Question on Wedgies

Wow, not a lot of activity on the ol' newsgroup lately...

Question that has been gnawing at me. Quite simply, why does wedging
the period val with the generator val always yield the wedgie? No
matter if it is 2&gen, 3&gen, 5&gen or 7&gen. (Sorry I don't have the
post with me here at work). I know that it yields the correct answer,
but my question is, how does this work?

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:

> Question that has been gnawing at me. Quite simply, why does wedging
> the period val with the generator val always yield the wedgie?

Any two vals which generate the entire group of vals of some linear
temperament will do this, and specifying the period and generator
together gives all of the notes of a linear temperament, so they must
generate the entire group of vals. One way to see what is going on is
to think about what happens when you wedge two vals a and b compared to
a+kb and b, where k is any integer:

(a + kb)^b = a^b + k b^b = a^b

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> > Question that has been gnawing at me. Quite simply, why does
wedging
> > the period val with the generator val always yield the wedgie?
>
> Any two vals which generate the entire group of vals of some linear
> temperament will do this, and specifying the period and generator
> together gives all of the notes of a linear temperament, so they
must
> generate the entire group of vals. One way to see what is going on
is
> to think about what happens when you wedge two vals a and b
compared to
> a+kb and b, where k is any integer:
>
> (a + kb)^b = a^b + k b^b = a^b

I see. So b^b is 0?

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:

> I see. So b^b is 0?

Since u^v = -v^u, b^b = -b^b, and hence must be the zero bival (all
coefficients zero.)