[Search for users]
[Overall Top Noters]
[List of all Conferences]
[Download this site]
Title: | Mathematics at DEC |
|
Moderator: | RUSURE::EDP |
|
Created: | Mon Feb 03 1986 |
Last Modified: | Fri Jun 06 1997 |
Last Successful Update: | Fri Jun 06 1997 |
Number of topics: | 2083 |
Total number of notes: | 14613 |
1611.0. "Claimed elementary trisection and cube duplication proofs" by CADSYS::COOPER (Topher Cooper) Mon May 18 1992 15:04
From: [email protected] (David H. Bailey)
Newsgroups: sci.math
Subject: Elementary proof of impossibility of trisection
Date: 15 May 92 20:57:30 GMT
Organization: NAS, NASA Ames Research Center, Moffett Field, CA
In an earlier posting I lamented the regrettable fact that in the
orthodox mathematics curricula, many of the "famous" classical
theorems are proven only after long chains of preliminary theorems,
and only in advanced undergraduate or graduate math courses. As a
result, these great ideas are utterly inaccessible to non-
mathematicians, and even professional mathematicians have difficulty
reconstructing the conventional proofs. In that posting I included an
elementary, self-contained proof of the fundamental theorem of
algebra. A number of you have requested other examples.
I present here a proof of the impossibility of trisecting an angle or
of "duplicating" the cube by straightedge and compass. Although this
proof is significantly longer than the one I previously cited for the
FTA, it is still elementary and self-contained. Contrary to custom,
no familiarity with group theory, the Euclidean algorithm, polynomial
rings, vector spaces, field theory, Galois theory, real analysis or
complex analysis is assumed. The only results employed from "advanced
math" are two elementary facts about vector space dimensions, which
are proven below (lemmas 1 and 2). A bright first year college
student can probably follow this proof with a little effort.
I have not seen an elementary proof like this in any textbook, but I
hesitate to claim much credit for it, since it is entirely
straightforward. The question is this: why does the high priesthood
of the mathematical establishment lead us to believe that these
theorems cannot be shared with the unwashed?
David H. Bailey
NASA Ames Research Center
----------------------------------------
Definitions:
For the purposes of this proof, we will define a "field" as a subset
of the real numbers that is closed under the four arithmetic
operations (except for division by zero). The field B is an
"extension field" of A provided that A < B (i.e. A is a subset of B),
and every element b of B can be written b = a_1 x_1 + a_2 x_2 + ... +
a_n x_n where x_1 = 1, a_i are in A, x_i are in B, and where 0 has a
unique representation (all a_i = 0) in terms of these x_i. It follows
immediately from this last condition that all b in B have a unique
representation of this form (consider the difference of two different
representations of a number b). The sequence X = (x_1, ..., x_n) will
be termed a "basis" for this extension, and the "degree" of this
extension is n.
Lemma 1: Any two bases for an extension field have the same degree.
Proof: Suppose we have two bases for B over A, namely X = (x_1, x_2,
..., x_m) and Y = (y_1, y_2, ..., y_n), with m < n. Recall that by
definition, x_1 = y_1 = 1. Note that x_2 can be written in terms of
Y, and at least one coefficient a_k must be zero. For simplicity,
assume a_2 is nonzero. Then we can form a new basis W = (x_1, x_2,
y_3, ..., y_n), for if zero has a nontrivial representation with W,
this can be reduced to a nontrivial representation with Y. In this
manner, we can continue to substitute the x_i for the y_i, producing
the basis Z = (x_1, x_2, ..., x_m, y_m+1, ..., y_n). But y_n can be
written in terms of the basis X, say y_n = r_1 x_1 + r_2 x_2 + ... +
r_m x_m. Now we conclude that zero can be nontrivially represented
with the basis Z as 0 = r_1 x_1 + r_2 x_2 + ... + r_m x_m - y_n, which
is a contradiction. Thus m = n, and we may use the notation deg(B:A)
= n. [Some minor details have been glossed over here.]
Lemma 2: If A < B < C, then deg(C:A) = deg(B:A) * deg(C:B).
Proof: Let X = (x_1, ..., x_m) be a basis for B over A, and let Y =
(y_1, ..., y_n) be a basis for C over B. Note that in terms of X and
Y, any element c in C can be written uniquely as c = sum_j b_j y_j =
sum_j sum_i a_ij x_i y_j. In particular, 0 is represented uniquely in
this form (all a_ij = 0). Thus the mn numbers {x_i y_j} constitute a
basis for C over A, and deg(C:A) = mn.
Main Proof:
It is easily verified that by the rules of compass and straightedge
construction, one can perform the four standard arithmetic operations,
plus square root extraction. By solving the general algebraic
equations for lines and circles, one can verify that the intersection
coordinates are either contained in the same field F from which the
lines and circles were constructed, or else they are in a field G that
is F extended by the square root of a positive element of F. In the
latter case, G is the set of g that can be written g = a + b sqrt(c),
where a, b and c are in F, and where sqrt(c) is not in F. It is easy
to verify that G is a field [the reciprocal of g is (a - b sqrt(c)) /
(a^2 + b^2 c)] and that zero is uniquely represented as 0 + 0 sqrt(c).
Thus deg(G:F) = 2. Let Q denote the field of ordinary rational
numbers. After a finite sequence of construction steps, the smallest
field containing all the numbers x and y that have been constructed is
some F_n, where Q < F_1 < F_2 < ... < F_n. Since deg(F_j+1:F_j)= 2,
it follows that deg(F_n:Q) = 2^n for some n.
