tlhIngan-Hol Archive: Wed Oct 11 18:37:05 2000
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RE: math questions / speculations (longish)
- From: "De'vID" <[email protected]>
- Subject: RE: math questions / speculations (longish)
- Date: Wed, 11 Oct 2000 21:37:19 -0400
- Importance: Normal
De'vID:
>[... math equations ...]
quljIb:
> Let me (a Beginner) take these one by one.
>
> A + B --> B boq A = "A aligns with B."
> A - B --> A boqHa' B = "B dis-aligns with/from A."
> A x B --> A-logh boq'egh B = "B aligns with itself A times."
>
> So far, so good.
Let it be noted that A and B may be switched for addition and
multiplication, just so you don't read too much into the form
that I put the equations in. I abstracted the equation forms from the
examples, but as has been pointed out by others, in the actual
description of the grammar, A and B can be switched without changing
the meaning. So don't get too hung up on the order of the operands
(unlike me {{=) ).
> A / B --> B-logh boqHa''egh A = "A dis-aligns with/from itself B
> times."
>
> That doesn't parse well to me. Can someone explain it? Or is it simply
> something we must accept?
Maybe the /-Ha'/ is "undo"ing the /boq'egh/?
B-logh boq'egh A; chen (A*B).
B-logh boqHa''egh (A*B); chen A.
The /boqHa''egh/ seems to "undo" the previous action of /boq'egh/.
However the subject of /B-logh boqHa''egh C/ is not C, but the
factor (C/B) that remains.
/B-logh boqHa''egh (A*B)/ "(A*B) undoes the {action of having (something)
aligned B times with itself}; that (something) being A"... The /-'egh/
seems to refer to the unspecified A ( = (A*B)/B) and not to (A*B).
taQ. jIQIjlaHbe'. Are there any other examples of /-Ha'/ together
with /-'egh/? Which order are they applied in?
--
De'vID
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