C OMPOSITIO M ATHEMATICA
R. D UNCAN L UCE
Periodic extensive measurement
Compositio Mathematica, tome 23, n
o2 (1971), p. 189-198
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189
PERIODIC EXTENSIVE
MEASUREMENT 1, 2
by R. Duncan Luce
COMPOSITIO MATHEMATICA, Vol. 23, Fasc. 2, 1971, pag. 189-198 Wolters-Noordhoff Publishing
Printed in the Netherlands
1. Introduction
The classical model for
physical
measurementincludes,
asprimitives,
a
non-empty
setA,
a closedbinary (concatenation) operation
o, andan
(ordering)
relation~,
and the axioms are(necessary and)
sufhcientto establish a
homomorphism
into(R,
+,~ ~ [1, 2, 3, 4, 7].
This modelaccounts for the measurement of many
important physical quantities
such as mass,
length, duration,
electricalresistance,
etc., but it is unsuitable for manyothers,
among themperiodic quantities
such asangle
and clocktime. The purpose of this note is to
present
axioms in terms of the usualprimitives (A, o, ~~
that result in ahomomorphism
into~[O, K],
+
(mod K), ~~.
At least two earlier axiomatizations of
angles
exist. Zassenhaus[8]
treated them as abstract entities which are oriented and among which
judgments
ofequality
can be made. Inaddition,
he assumed thatsome
pairs
ofangles,
but notall,
can be concatenated to form new ones;he
imposed
this restriction becauserepeated
sums ofpositive angles ultimately
lead tonegative
ones. This axiomatization differs in several ways from thosetypical
of other extensivequantities. First,
an orderrather than
just
anequivalence
isusually postulated. Second,
unless the intendedinterpretation actually
forces an inherent restriction on con-catenation
(e.g.,
thedisjointness
of events inherent in theadditivity
of
probability),
it is usual to let the concatenationoperation
be closed.Third,
orientation isignored
wheneverpossible;
this is true, e.g., of the usual axioms oflength
and duration measurement. Inaddition,
Zas- senhaus’ axioms do notadequately highlight
theperiodic
nature of an-gular
measurement.1 This work was supported in part by NSF grant GB-6536 to the University of Pennsylvania; it was initiated while 1 was an Organization of American States Pro- fessor at the Pontificia Universidade Catôlica do Rio de Janeiro, Brasil.
2 1 wish to thank the referee for suggesting the reformulation in terms of cyclic orders, for pointing out an error in the original proof of Lemma 8, and for other useful suggestions.
The other paper, Lenz
[5],
has a différent set of limitations from ourpoint
of view. The main one is that hisprimitives
are far more elaboratethan is usual in the
theory of measurement;
inparticular,
hepostulates
agood
deal of Hilbert’sgeometrical
structure, definesangles
in terms ofit,
and then adds sufhcient axioms to
yield
theperiodic representation.
Onecan view this note as an
attempt
to isolate the basic abstract structure un-derlying
Lenz’ construction. In somerespects,
ourproof is
similar tohis;
the most obvious
parallel
is the naturalembedding
of thegiven
structureinto an
ordinary
extensive one. A noticeable difference is the form of the Archimedean axiomwhich,
in the absence of anygeometrical
structure,must be
quite strong,
as in other theories of extensive measurement.2. Axioms and statement of
représentation
theoremThroughout
this note, A is anon-empty
set,~
abinary
relation onA,
and o a closed
operation
on a.By ~
we meanboth ~ and ~ 3,
andby
~ we meanboth ~
andnot ~.
DEFINITION 1. The
triple ~A, ~ , o ~ is a periodic
extensive structureiff, for
all a,b,
c EA,
1.
~A, ~~ is a
weakorder, i.e., ~ is
transitive and connected in thesense
that, for
all a, b EA,
eithera ~
b orb ~
a.2.
~A, ~, o ) is a
weak abeliansemigroup 4.
