Talk:Roman arithmetic

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[edit] CPU Arithmetic

Shibboleth is apparently correct about multiplying and dividing in modern machine language (I haven't tried it much since about 1980), although I wish he had explained it a little to Mr. Hardy. The mul, div, imul, idiv, fmul and fdiv commands at x86 instruction listings are apparently intended for multiplying and dividing. Art LaPella 02:38, Sep 4, 2004 (UTC)

The CPU in modern computers generally is only able to directly perform these two simple operations. was deleted from the article.

The machine language may have the instructions, but that does not mean the CPU actually multiplied or divided as a single step. In the ALU/FPU of many CPUs, the actual multiplication is a series of shifts or additions performed by the processor. The same holds true for division. The orginal text did not state that CPUs lacked instructions to perform more than addition or subtraction (which was true with the Z80's instruction set). I wish Shibboleth had taken the time to discussed it here first, rather than just making the deletion from the article. --Denise Norris 04:31, Sep 4, 2004 (UTC)

[edit] Historical evidence

Is there any evidence that the algorithms presented here were, in fact, the ones used by ancient romans? --Mathish 01:08, 13 November 2005 (UTC)

I would also like to know if these algorithms are historical. There is an alternate method using cancellation for addition of values with subtractive notation, and multiplication can be done with a (presumably modern, albiet simple) multiplication table rather than the iterative solution presented here. --Dcorrin 16 November 2005

I found this guide to Roman Multiplication, but I don't know his sources. It seems like a much easier method of multiplying MDCCLXIII by CCXVII, and other large quantities. http://www.phy6.org/outreach/edu/roman.htm ih8evilstuff 15:52, 6 June 2006 (UTC)

If the Romans used it, surely they must have known why it works. Even if they didn't know the binary number system, they knew that a x b = a/2 x 2b, right? It's essentially still repeated addition, but just with a trick to do it faster.

While the multiplier is odd, remove one from the multiplier and add the multiplicand to the result. Use the aforementioned identity: halve the multiplier and double the multiplicand - this doesn't change the result. D.C. al fine.

You don't have to know binary to figure this out (it could be a nice stepping stone though, in either direction). Also note that 123 x 456 is a lot harder than 2 x 28044, the Romans must have known that too. Shinobu 12:40, 2 August 2006 (UTC)