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> Mathematicians call this new system a number “ring”; they can create an infinite variety of them, depending on the new values they choose to incorporate. Is

by ajuc 9y ago

> Mathematicians call this new system a number “ring”; they can create an infinite variety of them, depending on the new values they choose to incorporate. Isn't this a little wrong? As far as I remember ring is any set with multiplication, negation, and addition defined so that they satisfy a few conditions. No need for the "+ b * something" part. The usual integer numbers we use form a ring too, as well as booleans. I might be missing something, and it's irrelevant to the main subject of the article.

btilly 9y ago
It is right, and so is your memory. The number system that you generate by adding sqrt(5) to the integers is a ring because it satisfies the definition. And you can create an infinite variety of rings by adding new values to the integers.
- ajuc 9y ago
  I meant the article suggested it only becomes a ring after you add b*sqrt(5) part. But yeah, I'm nitpicking.
  - richard_todd 9y ago
    I’m not sure if this is what you are missing, but: Rings are closed, so you have to add an infinity of new numbers, not just the literal sqrt(5). a + b*sqrt(5) is a convenient way to denote that all those numbers are in your system. So in some sense, once you’ve decided to incorporate sqrt(5), it really is only a ring once you add that part. Obviously there are tons of rings that have nothing to do with this recipe... maybe that is your nitpick?
  - soVeryTired 9y ago
    b * sqrt(0)? But yeah, the integers form a ring in their own right. They call the new system a ring, but that doesn't mean the old system wasn't also a ring. Also not all rings are of that form - as you say a ring is a very general algebraic structure.
gizmo686 9y ago
"Number ring" has a specific meaning [0]. I am not sure if it is equivalent to the integers adjoined with some element though. >In fact the usual integer numbers we use form a ring too. Depending on definitions, the integers can be viewed as the ring of even numbers adjoined with the element 1. This does require that we do not define rings to necessarily contain 1. In my experience this definition is is becoming more of a historical footnote though. There is also a natural generalization of adjoining multiple elements to the natural numbers (which still results in a ring) In this case, the natural numbers would just be a special case of adjoining 0 elements. The ring of real numbers, in contrast, cannot be constructed by adjoining any finite set of elements to the integers. The ring of integers mod n is also a ring, but does not contain the integers as a subring (and therefore cannot be thought of as the integers adjoined with any set (finite or infinite) of elements). The polynomials with coeficients mod n form a non-finite ring which does not contain the integers. I suspect the point that the author was attempting to make was just that Z[√5] formed a type of structure that mathematicians are familiar with. [0] http://mathworld.wolfram.com/NumberRing.html http://mathworld.wolfram.com/NumberRing.html
- ajuc 9y ago
  Oh, right, that's what I missed, number ring isn't just a ring of numbers. Thanks.
infinity0 9y ago
It's a ring, but not all rings have to be in that form. I can see that you think they weakly imply the latter, which really is just a bug in our (humans') natural language processing skills. :)

by ajuc 9y ago

btilly 9y ago

It is right, and so is your memory. The number system that you generate by adding sqrt(5) to the integers is a ring because it satisfies the definition. And you can create an infinite variety of rings by adding new values to the integers.

ajuc 9y ago
I meant the article suggested it only becomes a ring after you add b*sqrt(5) part. But yeah, I'm nitpicking.
- richard_todd 9y ago
  I’m not sure if this is what you are missing, but: Rings are closed, so you have to add an infinity of new numbers, not just the literal sqrt(5). a + b*sqrt(5) is a convenient way to denote that all those numbers are in your system. So in some sense, once you’ve decided to incorporate sqrt(5), it really is only a ring once you add that part. Obviously there are tons of rings that have nothing to do with this recipe... maybe that is your nitpick?
- soVeryTired 9y ago
  b * sqrt(0)? But yeah, the integers form a ring in their own right. They call the new system a ring, but that doesn't mean the old system wasn't also a ring. Also not all rings are of that form - as you say a ring is a very general algebraic structure.

gizmo686 9y ago

"Number ring" has a specific meaning [0]. I am not sure if it is equivalent to the integers adjoined with some element though. >In fact the usual integer numbers we use form a ring too. Depending on definitions, the integers can be viewed as the ring of even numbers adjoined with the element 1. This does require that we do not define rings to necessarily contain 1. In my experience this definition is is becoming more of a historical footnote though. There is also a natural generalization of adjoining multiple elements to the natural numbers (which still results in a ring) In this case, the natural numbers would just be a special case of adjoining 0 elements. The ring of real numbers, in contrast, cannot be constructed by adjoining any finite set of elements to the integers. The ring of integers mod n is also a ring, but does not contain the integers as a subring (and therefore cannot be thought of as the integers adjoined with any set (finite or infinite) of elements). The polynomials with coeficients mod n form a non-finite ring which does not contain the integers. I suspect the point that the author was attempting to make was just that Z[√5] formed a type of structure that mathematicians are familiar with. [0] http://mathworld.wolfram.com/NumberRing.html http://mathworld.wolfram.com/NumberRing.html

ajuc 9y ago
Oh, right, that's what I missed, number ring isn't just a ring of numbers. Thanks.

infinity0 9y ago

It's a ring, but not all rings have to be in that form. I can see that you think they weakly imply the latter, which really is just a bug in our (humans') natural language processing skills. :)