one more upvote till i cap, but I think for beginners the proof with sequences is more instructive

capped thank you all :)

@DominicMichaelis Yeah, that was the idea =)

gee i am still in the top 3 this week

wolframalpha.com/input/?i=whats+the+meaning+of+life

@DominicMichaelis I added something else!

$d(x,A)\leq d(x,y)+d(y,A)$ for any $x,y\in X$.

yeah i got the same idea

but you need to make it two times or mention the symmetry

@DominicMichaelis Make what two times?

@Ethan Yeah, that's from a book I think.

I read half of the book, for an English class

I don't really like reading novels

@petertamaroff you need to take the absolute values for proofing it is lipschitz

@DominicMichaelis proving!!!

@DominicMichaelis Yeah, sure.

oh sry

help

lol where are you geting these wacky gcd problems

@Ethan Burton.

burton?

no it's from a national exam

Marvis gave an answer

@pourjour sence $gcd(a,b)=gcd(a,b+a)$

add 2n+1 to both sides and prove it inductively

$n^2+(2n+1)=(n+1)^2$

or somthing like that

@JasperLoy Did you see my latest answer?

there habe been 572 first posts today to get reviewed O_o

@DominicMichaelis What?

Where did you see that?

here

i think that is pretty much isn't it ?

@WillJagy: Here I will go: icmat.es/NTHA/WS-HAPDE

@anon

@JonasTeuwen, good. I saw something about the full-year program in Spain. You have been to Madrid before?

@WillJagy Can I ask you something about the wording of an exercise?

@PeterTamaroff, alright.

@WillJagy The author says: "Let $z$ be an element of a ring for which there exists $w\neq 0$ such that $zwz=0$. Show that $z$ is either a left zero divisor or a right zero divisor."

Now, I must assume $z$ is a left zero divisor and show it can't be a right zero divisor, yes?

And viceversa.

Or I must show it is a zero divisor first?

@Peter, backwards. Assume $z$ is NOT a left zero divisor and show that it must then be a right zero divisor.

@JonasTeuwen, "Some sessions of short talks will be scheduled to accommodate those participants wishing to present their own recent results. The final list of short talks will be decided during April."

@WillJagy I am sure you're right, but why?

@WillJagy If I want to show that $u\in A\triangle B$, then I must show $u\in A$ yet $u\notin B$ or $u\in B$ yet $u\notin A$, yes?

Doesn't matter what the setting and desired conclusions might be. The setup is blah blah blah show either A or B must be true. If you assume A you have added no knowledge. If you assume A is false, and use that to show B must be true, you have added knowledge. Sometimes, you can show either A or B without assumptions and contradictions, but this problem seems suited for it.

@WillJagy Yes, but I know the organizers (and they are good friends with my advisors).

Also, I've got the mail myself.

@WillJagy Oh, I see! The proof is easy now. Thanks you, sire.

The other one in Romania is too big, and according to my advisor 'not many nice names'.

@JonasTeuwen, I'm glad this is working out, including what seems good contacts in Madrid and permission/approval from your adviser. Again, can you give a talk, or is that not important at this time?

Let's see, do you mean they emailed you an invitation?

@PeterTamaroff, good.

@WillJagy Now onto: "If $1-ab$ is invertible in a ring, so is $1-ba$." You work on Number Theory, yes?

@WillJagy I can.

@WillJagy Yes, invitation to go, but I can do one of the contributed ones.

Was planning on doing that last June but my voice completed zapped away.

@Jonas, I understand. And giving a talk is a nervous thing. So, as I said, I gave no talk in San Diego and made no promises ahead of time, although I made sure to inform the one guy who might have been, um, taken aback by seeing me there. And, as you may have observed, the most important interactions are between talks, during lunch, and so on. Well, giving a talk is worth something, it shows people you want to let them in on your stuff without undue effort.

Yes, been to a couple of conferences and gave lectures.

I would not be so nervous as I probably know it the best. Well, maybe... not there.

Well, perhaps still as the other one is kind of my advisor.

@PeterTamaroff, I don't remember, in a ring that may have zero divisors but does have a 1, is a left inverse the same as a right inverse? It is in a group. Let me get some books.