Now consider the angle 60 degrees. Clearly this angle can be
trisected if and only if the number x = cos(20o) can be constructed.
By applying the trigonometric identity for cos(3x), it follows that x
is a root of the polynomial P(t) = 8 t^3 - 6 t - 1. Since P cannot be
factored with integer coefficients (by examining the handful of
possibilities), P cannot be factored with rational coefficients
either. Therefore x and the other roots of P are not in Q, nor do
they satisfy any quadratic equation with rational coefficients.
Let H be the set of all h that can be written h = a + b x + c x^2,
where a, b and c are in Q and where x = cos(20o). Closure of H under
addition and subtraction is immediate. Closure under multiplication
can be seen by utilizing P(x) = 0 to eliminate terms with x^3 and x^4.
Closure of H under division by nonzero h can be seen as follows:
If c is zero and b is nonzero, assume for simplicity that b = 1, i.e.
h = a + x. Then 0 = P(x) = h q + r, where q = 8x^2 - 8ax + 8a^2 - 6,
and where r = -8a^3 + 6a - 1. Note that r = P(-a) cannot be zero,
since P has no rational roots. Thus we can write 1/h = -q / r. If c
is nonzero, assume for simplicity that c = 1, i.e. h = a + bx + x^2.
Then 0 = P(x) = h q + r, where q = 8(x - b) and r = (8b^2 - 8a - 6)x +
8ab - 1. Again, r cannot be zero because then q would be zero and b
would be a rational root of P. Therefore 1/h = -q * (1/r), where 1/r
is computed as in the c = 0 case.
Thus H is a field. And since x does not satisfy any quadratic
polynomial with coefficients in Q, 0 can be represented as a + b x + c
x^2 only with a = b = c = 0. Thus H is an extension field over Q, and
deg(H:Q) = 3. If we assume that x is constructible, then x (and
therefore all of H) is contained in some field F_n, where deg(F_n:Q) =
2^n. But in that case deg(H:Q) = 3 must divide 2^n by Lemma 2 above,
which is a contradiction. A similar argument applied to P(t) = t^3 -
2 shows that the cube root of 2 cannot be constructed.
T.R | Title | User | Personal Name | Date | Lines |
---|
1611.1 | but you all knew that, didn't you? | SGOUTL::BELDIN_R | All's well that ends | Thu May 21 1992 18:25 | 11 |
| >The question is this: why does the high priesthood
>of the mathematical establishment lead us to believe that these
>theorems cannot be shared with the unwashed?
>David H. Bailey
>NASA Ames Research Center
'cause mathematicians (even textbook writers) don't write for students,
but for their peers and superiors (who control their employment).
|
1611.2 | on a more general problem in education...? | STAR::ABBASI | i^(-i) = SQRT(exp(PI)) | Fri May 22 1992 01:05 | 26 |
| also because it is a lost art to explain hard things in simple terms.
a very good example of explaining hard things simply, is a book called
"The Evolution of physics" by Albert Einstein and Leopold Infeld.
the whole book (300 pages) do not have one single equation, yet after
you read it, you've understand why many things in physics work the way
they do, it is one of the best books I've ever read, in terms of
taking a complex matters and explaining things step by step in natural
ways. any one who is interested in physics should read this book. it
was a joy for me to read it many years ago..
this whole subject do not just apply to math, it applies to many other
things in education, such as teaching, I've only had rare occasions to
take a course from a teacher who talk to students as students , not as
fellow researches, many dot care if the student learns, they just want to
show how much they know...or they just come in and recite what is in
the text book and leave..
any way, iam digressing here...
/nasser
|
1611.3 | a non-cynic's viewpoint | AUSSIE::GARSON | | Sun May 24 1992 20:53 | 11 |
| re .0
>The question is this: why does the high priesthood
>of the mathematical establishment lead us to believe that these
>theorems cannot be shared with the unwashed?
The shortest proof is not always the most didactic. Furthermore some
intermediate results may prove useful elsewhere. A more difficult but
more general proof may give rise to additional results. A proof by a
different route may show a connection between two branches of
mathematics that was not previously evident.
|
1611.4 | it's not cynical to calibrate against the audience | SGOUTL::BELDIN_R | All's well that ends | Mon May 25 1992 09:17 | 15 |
| re .3
Of course, there are legitimate reasons for introducing "complexities"
to improve the didactic process. However, in my experience, most of
them assume that the student can hang on to the common thread, as well
as the details, of an argument. This is not the case with most
students in most math classes. Most have a hard time with just the
logical chain. It is not realistic to expect most students to follow
the implicit or high level argument. So, I, like Mr. Bailey, prefer
simple arguments with very little presumed infrastructure.
fwiw,
Dick
|
1611.5 | It ain't necessarily so | VMSDEV::HALLYB | Fish have no concept of fire. | Tue May 26 1992 15:06 | 19 |
| > So, I, like Mr. Bailey, prefer
> simple arguments with very little presumed infrastructure.
But sometimes it's just the other way around -- the "simple" arguments
are simple precisely because they contain substantial infrastructure
which in itself is complex or limiting.
Two examples that come to mind are geometrical proofs that rely on
the parallel postulate, even though the theorem does not require it.
And in Set Theory there are proofs that rely on the Axiom of Choice
although the resulting theorem does not require it.
In both cases you get a "better" proof when you construct a theorem
that does not rely on such a fundamental postulate, because then you
have proved more. Your theorem is then valid across a wider range
of systems. This often results in a longer and more complex proof,
but a proof that is worth more.
John
|