🚀 go-pugleaf

In article <1992Nov20.190744.6915@meteor.wisc.edu> tobis@meteor.wisc.edu (Michael Tobis) writes:
>I note that Abian has not reposted the analysis taht originally caught
>my attention, where the fallacy is equally egregious but much more obvious.
>
>As I recall, he obtained an expression (1 + z + z^2) and "proved" it equal
>to zero by synthetic division yielding two "different" results. The obvious
>problem with this approach is that it results in 1 + z + z^2 = 0 FOR ALL Z.
>The flaw is similar to the one you point out, i.e., one of the series is
>divergent.

You're misremembering some details of the proof, I've attached it below.

Your objection about divergent series does not hold water.  To prove
that 1+z+z^2 has a zero Abian assumes it does not and argues that under
this hypothesis one is permitted to exand 1/(1+z+z^2) out in a Laurent
series as

	1/(z^2) - 1/(z^3) + 1/(z^5) - 1/(z^6) + ...

from which he obtains a contradiction.  Your argument seems to be that
1/(1+z+z^2) can't be so expanded.  Whether or not it can be, you are
forgetting that any reductio ad absurdum *must* contain some absurdity
in order to succeed.  Pointing out a falsehood during a reductio ad
absurdum does not demonstrate a flaw in the proof.  If his expansion is
valid for a complex function with a derivative defined everywhere you
cannot complain about such a valid use, regardless of how absurd it may
seem to you.

When I first saw the proof my reaction was that the bug must be that a
sum of singularities such as 1/(z^2) - 1/(z^3) + 1/(z^5) - 1/(z^6) + ...
need not itself be a singularity.  Then I somehow talked myself out of
this in a weak moment.  Now that I look again I don't see how it
follows that this sum of singularities need be a singularity.  Unless
Abian can demonstrate this, I would say *that* was a real flaw in
his argument.  Mr. Abian?

I haven't checked whether the division itself is valid, but this
becomes a moot point if my objection is sustained.

For the record here's Abian's proof (that 1+z+z^2 has a zero) again.

>  PROOF.  Assume on the contrary, and without loss of generality, let
>     1 + z + z^2  have no root, i.e.,  1 + z + z^2  never vanishes.
>Long divide  1  by  1+z+z^2  in two ways - in descending and ascending 
>powers of z,  i.e., 
> 
>    1/ (1+z+z^2)  =   1 - z + z^3 - z^4 + z^6 - ...
> 
>    1/ (z^2+z+1)  =   1/(z^2) - 1/(z^3) + 1/(z^5) - 1/(z^6) + ... 
>   
>The two expressions on the right side of =  must be identical. But 
>they are not.  Contradiction.  Hence our assumption is false and the 
>Theorem is proved.
-- 
Vaughan Pratt	  		A fallacy is worth a thousand steps.