Comments on: Addiplication http://bit-player.org/2007/addiplication An amateur's outlook on computation and mathematics. Thu, 17 May 2012 08:50:56 +0000 http://wordpress.org/?v=2.6.3 By: Kennedy http://bit-player.org/2007/addiplication#comment-1493 Kennedy Mon, 03 Sep 2007 17:30:46 +0000 http://bit-player.org/?p=105#comment-1493 That's HTML entity ampersand lowast semicolon. That’s HTML entity ampersand lowast semicolon.

]]> By: Kennedy http://bit-player.org/2007/addiplication#comment-1492 Kennedy Mon, 03 Sep 2007 17:29:55 +0000 http://bit-player.org/?p=105#comment-1492 ∗ is a pretty close facsimile of Tanimoto's star operator. ∗ is a pretty close facsimile of Tanimoto’s star operator.

]]> By: brian http://bit-player.org/2007/addiplication#comment-1491 brian Mon, 03 Sep 2007 12:44:00 +0000 http://bit-player.org/?p=105#comment-1491 Uh oh. As far as I can tell, the issues raised by Jonathan Katz and Barry Cipra are problems with my interpretation of Tanimoto's paper, not problems with the paper itself. To address Barry's point first, it appears that the correct relations are as follows: if  <em>a</em>+<em>b</em>  ≤  <em>a</em>×<em>b</em>,   <em>a</em>+<em>b</em>  ≤  <em>a</em> ◊ <em>b</em>  ≤  <em>a</em>×<em>b</em>; if  <em>a</em>+<em>b</em>  ≥  <em>a</em>×<em>b</em>,   <em>a</em>+<em>b</em>  ≥  <em>a</em> ◊ <em>b</em>  ≥  <em>a</em>×<em>b</em>; As for the much simpler construction by Jonathan Katz: Tanimoto writes "the binary operation can be regarded as an intermediate operation between addition and multiplication" but never claims it is unique in this respect. For that matter, I made no claim of uniqueness either, but I certainly missed the point that constructing such an intermediate operation is not in itself particularly remarkable. Katz's example seems an especially nice one. Of course one could also use the geometric mean of a+b and a×b, or even the AGM of the sum and the product. Uh oh. As far as I can tell, the issues raised by Jonathan Katz and Barry Cipra are problems with my interpretation of Tanimoto’s paper, not problems with the paper itself.

To address Barry’s point first, it appears that the correct relations are as follows:

if  a+b  ≤  a×b,   a+b  ≤  a ◊ b  ≤  a×b;

if  a+b  ≥  a×b,   a+b  ≥  a ◊ b  ≥  a×b;

As for the much simpler construction by Jonathan Katz: Tanimoto writes “the binary operation can be regarded as an intermediate operation between addition and multiplication” but never claims it is unique in this respect. For that matter, I made no claim of uniqueness either, but I certainly missed the point that constructing such an intermediate operation is not in itself particularly remarkable. Katz’s example seems an especially nice one. Of course one could also use the geometric mean of a+b and a×b, or even the AGM of the sum and the product.

]]> By: Barry Cipra http://bit-player.org/2007/addiplication#comment-1490 Barry Cipra Sun, 02 Sep 2007 03:43:25 +0000 http://bit-player.org/?p=105#comment-1490 Another problem: Tanimoto shows that 1◊x=x for all (positive) x, which violates your stipulation if  a+b  >  a×b,   a+b  >  a ◊ b  >  a×b (here a=1, b=x). Another problem: Tanimoto shows that 1◊x=x for all (positive) x, which violates your stipulation

if  a+b  >  a×b,   a+b  >  a ◊ b  >  a×b

(here a=1, b=x).

]]> By: Jonathan Katz http://bit-player.org/2007/addiplication#comment-1489 Jonathan Katz Sun, 02 Sep 2007 00:48:57 +0000 http://bit-player.org/?p=105#comment-1489 Maybe I'm missing something, but isn't there a trivial solution? Let me use . instead of your symbol, and AM for arithmetic mean. Then why not define: a.b = AM(a+b, a*b) ? Maybe I’m missing something, but isn’t there a trivial solution? Let me use . instead of your symbol, and AM for arithmetic mean. Then why not define: a.b = AM(a+b, a*b) ?

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