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Topic: when is sqrt(a/b) not the same as sqrt(a)/sqrt(b) ?
Replies: 8   Last Post: Mar 5, 2014 3:34 PM

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Axel Vogt

Posts: 1,068
Registered: 5/5/07
Re: when is sqrt(a/b) not the same as sqrt(a)/sqrt(b) ?
Posted: Mar 5, 2014 3:34 PM
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That - in my opinion - has not much to do with CAS per se. But more with some
consistent mathematical conventions. Where one also may want to look beyond
sqrt, avoiding similar discussions for cube roots etc.

It is a classical result that having 'enough' square roots one has logarithms,
which is a theorem of Hurwitz (Remmert, Funktionentheorie 1, "Die Kraft der
Quadratwurzel" for a proof, pointing to H. and his article Math. Ann 70 (1911),
p. 33 - 47). The converse 'powers if having log' is common in lectures.

That is for functions and thus much more general than discussed here. But it
shows that the proper setting is to consider Log. And Exp.

I have not looked further in Remmert (much material), but would suggest to
have a look into H. Cartan's 'little' book for a concise treatment, where one
finds exp & log (and variations/selections), covering a more formal view.

To come back to sqrt: the initial question was over the complex numbers.
One can not understand 'solutions' for that without basic complex analysis.

And in a brute way: not, sqrt is not a holomorphic function. It is not even
a continous function in the plane. It is just 1 selection (IIRC correctly
there was the notion of holomorphic correspondance [like in Topology] and
srqt becomes a fct after branch cut with semi-contious extension into negatives,
if one wants to avoid Riemannian surfaces).

In school (which ones?) that is operationally reduced to "just do it in C++"

And I find it good that Maple et al refuse to follow that.

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