Thomas

hello

And again, I don't think the OP would find it helpful to include this in an answer. I even think the group theory answer is a bit of a stretch. The OP has currently accepted an answer that says more or less nothing, so ... But, please provide your answer. Maybe if the OP would provide some more context/background it would be easier to provide a helpful answer.

Hi palsch. Was my answer helpful?

I don't think the OP really wants to know about how to construct the real numbers. I don't think the OP is looking for that kind of depth.

That is a good question. And it might/ might not be helpful to the OP if you wrote an answer on the construction of the real numbers. My only point was that viewing the real numbers as a group, $-0$ is the inverse of $0$. Now no matter what, you will always end up having to talk about what the real numbers are. But I don't think this is helpful to the OP. But then again, I don't know.

Ok, so the real numbers under addition is a group. Here $-0$ is just the inverse of the element element $0$. $0$ is the identity element in the group.

And so you know what a group is?

Have you studied abstract algebra?

I don't have a question. I don't understand your comment.

?

Obviously the group theory answer requires more of the OP and that is why I wrote that I wasn't sure that it was helpful.

so?

I present two "view points" in my answer. The first is the "definition". One can simply answer the question by saying that by definition $0$ is a solution to the equation. The second view point is the one from group theory where we have a precise definition of what $-x$ means in an additive group. There are probably other ways to think about it.

@user170039: The second viewpoint is just from group theory where we consider the group of real numbers.

@user170039: The definition of $-x$ is an element $y$ such that $x + y = 0$. This is exactly satisfied by what we usually denote by $-x$.

Yes, the new thing was (is) that it makes PGP easy to use. That is, the encryption is done in the browser and the keys are stored locally (I think).

But I am wondering if there are any compromises.

I had heard that the "new" thing is that it makes GPG (PGP) easy.

I hadn't heard about SCRYPTmail.

I guess I am just wondering how trusted it is. Is it trusted like Truecrypt is (was)?

I don't have a goal (maybe that is a problem). I had read recently about Protonmail and I was wondering how revolutionary it actually is. I am wondering partly because of he thing with Hushmail (Canada?) and there was something about that not working very well after all. So it is just pure interest,

So maybe my question is: How does Protonmail compare? (Would that be on-topic?)

(How does Protonmail compare?)

At one point there was something called Hushmail. What is the story with that?

I didn't see the deleted message

@kalina: Thanks for that link. Would it make sense to ask for any updates to that same question? (Given that they recently went public)

@TildalWave: Thanks.

Ok, so between two Protonmail accounts at least.

@kalina: From what I understand they actually encrypt all email.

@k

Hello. I wanted to ask about Protonmail and whether or not it is considered a secure/safe/trusted email service. Would such a question be on-topic?

Ok

@AviD: Yes, I mean: did you write it?

@AviD: Are you behind the page: attrition.org/errata/charlatan/steve_gibson ?

Are you behind the site?

@AviD: Yes, I have already seen that page (it still seems a bit outdated though).

Ok, thanks. I wasn't sure about the commas when writing a letter.

Hello. I have a quick question. When sending a letter to three people, does one put a comma between each name? For example, is: "Dear Bob, Joe, and Chuck" correct?

Now an ideal of a ring $R$ is a subgroup $I$ and because of this you will always have $0\in I$ for any ideal $I$.

With these problems you have to be attentive to your definitions. Now a ring $R$ is an Abelian group $(R, +)$ with one more operation (that we usually denote by multiplication). Because of this we say that $0 \in R$ to point out that $R$ has an additive identity element.

I posted an answer for you.

I don't know that book.

What book are you using?

My questions is: What does it mean that $I$ is an ideal of $J$?

If your definition of a ring does not require the existence of a $1$ (multiplicative identity), then an ideal is actually a subring.

This all depends a bit on whether or not your rings have a multiplicative identity. If you say that $I$ is an ideal in $J$, that makes it sounds like $J$ is a ring. If a ring has a $1$, then $J= R$, so I am guessing that your rings do not have a mutiplicative identity.

Ok, I thought something was strange with your question. My second question is: In your definition of a ring, do you have a multiplicative identity?

But $\mathbb{Z}\times 0$ doesn't contain the multiplicative identity from $\mathbb{Z}\times\mathbb{Z}$.

So you mean the additive identity?

If $I$ contains the (multiplicative) identity, then $RI = R$?