Does anyone feel nostalgic for that old song from Lamb Chop’s Play-Along: “This is the song that doesn’t end…”
Wow! It has been fun to watch this discussion flow, even though I freely admit that some of it has gone over my head. (I’ve admitted before that my retention of college-level mathematics is poor.)
For anyone who is still trying to keep up with the discussion, here are a few new blog posts:
Devlin’s Right Angle, Part III
Devlin’s Right Angle, Part IV
Devlin’s Right Angle, Part V
The article that launched a thousand posts…
Well, it’s nice people are writing about me (blush
Here’s my challenge though- for whole numbers only – show me that Peano’s definition of multiplication via recursive addition is “totally and utterly false” as Devlin has stated. Use a supercomputer if desired.
That’s it –
Now,to Text-Savvy, as to the equation a+a+a+a = 6 — considered as a Diophantine equation it has no solution. If you need more help please ask directly. LOL!!!
Oh, and by the way, I have it from Devlin directly (email) that
” sorry, mathematics is not a matter of opinons,
it’s either right or wrong. and multiplication
is not repeated addition on any domain.”
So – there it is – Devlin is saying:
1. On ANY domain (say, pairs of non-negative integers) that repeated addition (meaning the usual and common definition given in schools, or the absolutely equivalent formulation by Peano) is NOT multiplication –
and
2. He’s talking mathematically (he says he’s not taking philosophically, not pedagogically, not guru-in-the-sky-new-age.)
So – show me that on the domain of pairs of non-negative integers either the usual classroom definition as repeated multiplication (use pedia for a neutral party definition!) or Peano’s definition is “totally and utterly false”.
Please make it a mathematical proof – not a sidestep, a semantic quibble, an appeal to some authority, or a stamp-the-foot insistence.
And, at the risk of being utterly and totally clear on my position, and as has been said before, when I say “A” *is* “B” – where “A” and “B” are functions on some domain, I mean precisely what the usual definition for function equality states – that for every element of the domain the functions “A” and “B” have the same functional values. All this was put plainly to Devlin before he made the assertion that:
no – they are *not* the same function.
So substitute for “A” the usual old definition via repeated addition, and for “B” the “multiplication” that Devlin is talking about, and the domain here is given as non-negative integers.
Prove them not the same.