Math Forum / Mathematics / Undergraduate Math / July 2007

On Fri, 20 Jul 2007 06:17:01 EDT, riderofgiraffes
<mathforum.org_am@solipsys.co.uk> wrote in
<news:27572844.1184926651954.JavaMail.jakarta@nitrogen.mathforum.org>
in alt.math.undergrad:

[...]

You don't need to define subtraction; just define addition
and the negative.

I've taught it; it's not *that* bad.  Define a Dedekind cut
in Q to be a non-empty, proper subset C of Q with the
following properties:

  (1)  if p < q in C, then p in C (i.e., C is
       downward closed), and
  (2)  C has no maximum element.
 
At some point you have to show that for any cut C and any
positive rational e there are p in C and q in Q\C such that
q - p <= e, but this is easy.  If it failed for some e > 0,
then for every p in C we'd have p + e in C, and it would
follow from the Archimedean property of Q and the downward
closure of C that C = Q, which is impossible.
 
For cuts C and D let S(C, D) = {c + d : c in C, d in D}.
Clearly S(C, D) is non-empty.  Choose p in Q\C and q in Q\D;
clearly c + d < p + q for all c in C and d in D, so p + q is
not in S(C, D), which is therefore a proper subset of Q.  

Now fix c in C and d in D, and suppose that p in Q satisfies
p < c + d.  Let q = c + d - p; q is a positive rational, so
d - q is in D, and p = c + (d - q) is in S(C, D), which is
therefore downward closed.

It's possible that S(C, D) has a maximum element m.  If so,
let C + D = S(C, D)\{m}; this is clearly a cut.  If not,
just let C + D = S(C, D).

Let Z = {q in Q : q < 0}; this is clearly a but, and for any
cut C we have C + Z = S(C, Z) = C, so Z is the zero cut.

For any cut C let M(C) = {-q : q in Q\C}.  If M(C) has a
maximum element m, let -C = M(C)\{m}, and otherwise let -C =
M(C); it's easy to check that -C is a cut.  If -C = M(C),
then S(C, -C) = {c + d : c in C, -d in Q\C} =
{c - d : c in C, d in Q\C}; this is clearly a subset of the
negative rationals, and the result mentioned immediately
after the definition of cut above easily implies that it is
precisely the set of negative rationals, so that C + (-C) =
S(C, -C) = Z.  Otherwise M(C) = {-q : q in Q\C} has a
maximum element m, and it's not hard to see that C + (-C) =
S(C, -C) = S(C, M(C)) = {c + m : c in C} = Z.

Unnecessary: you need only define multiplication and the
reciprocal.  (Admittedly this is a bit messier than addition
and the negative, but the mess has more to do with the
nature of multiplication than with any endpoint problems.)

It's no improvement.  Suppose that (A, B) and (C, D) are
cuts in this new sense.  At some point you'll have to define
(A, B) + (C, D).  Let (L, R) be this cut; how will you
define L and R?  It doesn't work simply to let L =
{a + c : a in A, c in C} and R = {b + d : b in B, d in D}:
if x is an irrational number, (A, B) corresponds to x, and
(C, D) corresponds to -x, this definition will give you L =
{q in Q : q < 0} and R = {q in Q : q > 0}, and (L, R) won't
be a cut.  More generally, this attempted definition of the
sum will fail whenever (A, B) and (C, D) correspond to
irrational numbers whose sum is rational.  You'd actually
have to define (A, B) + (C, D) as follows.

First define L and R as I did in the previous paragraph.  If
L U R = Q, set (A, B) + (C, D) = (L, R); otherwise set
(A, B) + (C, D) = (Q\R, Q\L).  Then you'll have to prove
that in each case you actually do get a cut, which isn't
entirely trivial.

Brian

[snip]

[snip]

Sorry to have joined this thread late. You've already gotten good
replies from Dik Winter in sci.math and Brian Scott in
alt.math.undergrad. But I think you'd be interested in what John
Horton Conway had to say, and so I quote a passage from his "The
Surreals and the Reals". Of course I agree with most of what he says;
otherwise I won't have taken the trouble to quote it. I've inserted
one comment in brackets; my other two comments are given below the
quotation. Conway's article appears on pp. 93-103 of _Real Numbers,
Generalizations of the Reals, and Theories of Continua_, P. Ehrlich,
ed. (Kluwer, 1994). From page 95:

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[One solution to that problem is given in my
" Multiplying Dedekind cuts "naturally" "
<http://groups.google.com/group/sci.math/msg/ecc01c35f7d1d54a>.
But I think that a better solution is what Conway gives
in the next paragraph.]

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Comment 1:

The unorthodoxy (A) mentioned by Conway has a counterpart unorthodoxy
(B) in which, in the event that the cut corresponds to a rational
number q, _neither_ L nor R contains q itself. (Brian has already
mentioned both of those unorthodoxies in this thread.) Of course, then
L and R do not necessarily partition Q+.

Which is preferable, A or B?

From a philosophical standpoint, I can see arguments both ways. With
A, rational cuts are structural different from irrational cuts. One
might perceive this as an advantage, saying that "rational" real
numbers are _special_ reals and the different structure of the
rational cut properly reflects that special status. On the other hand,
from an egalitarian viewpoint, one could argue that all real numbers
should be structurally the same and so B is preferred.

From a mechanical standpoint, there is also a difference. Suppose we
multiply the cut for sqrt(2) with itself. Using A, the product is not
"naturally" a cut at all because neither L nor R contain 2, and so we
must go in surgically and attach 2 to both L and R. Using B, however,
no surgery is needed, the product is naturally what is needed.

I hope my presentation of the arguments was unbiased. But I'm not
neutral -- I prefer B.

Comment 2:

I also favor another unorthodoxy: dropping the requirements that L and
R, as subsets of Q+, be nonempty and proper. I consider those
requirements to be artificial. Once they are dropped, we construct the
nonnegative extended reals, [0, +oo], with no more trouble than we
would have had in constructing just the positive reals, had the
requirements not been dropped. And of course, having constructed
[0, +oo], you have also constructed (0, +oo). My point is that, by
dropping artificial restrictions, you have also gotten something extra
which is useful, and without any additional effort.

David W. Cantrell