3. a ~ b iff
either(i) a o c ~ b o c ~
c,(ii)
c ~a o c ~ boc,
or(iii) b o c ~ c ~ a o c.
4.
If a >- b,
then there exists apositive integer
n such that a ~ na andnb ~ b,
where la = aand
na =(n -1) a o a.
Axioms 1 and 2 are
unexceptional
in thetheory
of extensive measure-ment and have an obvious
interpretation
forangles.
Axioms 3 and 4 reflect the fact that when a and b are
angles
whose sum,a o
b,
is less than onecycle,
then a ob ~
a,b;
whereas when a o b isgreater
than onecycle,
then a,b >- a
o b(see
Lemma 2 and Def.3). Thus,
in Axiom3,
if a o c and b o c are both less than acycle (case i)
or are bothgreater
than a
cycle (case ii),
theorder a ~
b isunchanged by adding
c both to.a and to
b;
but when a o c isgreater
than onecycle
and b o c isless,
theorder is reversed
(case iii).
And Axiom 4 says that if oneangle
exceedsanother,
then there is someinteger n
such that na( =
sum of ncopies
ofa) just completes
acycle
but nb fails to do so. This is atype
of Archime- dean Axiom.3 Given Axiom 1, it is trivial to show that ~ is an equivalence relation.
4 That is, o is a closed binary operation such that a o (b o c) ~ (a o b) o c and aob-boa for all a, b, c E A.
191
THEOREM 1.
~A, ~ , 0)
is aperiodic
extensive structure withidentity
eiff for
any real K > 0 there exists aunique function ~ from
A into[0, K)
such that
(i) a ~
biff ~(a) ~ ~ (b),
(ii) ~(a
ob) = ~(a) + ~(b) (mod K), (iii) ~(e)
= 0.The most common values of K for
angles
are1, 203C0,
and360,
in whichcase the unit is
called, respectively,
thecycle, radian,
anddegree,
and forclock time 12 and 24 are the familiar values.
The axioms of Definition 1 can be reformulated in terms of the follow-
ing concept
due toRieger [6] ;
see Fuchs[3],
p.62,
and Lenz[5].
DEFINITION 2. A
trinary
relation C on asemigroup ~A, 0)
is acyclic
order
iff, for
all a,b,
c, d EA,
1.
Exactly
oneof C(a, b, c)
orC(a,
c,b)
holds.2.
C(a, b, c) implies C(b,
c,a).
3.
C(a, b, c)
andC(a,
c,d) imply C(a, b, d).
4.
C(a, b, c) implies C(a
od,
b od, c o d)
andC(d o
a, d ob,
d oc).
If ~A, 0)
is abelian and has anidentity
e, C is called Archimedeaniff
5.
C(e,
a,b) implies
there exists apositive integer
n such thatC(e,
a,na)
andC(e, nb, b).
If
we think of the elements of A aslying
in acircle,
we caninterpret C(a, b, c)
to mean that a,b,
and c are clockwise in the order a,b,
c.THEOREM 2.
If ~A, ~, 0) satisfies
Axioms 1- 3of Definition
1 andif
C on
AI -
isdefined by
where a is the
equivalence
classcontaining
a, then C is acyclic
order.If,
inaddition, ~A, o~
has anidentity
e and Axiom 4holds,
then C is Archimedean.Conversely, if
C is acyclic
order on an abeliansemigroup ~A, o~
withidentity
e andif ~
on A isdefined by
then
~A, ~ , o~ satisfies
Axioms 1- 3of Definition
1.If
C isArchimedean,
then
~A, ~, o~ satisfies
Axiom 4.Combining
these twotheorems,
it is easy to show thefollowing.
COROLLARY.
If (A, o~
is an abeliansemigroup
withidentity
e and Cis an
Archimedean cyclic order, then for
any real K > 0 there is aunique
function ~ from
A into[0, K)
such that(i) C(a, b, c) iff 0(a)
>~(b)
>~(c), 0(b)
>~(c)
>~(a),
or4J( c)
>~(a)
>0(b).
