I forget the Induction Joke ...
- From: "'foolsrushin.'" <dolomite8@xxxxxxxxxxx>
- Date: Wed, 18 Jul 2007 14:52:11 -0700
'The Qualifying Examination
Richard Roth, University of Colorado
Drama in one act with 4 characters: The Grand Alpha, The Grand Beta,
The Grand Omicron, and The Candidate.
As the curtain rises, Alpha, Beta, and Omicron are seated in a
classroom of a large university and the Candidate comes in.
ALPHA: Who enters?
CANDIDATE: I am the Candidate.
ALPHA: State your purpose.
CANDIDATE: I come in pursuit of mathematical knowledge. I am prepared
in the fundamental fields, Algebra, Analysis, Anti-derivation. You may
question me.
ALPHA: The candidate will please define what is meant by a continuous
denominator.
CANDIDATE: Consider the set of all doubly evocative singly homologous
functions on the unit sphere. Introducing a continuous group structure
in the usual way we may define the Skolem uniformity of automorphic
cycles to be the theta relation on all sets of measure zero and the
zeta function on left ideals whose valuation is Gaussian, uniformly on
compacta. Then given any cardinal predicate, the continuous
denominator is the corresponding normal quaternion for which the
problem vanishes almost everywhere.
BETA: Could the candidate please give an example of a non-Skolem
uniformity?
CANDIDATE: I believe the inversion of the reals under countable
intersections is non-Skolem... at least almost everywhere.
BETA: That's correct. Now could you...
OMICRON: (Interrupting) I wish to contradict. It isn't a non-Skolem
uniformity since the third axiom concerning the density of the seventh
roots of unity is not in fact satisfied.
BETA: Ah, yes, but you see, in my paper on toxic algebras... 1957...
Journal of Refined Mathematics and Statistical Dynamics of the
University of Lompoc... I showed that the third axiom need not be
satisfied if the basis is countably finite and the metric is
Noetherian, hence...
ALPHA: (Interrupting) Ahem, excuse me. The candidate will please prove
the hokus-locus theorem on uniform trivialities.
CANDIDATE: By the Heine-Borel Theorem we reduce the Hamilton-Cayley
equation to the canonical Cauchy-Riemann form. The Bolzano-Weierstrass
property then shows that the Radon-Nikodym derivative satisfies the
Jordan-Holder relation. Hence by the Stone-Weierstrass approximation
we can get the Schroeder-Bernstein map to be simply separable. The
Lebesgue-Stieltjes integral then satisfies the Riemann-Roch result
when extended by the Hahn-Banach method almost somewhere.
BETA: Please define a compact set.
CANDIDATE: A set is compact if every covering by open sets has a
finite sub-opening. I mean every opening by finite sets has a compact
subcovering. Er... rather, every compact by an open finite has a
subcover. I mean a finact combine subopen if setcover set everything.
That is, almost some of the time.
ALPHA: Leave that for a moment. Instead could you give us an example
of a compact set.
CANDIDATE: Uh, you consider the real line and take any bounded subset,
I mean closed subnet, er, I mean complete subsequence... bounded
elements...
BETA: For example, is an interval compact?
CANDIDATE: Yes... er, I mean no... that is sometimes... almost
everywhere?... if it is finite... or rational, I mean the irrationals
-- given a Dedekind cut -- er, all the numbers less than square root
of 2 have a limit, that is...
OMICRON: Never mind. Look... is square root of 2 rational or
irrational?
CANDIDATE: It's rational... I mean it's not rational... n2 = 2m2 and
all that... n less than m or I mean prime to 2... they're all integers
of course.
ALPHA: What do you mean by integers?
CANDIDATE: Well... there's Peano's postulates or axioms and there's
this element 1 and s(1) is 2 and s(s(1)) and so forth. I think almost
everywhere and uh... yes.
BETA: We have a feeling that you are not quite sure of the material.
For example, how much is 2 added to 2?
CANDIDATE: Well, we have a binary operation +, defined by induction
and we let 2 denote...
BETA: Never mind the proof... Just tell us the ordinary name of the
integer which results from adding the integer 2 to the integer 2.
CANDIDATE: Er... uh... that sounds familiar. I remember: 2 generates a
prime ideal in a Dedekind domain, which is ramified when...
ALPHA, BETA, and OMICRON in chorus: How much is 2 and 2? You learned
it in the first grade?
CANDIDATE: Yes, oh yes... I just can't think... I really know it...
let me see... the first grade, you say. That's right... 2 plus 2 is...
Now first one plus one is two, one plus two is three, 8 times 8 is
65... Stuff like that. 2 plus 2 is 2 plus 2 is 2 plus 2 is....
ALPHA: That is quite enough. The examination is over. The candidate
will write his name on the board while the committee deliberates on
its decision.
(The candidate, chalk in hand, stands facing the blackboard, writes a
few letters on the board, erases them, looks blankly around the room
as the curtain falls.)
Note: This is a shortened version of a piece written in January 1960,
when the author was a graduate student at the University of
California, Berkeley.) '
--------------------------------------------------------------------------------
--
'foolsrushin.'
PS: The induction joke is that it does not work! - though we all know
the conclusions!
.
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