> The dictionary says that u/y inserts u between the items of y. For the
> case where 1=#y there is only one item of y and so there is no place
> to insert u, so u is inserted nowhere.
How does it follow from nowhere-inserting u that u/y returns y as
it is? How does it follow that it returns anything to do with y?
Or anything at all? It doesn't follow. Nothing in the definition
leads to such a conclusion. The definition is underspecified.
Returning an empty or identity value or even an error is as complying
with such a definition as returning y.
(Note: whether any of these interpretations is meaningful and useful for
any specific purpose is irrelevant here: we are discussing what is deducible
from /'s definition, that's all.)
Besides, consider the following hypothesis. Let us pretend that the DoJ
didn't define u/y for an empty y. Then, if it were true that from the absence
of the possibility to apply u you could still infer /'s behaviour, you should
be able to tell what that behaviour would be. So, could you tell what
would u/$0 have to return then, and why?
For me, common sense means that the verb in the sentence
`applies the dyad u between the items of y'
should indeed be understood as `applies' (not including `sometimes it
doesn't'), and precisely `between'. Assuming what happens when there
is no `between' is not a formal consequence of the sentence.
> http://algebra.math.ust.hk/determinant/01_geometry/lecture1.shtml
I fail to see the relevance of this text to the discussion on /'s definition.
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