of "function". It's just a useful way to think about functions). If f is a function from X into Y, then for each x in X, the member of Y that f associates with x is denoted by f(x). (X is called the domain of f. Y is called the codomain of f. The statement "f is a function from X into Y" is written as f:X→Y).

A unary operation on your set L is a function u from L into L. So for each x in L, u(x) is in L.

Examples: The function that takes each non-zero real number x to 1/x. This is a unary operation on ℝ-{0}. If you need to see a unary operation on L, consider e.g. the function u:L→L defined by u(2)=3 and u(3)=2.

From what I understand a binary operation would be (3,3)+(into?)(2,3)
This would equal (5,6), but this is not an element of any of these two sets.(only true in every case for an infinite set right?)

I don't understand what you're saying here. To specify a binary operation on L, let's call it b, you have to specify b(x,y) for each ordered pair (x,y) in L×L. Example: b(2,2)=2, b(2,3)=3, b(3,2)=3, b(3,3)=2.

I just meant to say in just R^1 for example, we can do a multiplication. For this multiplication, no cartesian product needs to be done.

The multiplication operation on ℝ is a binary operation on ℝ, i.e. a function from ℝ×ℝ into ℝ.

But is addition defined?

There are many ways to define the integers. One way to do it is to say that if (Z,+,·) is a ring such that the set Z and the binary operations + and · satisfy certain conditions, the members of Z are called integers. Another way is to explicitly show that the ZFC axioms implies that such a ring exists. If you choose the latter option, the addition operation must be defined.

So, for division, Z is not closed and isn't a ring(for division)?

Right, if you view Z as a subset of the real numbers, it makes sense to say that it's not closed under division. If you think of Z as just a ring, division is simply undefined. It's still a ring, but not a division ring.

 

[Stephen Tashi]

And the definition of ring acc. to wiki is: "An algebraic structure consisting of a set together with two binary operations usually called addition and multiplication.

The tradition in mathematical definitions is not to interpret the statement that there are two of something to mean that there are exactly two or nor more than two of the thing. The definition you cited means there are at least two binary operations in a ring.

When we wish to specify that there are exactly two of a thing, we are supposed say "exactly two" or "two and only two" etc.

So it's technically correct to say "There are two primes less than 100" even though there are more. (It's not wise to say such a thing. It's best to be clear.)

 

[bentley4]

A unary operation on your set L is a function u from L into L. So for each x in L, u(x) is in L.
Examples: The function that takes each non-zero real number x to 1/x. This is a unary operation on ℝ-{0}. If you need to see a unary operation on L, consider e.g. the function u:L→L defined by u(2)=3 and u(3)=2.

Suppose again the set L={2,3}. For u:L→L defined by u(2)=3 and u(3)=2 then u is a unary relation(operation).
In this case there doesn't exist any unary operation u right?

I don't understand what you're saying here. To specify a binary operation on L, let's call it b, you have to specify b(x,y) for each ordered pair (x,y) in L×L. Example: b(2,2)=2, b(2,3)=3, b(3,2)=3, b(3,3)=2.

So an example of b:LxL→L is b(x)=2 How can a relation of pairs give a number? Can you give an example in this set?
An example of b:L→LxL is b(x)=(2,3) How can a relation of numbers give an ordered pair as a result?

The multiplication operation on ℝ is a binary operation on ℝ, i.e. a function from ℝ×ℝ into ℝ.

But a multiplication could just as well be a ternary operation right? If not, which are these operations called then?

There are many ways to define the integers. One way to do it is to say that if (Z,+,·) is a ring such that the set Z and the binary operations + and · satisfy certain conditions, the members of Z are called integers. Another way is to explicitly show that the ZFC axioms implies that such a ring exists. If you choose the latter option, the addition operation must be defined.

Ok, thnx.
For the case of S=LxL={(2,2),(2,3),(3,2),(3,3)}.
This set has no identity element for addition or multiplication.
Is not closed for addition and multiplication.
Is associative for addition and multiplication.
is commutative for addition and multiplication.
Is this correct?

 

[Fredrik]

Suppose again the set L={2,3}. For u:L→L defined by u(2)=3 and u(3)=2 then u is a unary relation(operation).
In this case there doesn't exist any unary operation u right?

What do you mean? I just told you that this u is a unary operation on L. Is the word "other" missing from your question? As in "there doesn't exist any other unary operations on L, right?". In that case, the answer is that there are exactly four unary operations on L. Can you find them on your own?

Don't confuse relations with operations. For example, a binary operation on a set S is a function from S×S into S, but a binary relation on S is a subset of S×S. A unary relation on S would be a subset of S.

So an example of b:LxL→L is b(x)=2

If you mean "b(x)=2 for all x in L×L", then yes, this function b would be a binary operation on L.

How can a relation of pairs give a number? Can you give an example in this set?
An example of b:L→LxL is b(x)=(2,3) How can a relation of numbers give an ordered pair as a result?

I don't understand what you're asking.

But a multiplication could just as well be a ternary operation right? If not, which are these operations called then?

The definition of the real numbers includes two binary relations, one of them called "multiplication". You can of course use it to define a ternary relation, e.g. t(x,y,z)=x(yz).

For the case of S=LxL={(2,2),(2,3),(3,2),(3,3)}.
This set has no identity element for addition or multiplication.
Is not closed for addition and multiplication.
Is associative for addition and multiplication.
is commutative for addition and multiplication.
Is this correct?

It depends on what binary operations on you have in mind when you say "addition" and "multiplication". I don't even know if what you have in mind are binary operations on L or on S.