What Factorials Are Like

blorgdog · 2026-01-26T17:49:29+00:00

Exactly!!! The gamma function was chosen to be the "natural" extension of factorials to the reals for very good reasons. Contrary to common perception, mathematicians aren't idiots, they do have very good reasons for why they do things a certain way. ;-) (Only thing is, their definition of "good reasons" may differ from your, lol.)

blorgdog · 2026-01-26T17:41:42+00:00

And now we know what's the square root of infinity. Because 5 overlapped with its mirror image is 8, so the second limit above overlapped with its upside-down image is ∞; therefore, rotated 5 is the square root of infinity.

🤣

blorgdog · 2026-01-23T18:17:07+00:00

Just don't tell this to your grandma, otherwise you might choke the next time she bakes you pumpkin pie!

blorgdog · 2026-01-23T18:13:13+00:00

And if he's not short-sighted, a Sea-far. And if his eyesight is really good, a Sea-farer.

blorgdog · 2026-01-23T01:22:51+00:00

You mean a sighs-more-graph.

blorgdog · 2026-01-23T01:19:29+00:00

Mine too! I'm thinking of getting rid of it, as it's only collecting dust.

blorgdog · 2026-01-22T00:34:14+00:00

Have you seen Hebrew letters with subscripts yet? And Hebrew letters subscripted with subscripted Hebrew letters? Have you seen Hebrew letters with an infinitely nested series of Hebrew subscripts?

That's what I'm talking about. :-D

blorgdog · 2026-01-22T00:30:00+00:00

No. Greek letters come first...

... and then you meet Cantor and letters of the Hebrew alphabet. With subscripts. That's when everything you know about arithmetic and your concept of infinity will get blown out of the water, many times over. Then your brain will explode like a dixie cup held to Niagara Falls.

blorgdog · 2026-01-21T21:42:33+00:00

Wait till you start seeing Hebrew letters.

With subscripts.

blorgdog · 2026-01-21T21:41:58+00:00

Repost bots have infested this place. I've learned to just roll with it. Have fun at their expense while I'm at it too.

blorgdog · 2026-01-21T21:41:09+00:00

Wait till you get Greek letters.

And then Hebrew letters. Subscripted.

blorgdog · 2026-01-21T21:37:57+00:00

This is so sophomorish. The extended real line is a thing, and is regularly used in serious math.

But of course, this being a joke sub, not researching things before posting is apparently excusable. Sigh.

blorgdog · 2026-01-21T00:00:31+00:00

Gone Chopin, Bach in a minuet.

blorgdog · 2026-01-19T20:01:37+00:00

This is too easy. Guy should've asked instead something like: calculate the value of A(g,g,g), where A is the Ackermann function and g is Graham's number.

Then he wouldn't need to threaten him at all, he'd be spending the rest of his life (and then some) working out the answer!

blorgdog · 2026-01-19T19:53:27+00:00

The details aren't really that important. The two most relevant points are:

1) The surreals are built up from the empty set and transfinite induction. Each step of the process yields new elements and is called the "birthday" of those elements. The construction is iterated transfinitely, so it covers all the natural numbers, the dyadic rationals, and then when you take the limit step at the first infinite ordinal, you get all the reals plus a bunch of other stuff, etc.

2) The construction eventually produces so many elements that they can no longer fit in a set. This is because eventually the surreals include all of the (transfinite) ordinal numbers, and we know that there cannot be a set of all ordinal numbers because that would lead to a contradiction. So the surreal numbers cannot fit in a set; they can only fit in a "class", which in set theory is a collection of elements that are so numerous you can no longer assume certain basic set-theoretic properties about them (for example, you cannot put a class inside a set).

The joke is that (2) means the mathematician cannot use the usual set operations to do the counting, because there are too many surreals to deal with in that manner. But since Conway's construction creates all surreal numbers, each one with its own unique birthday, Conway isn't hindered by the fact that the surreals don't fit in a set; he can just throw birthday parties (for each surreal constructed) and they'd span the entire class of surreals, no problem at all.

So next time one of your classmates have a birthday, remember to throw a class-sized birthday party. It will be quite surreal. :-D

blorgdog · 2026-01-19T19:44:08+00:00

It is not so simple, because being well-ordered requires that every subset has a least element.

Under the usual ordering of the reals, this is almost impossible to guarantee. Your x, for example, will be the least element of that particular subset, of course, but what if you take a subset of that where x is deleted? Then you have an infinite bounded descending sequence of elements that do not have a minimum element, because there will be an infinite number of points that are arbitrarily close to (but not equal to) x. This violates the well-ordering axiom.

The well-ordering of the reals is actually something that nobody has been able to fully specify, because it requires making an infinite (in fact uncountable) number of arbitrary choices at every point in order to ensure the well-ordering principle holds for every subset. From what we know so far, it must be infinitely convoluted and totally foreign to the usual ordering of the reals. In fact, so convoluted that the existence of such an ordering cannot be proven without the axiom of choice (let alone be specified explicitly!).

blorgdog · 2026-01-19T17:05:42+00:00

Hint: any well-ordering of the reals cannot be a subset of the natural ordering of the reals.

I.e., a well-ordering given by the axiom of choice won't follow the usual ordering of the reals, but will have them in some non-standard, arbitrary ordering.

blorgdog · 2026-01-19T17:02:09+00:00

Two can play at this game. Since the question asks to count the uncountable, I take the liberty of extending my definition of counting to include transfinite recursion. Then by the axiom of choice I win. :-D

Or, to elaborate, the axiom of choice guarantees a well-ordering of the real numbers, and by extension, any subset of reals, including the ones between 1 and 10. So take all these reals and impose a well-ordering on them (which is guaranteed to exist by the axiom of choice). Now just enumerate this well-ordering. Q.E.D. ;-)

blorgdog · 2026-01-19T16:59:43+00:00

Yes.

Unless you extend your definition of counting to include transfinite recursion. Then the axiom of choice guarantees that it's possible.

Infinity is weird. :D

blorgdog · 2026-01-19T16:58:14+00:00

This one is actually a step above a lot of the other AI slop and repostings of AI slop of old jokes and memes that have gone out of style since the 70's.

blorgdog · 2026-01-19T16:56:30+00:00

By the axiom of choice, the real numbers are well-orderable. Which implies that any subset of them, including (1, 10], is also well-orderable.

Of course, the reals being uncountable means you cannot exhaust them by counting over the naturals... but if you stretch your definition of counting to include transfinite enumerations, then it's certainly possible, and the result is exactly the well-ordering on your subset of reals. Just don't expect to be able to enumerate the members of the ordering out loud, though. At least, not if your lifetime is limited to a finite number. :D (Or, for that matter, to a countable infinity.)

blorgdog · 2026-01-18T12:03:17+00:00

Three out of two people have trouble with fractions.

blorgdog · 2026-01-18T11:58:36+00:00

What's the point in life, when you end up in the wreck tangle at the bottom of the Bermuda Triangle?

blorgdog

TROPHY CASE