I am more 'worried' about the possibility of not being able to travel due to recovering from surgery (which seems to be an open option again...)

@WillJagy Yes, I think so.

Suppose $rr'=1$ and $r''r=1$. Then $r'=1r' =r''rr'=r''1=r''$ @WillJagy

@Jonas, it might be best not to promise a talk, then, or make it clear to the organizers that things depend on your health and a possible procedure. I had everything set up for March 2010 in Florida, including a time to speak. Then my father died, about three days before my scheduled flight. The organizers were very nice about it. in essence, they undertood my problem, and would have made any needful changes.

@Peter, good. However, does the existence of a left inverse imply the existence of a right inverse?

@WillJagy No, I don't think so. I am not sure, however. I think anon said otherwise.

@JonasTeuwen How long do they recommend you take to recover?

@WillJagy Yet, invertible means both left and right inverses exist (and are equal).

@Peter, alright, so that is the definition in this setting. Good.

A lifetime.

(never)

But, on the other hand, the surgery can be from 2 days to 6 months, depending on the amount of damage.

@AlexBecker I am trying to show that if $1-ab$ is invertible in a ring, then so is $1-ba$. I was doubting also wether an element of a ring may have a unique left inverse but no right inverse, or no unique right inverse.

what is the difference between error and mistake?

@pourjour I take them as synonyms.

@PeterTamaroff (left/right) Inverses are always unique.

@PeterTamaroff An element of a ring can easily have a left inverse but not a right inverse, though, assuming it is non-commutative.

@AlexBecker How do you prove that?

@PeterTamaroff Hold on, I'm trying to think of an example.

I can see that if $R$ is a domain $rr''=r'r''=1\implies (r-r')r''=0\implies r=r'$, say.

@PeterTamaroff Oh, you're asking about uniqueness.

@AlexBecker Right. Suppose $R$ is a ring $z\in R^*$, and $zw=1$. Suppose $zw'=1$. Does it follow $w=w'$?

@PeterTamaroff I'm rethinking my statement. You're right, that shouldn't follow.

@AlexBecker I'm trying to look for counterexamples using matrices.

Which are easy to model.

@PeterTamaroff There is no counterexample among matrices (at least over a field), as if $zw=1$ then $z$ is invertible so $w=z^{-1}=w'$.

@AlexBecker Right. Maybe over $\Bbb Z_n$, $n$ not prime =)

Anyways.

I'm more concerned about $1-ab$ invertible then $1-ba$ is invertible too.

@PeterTamaroff Right. Are you familiar with the standard formal nonsense for the inversion?

@AlexBecker Try me.

@PeterTamaroff Consider the infinite series $1-ba-baba-bababa-\cdots$

OK.

@PeterTamaroff This "should" be equal to $(1-ba)^{-1}$

@AlexBecker OK, you want pluses, not minuses!

@PeterTamaroff Oops, right, sorry

@AlexBecker No problem.

So we have $1+ba+baba+bababa+\cdots$

(Fun fact "baba" means "drool" in Spanish)

@PeterTamaroff Anyway, this should also be equal to $1+b(1+ab+abab+\cdots)a = 1+b(1-ab)^{-1}a$

"should" in the largest possible quotation marks

@AlexBecker How nice.

Purr-fect.

@AlexBecker Thank you.

Look at mah titties. I've got man boobs.

It's possible for AB=1 but BA=/=1 for countably-infinite matrices, fun fact. (Let B be the right-shift operator and A the left-shift operator.)

Aha.

What have you been smoking? I'd love to try.

I'll now make some popcorn and watch Breaking Bad!

@JonasTeuwen I'm listening to Norah Jones. Does it count?

@PeterTamaroff That adds in some 'fag', I guess.

You sir, are a ass hat.

'Sir! You have your head not up your ass!'

Ah, right, was doing functional calculus.

@JonasTeuwen I don't understand!

I failed miserably in my popcorn enterprise.

Dude! You're life much suck donkey balls.

NO. POPCORN.

@JonasTeuwen Dude.

I just made the most awesome popcorns you'll ever try.

@κρανίοπεριπολία Not at all.

I made caramel for them! @κρανίοπεριπολία

There you go.