Is the Set N Closed for Addition in English?

[johann1301]

If you take two arbitrary numbers from a set N - let's say N stands for the natural numbers - and add them together, the sum will always be an element of N.

In my language, there is a word for this, but i don't know what it is in english? If i translate it from norwegian, it would be something like;

"The set N is closed for addition"

Is "closed" the correct word?

 

[micromass]

Yes, closed is the right word. Other word is stable.

 

[johann1301]

Other word is stable.

You mean "stable" is a synonym for "closed"? (in this mathematical context)

 

[micromass]

You mean "stable" is a synonym for "closed"? (in this mathematical context)

Yes.

 

[johann1301]

Then i have a couple more questions...

I have heard that a set A is called a group if A is closed for both addition and subtraction, its called a ring if its altso closed for multiplication, and finally, its a body if its closed for them all(division as well).

Are the words "group", "ring" and "body" the correct english words to describe a set?

 

[micromass]

Then i have a couple more questions...

I have heard that a set A is called a group if A is closed for both addition and subtraction, its called a ring if its altso closed for multiplication, and finally, its a body if its closed for them all(division as well).

Are the words "group", "ring" and "body" the correct english words to describe a set?

Group and ring are correct. A body is commonly used in many languages, but not english. In english it's called a field (or if you don't demand multiplication to be commutative; a division ring).

 

[johann1301]

Last question then... (i think)

I was given the problem to prove that the real numbers R is a field. The only way i can think of how to do this is by showing that √-1 does not show up in the calculation. Would this be a sufficient proof?

 

[micromass]

Last question then... (i think)

I was given the problem to prove that the real numbers R is a field. The only way i can think of how to do this is by showing that √-1 does not show up in the calculation. Would this be a sufficient proof?

What would your definition of $\mathbb{R}$ be?

 

[johann1301]

I don't have a mathematical/algebraic definition, but i would say its any number, as long its not an imaginary number. Or at least the coefficient of i, would be zero; a+bi=a+0i=a+0=a.

it could be;

zero
pie
tau
√2
2
-3
e

but not i...

 

[WWGD]

What would your definition of $\mathbb{R}$ be?

There is a list of axioms used to describe a field. You then only need to show these axioms are satisfied.

EDIT :Sorry, this was supposed to be a reply to Johann about showing the Reals are a field.

 

[Matterwave]

A linguistic tidbit: In English we usually say a set is closed under addition or subtraction or multiplication or what have you, and not closed for addition, etc.

 

[micromass]

I don't have a mathematical/algebraic definition,

If you don't have an actual definition, then it's impossible to prove it. Maybe they don't expect you to prove it but just to give some vague argument for why it could be true?

 

[Fredrik]

I don't have a mathematical/algebraic definition, but i would say its any number, as long its not an imaginary number. Or at least the coefficient of i, would be zero; a+bi=a+0i=a+0=a.

it could be;

zero
pie
tau
√2
2
-3
e

but not i...

My favorite definition of $\mathbb R$ is of the form "a field such that..." I we use that definition, there's nothing to prove. There's a similar definition of $\mathbb C$, I mean a definition that says that $\mathbb C$ is a field right at the start. This definition too makes the problem almost too easy. All you would have to do is to show that the sum and the product of two arbitrary real numbers is real. Everything else follows from the definitions that say that $\mathbb R$ is a subset of $\mathbb C$ and that $\mathbb C$ is a field.

The problem becomes much more interesting and difficult if you use one of the more explicit definitions of $\mathbb R$, for example the Dedekind-cut definition, or the equivalence classes of Cauchy sequences definition.

I think you really need to find out what definition of $\mathbb R$ you're supposed to use.

 

[FactChecker]

Last question then... (i think)

I was given the problem to prove that the real numbers R is a field. The only way i can think of how to do this is by showing that √-1 does not show up in the calculation. Would this be a sufficient proof?

No. Don't introduce complex numbers. That is extraneous. Instead, you can define the reals by their decimal representation. Then define addition, subtraction, multiplication, division by the usual algorithms. Prove that the definitions of a field are satisfied.

 

[micromass]

No. Don't introduce complex numbers. That is extraneous. Instead, you can define the reals by their decimal representation. Then define addition, subtraction, multiplication, division by the usual algorithms. Prove that the definitions of a field are satisfied.

You're in for a world of pain if you're going to define the real numbers by their decimal representations. It's really not a fun proof at all.

 

[FactChecker]

You're in for a world of pain if you're going to define the real numbers by their decimal representations. It's really not a fun proof at all.

I agree that it would be bad. But I can't think of another basic definition without defining it as the topological closure of rationals. And I think that would be a whole new can of worms. Are there any other options?

P.S. Even using the decimal representation, I don't know how to rigorously define addition of two numbers that require carrying digits from infinitely small ( like 5.555555... + 5.5555555...) So I guess I give up. Maybe the limits of rationals is the only rigorous way to do it.

 

[micromass]

I agree that it would be bad. But I can't think of another basic definition without defining it as the topological closure of rationals. And I think that would be a whole new can of worms. Are there any other options?

No, you're right. There is no easy definition. The decimal representation definition is definitely the most "intuitive", but is very difficult to work with.

P.S. Even using the decimal representation, I don't know how to rigorously define addition of two numbers that require carrying digits from infinitely small ( like 5.555555... + 5.5555555...) So I guess I give up.

It must somehow be possible to define it. I have no idea how though. And additionally, you have this awkward problem that you must define 1 = 0.9999... for some reason to make everything work. This is quite annoying.

 

[FactChecker]

It must somehow be possible to define it. I have no idea how though. And additionally, you have this awkward problem that you must define 1 = 0.9999... for some reason to make everything work. This is quite annoying.

I looked through my abstract algebra books and didn't see any that addressed this. I think it would require defining reals as limits of rationals. I see that your initial response, "What would your definition of R be?", got right to this issue. I didn't appreciate how crucial that was at the time.

 

[johann1301]

"R = The set of all real numbers, that is, all the numbers on the numberline."

This is the only definition which is given.

Im starting to think maybe this is the case:

Maybe they don't expect you to prove it but just to give some vague argument for why it could be true?

Due to what all of you are saying, it seems like a very complicated problem to solve. Way to difficult for a beginner in set-theory.

 

[johann1301]

(i think this thread might be more approriate i the homework section though, sorry about that)

 

[micromass]

Way to difficult for a beginner in set-theory.

Definitely true if you want to do things rigorously which I don't think they expect of you.

 

[Fredrik]

"R = The set of all real numbers, that is, all the numbers on the numberline."

If that's all you've been given, I think all you can do is to examine the field axioms and recognize them as statements that are satisfied by the real numbers.

 

[disregardthat]

Depending on the basis of assumptions in your given problem, the solution is simply to verify the field axioms of the real numbers. E.g., if x and y are real numbers, then x+y is a real number. I doubt that whoever gave you the exercise meant that you needed to define real numbers first.

 

[HallsofIvy]

Yes, one way to define the real numbers is as "the set of all equivalence classes of the set of all Cauchy sequences of rational numbers with two sequences, $\{a_n\}$ and $\{b_n\}$ equivalent if and only if $\lim_{n\to\infty} |a_n- b_n|= 0$. Notice that the sequences {1, 1, 1, ...} and {0.9, 0.99, 0.999, ...} are in the same equivalence class.

Equivalently use the set of all "decreasing sequence of rational numbers having a lower bound" or the set of all "increasing sequences of rational numbers having an upper bound".