## Updated Jarda numbers page

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• No, the numbers 1-10 haven t changed (except for the updated spelling of the Romanization). But I realized that now that Jarda s on a Sangari world, it needs
Message 1 of 3 , Feb 18, 2013
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No, the numbers 1-10 haven't changed (except for the updated spelling of
the Romanization). But I realized that now that Jarda's on a Sangari
world, it needs more words for duodecimal numbers. I decided to keep the
original octal system just to be different. (Maybe it's an East-West
cultural difference.) But when Jarda speakers need to count in
duodecimal, now they have the words to do it.

http://www.prismnet.com/~hmiller/lang/Jarda/numbers.html

Fortunately I already had words for all the basic numbers from 1-12. Now
all I need is words for higher powers of 12. Here they are:

ģŭn 12^2 (144)
kaf 12^4 (20,736)
mŏś 12^8 (429,981,696)

That should be enough for most practical purposes, since you can combine
these (źêvģŭnkafmŏś = 12^15, and that's a pretty big number).
• Minor question: What s to prevent _JaGkovRom ģağkôvṛôm (3*6+8) means 26 _ from being interpreted as 3*(6+8) i.e 3*14= 42? In Gwr, also octal, there
Message 2 of 3 , Feb 18, 2013
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Minor question: What's to prevent _JaGkovRom ģağkôvṛôm (3*6+8) means "26"_ from being interpreted as 3*(6+8) i.e 3*14= 42?

In Gwr, also octal, there would be a word division after "twenty".

Ah well, chaq'un a son gout.........

--- On Mon, 2/18/13, Herman Miller <hmiller@...> wrote:
http://www.prismnet.com/~hmiller/lang/Jarda/numbers.html
• ... For one thing, 6+8 would actually be written 8+6 (always with the larger number on the left if you re adding). But as I remarked on the web page, the
Message 3 of 3 , Feb 19, 2013
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On 2/18/2013 11:31 PM, Roger Mills wrote:
> Minor question: What's to prevent _JaGkovRom ģağkôvṛôm (3*6+8) means
> "26"_ from being interpreted as 3*(6+8) i.e 3*14= 42?

For one thing, "6+8" would actually be written "8+6" (always with the
larger number on the left if you're adding). But as I remarked on the
web page, the calculation is always from left to right. So you group the
first two numeric roots together, add or multiply, then group the result
with the next root, and so on, until you reach a classifier. For
3*(8+6), you'd need to separate the 3 from the 8+6 with the classifier
"ģê": "ģağģê ṛômkôv" [ɟaɣɟe ɻomkov].

In practice, multiplication is almost exclusively used with powers of 8
or 12 on the right. Forms like "ṛalkôṛ" (2*9) are atypical enough, and
"ģağģê ṛômkôv" would be considered weird (like the English equivalent
"thrice fourteen").
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