(ii) 4J(ao b) = ~(a) + ~(b) (mod K).
(iii) 0(e)
= 0.The
problem
ofcyclic
orders over non-abeliansemigroups,
whichis the natural extension of
Rieger’s
work oncyclic
orders over non-abe-lian groups, appears to be much more
complex.
Aspointed
outby
areferee,
it would beinteresting
to know under what conditions acyclically
ordered
semigroup
can be extended to acyclically
ordered group.3.
Preliminary
lemmasIn each of the
following
lemmas we assume that~A, ~, o~
is aperi-
odic extensive structure, that a,
b,
c EA,
andthat i, j, k, 1,
m, and n arepositive integers.
LEMMA 1. a-b
ifl a o c - b o c; and a ~ b iff
either(i) a c ~ b c ~
c,(ii) c » a o c » b o c, or
(iii) b c ~ c ~ a c.
PROOF. Axioms 1 and 3.
PROOF.
Suppose that,
on thecontrary, b >- a o b ~
a. Fromb >- a
ob,
Lemma 1
(with
c =a) implies
threepossibilities,
but thehypothesis
a
o b ~ a
excludes all buta o b ~ b o a ~ a o b o a ~ a. Similarly,
Axiom3
applied
to a ob ~ a (with
c =b) yields b ~ a o b o b ~ a o b. Thus,
a o b o b ~ a o b ~ a o b o a. Using
Axiom 2 and Lemma 1 to cancela o
b,
we concludethat a ~ b,
which is a contradiction.QED
Lemma 2 insures that the
following
notion is well defined.DEFINITION 1-
For
angles, r(a, b)
is 1 if the sum of the twoangles
is acomplete cycle
or more and is 0 if it is less than a
complete cycle.
PROOF. The
only possible
violation isr(a, c)
= 0 andr(b, c)
=1, i.e.,
a o c ~ c ~ b o c. By
Lemma1, b ~
a,contrary
tohypothesis. QED
193
LEMMA 4.
r(a, b) + r(a o b, c)
=r(b, c) + r(a,
b oc).
PROOF. There are four cases:
(i) r(a, b)
= 0 =r(a
ob, c). By
definition aob?: a,
b and(a o b)
oc ~
a o
b,
c.Using
Axiom2,
and so
by
Lemma2, r(a,
b oc)
= 0.Suppose r(b, c)
=1,
then a o b o c~
c ~ b
o c and so,by
Lemma1, b » a
ob,
which iscontrary
tohypothesis.
So
r(b, c)
= 0.(ii) r(a,b)= 1, r(a o b,c) = 0.Thus, a,b ~ a o b and a o b o c ~ a o b,c.
If
r(b, c)
=0,
then b o c~ b,
c.By
Lemma1, b » a
o bimplies
b o c ~a o b o c or a o b o c » c » b o c.
The latter isimpossible,
soby
Lemma2,
r(a, b o c) =
1. Ifr(b, c)= 1,
thenb, c » b o c.
Soa o b o c ~ c ~ b o c,
whence
r(a,
b oc)
= 0by
Lemma 2.(iii) r(a, b)
=0, r(a o b, c)
= 1. Similar to(ii)
(iv) r(a, b)
= 1 =r(a o b, c). Thus, a, b ~ a o b and a o b, c ~ a o b o c.
From
a » a o b » a o b o c,
Lemma 2implies r(a,
b oc)
=1,
and Lemma1
implies b » b
o c, soby
Lemma2, r(b, c)
= 1.QED
LEMMA 5.
If k
and n arepositive integers
such that n - k - 1 is apositive integer
anda ~ b,
thenr[(n-k-1)a, a]+r[(n-k)a, kb] ~ r(b, kb) + r [(n - k - 1)a, (k + 1)b].
PROOF.
Using
thecommutativity
of o, and so of r, and Lemmas 3and
4,
We next introduce a count of the number of
complete cycles
that ismade when the sum na is formed.
DEFINITION 4.
