meanwhile, back in the labyrinth ...

Steven Everett

Avid readers of this journal will remember that in MANIFOLD-4, we left Theseus and the Minotaur discussing the philosophical problems of the labyrinth. You will be aware of their lack of success in the rather more important issue of how to get out. To date they have not managed to travel more than the odd mile or so from the centre, and they now tend to spend their time in mathematical pastimes. We eavesdrop yet again at such a moment ....

Theseus: 1152!

Minotaur: Was that a factorial or an exclamation mark?

Thes: Oh, an exclamation mark - I'm nowhere as high as 1152!

Mino: What are you doing, then?

Thes: The four fours problem - trying to represent numbers by four 4s combined using mathematical symbols in some way. 1152 is 4!4!+4!4! - and I do mean !

Mino: (checking first) Yes, that's right. That doesn't seem bad at all - isn't 1152 terribly good?

Thes: Well. it would be if I had all the numbers up to 1152, but so far I've only got an unbroken sequence as far as 873.

Mino: What's the highest that you have?

Thes: 1152

Mino: Surely you can keep going to as high a number as you please? I mean. isn't 1152! just (4!4! + 4!4!)! and so on just adding factorials?

Thes: (slowly - which gives you some idea of the abilities of this pair) Yes.

Mino: 873 seems reasonable, though

Thes: It's not bad - Rouse-Ball won't be able to do any better than this until 1912.

Mino: Who's Rouse-Ball?

Thes: Oh, he's someone who hasn't been Bohrn (sic) yet!

Mino: What's sic mean?

Thes: Oh. that's to mark a horrible pun, because of what's coming later in this script.

Mino: I'm doing all right so far I'm as far as 12, but I can only use three 4s - either as 4+4+4 or as 4x4 - 4. How can I get to use four?

Thes: Well - you could use Ö4xÖ4 x 4 - 4 , but there's another way which is just to multiply one of your answers by 1, using onlv a single 4. One way of getting unity with a single four is 1 = [ÖÖ4] where the brackets mean take the integral part of.

Mino: Ah, I see - that's quite good.

(There is a pause - Minotaur is clearly working through the integers)

Mino: Can you do 1153?

Thes: No - not yet.

(Pause)

Mino: I can: look! 1153 = -Ö4 log₄log₄ÖÖÖÖÖÖ ... ÖÖÖ4

Thes: I'm sorry, I wasn't following that too closely - could you be more precise

Mino: Mmm ~ there were 1153 square root signs in case you didn't count.

Thes: It's still a little bit unclear, though.

Mino: Well: ÖÖÖÖÖÖÖ ... ÖÖÖÖ4 is 4 to the power 1/21153. O.K?

Thes: Yes ~ so that log₄ of it is just 1/21153

Mino: That's right, and log₄ of that is ...

Thes: ... -1153/2 which gives, when multiplied by -Ö4 ...

Mino: 1153!

Thes: You know something - I can do 1154

Mino: And 1155, and 1156, ...

Thes: We've pretty well sewn up the integers haven't we?

Mino: We have, rather - have we reached the sic joke yet?

Thes: Oh, yes - this is going to be invented by Niels Bohr in the 20th Century after Christ

Mino: Who's Christ going to be?

Thes: Ah, shaddupl What are we going to do now -- there's nothing left of the integers?

Mino: Well. we could try some irrational numbers p or e for example

Thes: Oh, that's transcendental, man! Anyway, I can do e, sort of:

Mino: What do you mean, sort of?

Thes: Well, (1 + 1/4.4!!!!!!...)^{4.4!!!!!!!!...} is a good approximation to e, but you have to take an infinite number of factorial signs.

Mino: I see - you're looking at e as the limit as n tends to infinity of (1 + 1/n)ⁿ. Gee, Theseus, you are clever - NOW what will we do?

Thes: p probably - but let's leave that as an easy exercise to the readerl

For those still lost in the labyrinth: p = Ö[(-Ö4/4)!]⁴