How do you commute?

[Griffin]

Originally Posted by Kazodev

Do Griffin integers commute?

Commutativity is a property of operations, not elements.

[Kazodev]

Originally Posted by Griffin

Commutativity is a property of operations, not elements.

Fine, do Griffin integers commute under the induced operation that is given by the implicit fiber structure of functor of Borel - Ricci skewfield? You may assume the Axiom of Choice.

[Griffin]

Why would the commutativity be limited to Griffin integers?

[Kazodev]

Originally Posted by Griffin

Why would the commutativity be limited to Griffin integers?

It wouldn't, was just curious if they do

[Griffin]

Originally Posted by Kazodev

It wouldn't, was just curious if they do

Are you working in Carhart-Banerjee Fields?

[Kazodev]

Originally Posted by Griffin

Are you working in Carhart-Banerjee Fields?

That's implicit by me assuming Axiom of Choice, you may asssume that sub-Riemannian metric carries over the topological properties of non-measurable subsets that are treated as special cases of those fields.

[Foamy]

Definitely with clean socks.

[Griffin]

Originally Posted by Kazodev

That's implicit by me assuming Axiom of Choice, you may asssume that sub-Riemannian metric carries over the topological properties of non-measurable subsets that are treated as special cases of those fields.

I suggest you read Gerald Danner's paper on non-classifiable manifolds in Hausdorf spaces.

Originally Posted by Bushchicken

I've finally decided to commute to work on a Harley. My wife laughed at the idea of an Actuary riding a motorcycle to work. Which, of course, begs the question...how do YOU get to work?

My guy was feeling guilty about the huge payments we're making on his Harley each month (in order to get it paid off ASAP), and was justifying it to himself in terms of gas savings...

"Well, we're paying $1000 a month for it, but I'm only using $10 a week in gas to get to work!"

I thought that was cute!

(He's not an actuary, but he is an engineer - pretty close.)

[JMO]

Originally Posted by Kazodev

That's implicit by me assuming Axiom of Choice, you may asssume that sub-Riemannian metric carries over the topological properties of non-measurable subsets that are treated as special cases of those fields.

Don't talk dirty, the moderators are watching.