Comments on: Addiplication http://bit-player.org/2007/addiplication An amateur's outlook on computation and mathematics. Fri, 29 Aug 2008 08:05:35 +0000 http://wordpress.org/?v=1.5.2 Comment on Addiplication by: Kennedy http://bit-player.org/2007/addiplication#comment-1493 Mon, 03 Sep 2007 17:30:46 +0000 http://bit-player.org/2007/addiplication#comment-1493 That's HTML entity ampersand lowast semicolon. That’s HTML entity ampersand lowast semicolon.

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

]]> Comment on Addiplication by: brian http://bit-player.org/2007/addiplication#comment-1491 Mon, 03 Sep 2007 12:44:00 +0000 http://bit-player.org/2007/addiplication#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 &#8194;<em>a</em>+<em>b</em>&#8194; &#8804; &#8194;<em>a</em>&#215;<em>b</em>, &#8195;&#8195;<em>a</em>+<em>b</em>&#8194; &#8804; &#8194;<em>a</em> &#9674; <em>b</em>&#8194; &#8804; &#8194;<em>a</em>&#215;<em>b</em>; if &#8194;<em>a</em>+<em>b</em>&#8194; &#8805; &#8194;<em>a</em>&#215;<em>b</em>, &#8195;&#8195;<em>a</em>+<em>b</em>&#8194; &#8805; &#8194;<em>a</em> &#9674; <em>b</em>&#8194; &#8805; &#8194;<em>a</em>&#215;<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&#215;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.

]]> Comment on Addiplication by: Barry Cipra http://bit-player.org/2007/addiplication#comment-1490 Sun, 02 Sep 2007 03:43:25 +0000 http://bit-player.org/2007/addiplication#comment-1490 Another problem: Tanimoto shows that 1◊x=x for all (positive) x, which violates your stipulation if  a+b  &#62;  a×b,   a+b  &#62;  a ◊ b  &#62;  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).

]]> Comment on Addiplication by: Jonathan Katz http://bit-player.org/2007/addiplication#comment-1489 Sun, 02 Sep 2007 00:48:57 +0000 http://bit-player.org/2007/addiplication#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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