W(l, a)
= 0PROOF. For any m > 0 and n = 1 the assertion is true
by
definition.By
induction and Lemma4,
LEMMA 7.
W(mn, a)
=mW(n, a) + W(m, na)
PROOF. For m =
1,
it isobviously
true.By
induction on m, andLemma
6,
For n ?
3, by
Lemmas 3 and 5(omiting meaningless
terms for n =3, 4),
LEMMA 9.
If
a ~ b and 1 is apositive integer,
then there exists apositive integer
k such thatW(k, a) ~ W(k, b) + l.
PROOF.
By
Axiom4,
there exists apositive integer
n such thatr [(n -1 )a, a]
= 1 andr [(n -1 )b, b]
= 0.By
Lemma 8195
Observe that if
r(c, jc)
= 1 forall j,
then( j -1 )c » jc » (j + 1)c.
Can-celing ( j -1 )c by
Lemma1,
c ~2c ~ 3c ~ ···. By
Axiom4,
this ter- minates in a finite number ofsteps
and so we may choose isufficiently large
thatW(i, cl
i - l.Choosing
this i for c = nb andletting k = in,
4.
Imbedding
in anon-periodic
structureIf a is an
angle
and m anon-negative integer,
then it is natural to inter-pret (m, a)
as theangle consisting
of mcycles plus
a. For the set of suchobjects,
we have thefollowing
natural notions ofordering
and concate-nation :
DEFINITION 5.
If ~A, ~, o~ is a periodic
extensive structure and Nis the set
of non-negatzve integers, define
THEOREM 3.
If ~A, ~, 0)
is aperiodic
extensive structure, then~A*, ~ *, *)
is anon-negative
extensive structure in thefollowing
sense[7 ] : for
all a,fi, y, ô
EA*,
1.
(A *, ~ *)
is weakorder;
2.
~A*, ~*, *)
is a weak abeliansemigroup;
3. 03B1
~* 03B2 iff
03B1*03B3~* 03B2*03B3;
4.
if 03B1 ~* 03B2,
then there exists apositive integer
n such thatn03B1*03B3 ~ * n03B2 * 03B4;
5.
03B1*03B2 ~ * 03B1.
PROOF. 1. This follows from Axiom 1 of Definition 1.
2. This follows from Axiom 2 of Definition 1 and Lemma 4.
3.
Suppose
that(m, a) ~* (n, b),
and consider the relation between(m, a) * (p, c) = (m +p + r(a, c),
ac)
and(n, b) * (p, c) = (n +p + r(b, c),
b o
c).
If~
*holds,
then either m > n or m = n and a ~ b. If m > n, then the conclusion is ~ * since eitherr(a, c) ~ r(b, c)
or a c~
c ~ b o c.If m = n and
a ~ b,
thenby
Lemma3, r(a, c) ~ r(b, c)
and so~
* fol-lows. To
complete
theproof
we needonly
show thatif a ~
b andr(a, c)
=
r(b, c),
then a c~ b o
c. Ifr(a, c)
=r(b, c)
=0,
then a oc ~ a,
c andb o
c ~ b,
c.By
Lemma1, a ~
bimplies a c ~ b c ~ c.
Ifr(a, c) = r(b, c)
=1,
then a, c ~ a o c andb,
c >- b o c.By
Lemma1, a ~
bimplies c » a o c » b o c.
If m > n, choose k so
large
that km >kn + 1,
thenby
Lemma8,
If m = n
and a >- b,
then choose k as in Lemma 9 andproceed
as above.5. If n >
0, then
because either
r(a, b)
= 1 orr(a, b)
= 0 whichimplies a 0 b?::
1.QED
5. Proof of Theorem 1.
If
~A*, ~ *, *~
is anon-negative
extensive structure, Theorems 2 and 3 of[7] ]
combine to say that there exists an additivehomomorphism which
isunique
up to asimilarity
transformation. For aperiodic
struc-ture
A, ~, o~
withidentity e
and a constant K >0,
letç*
be theunique homomorphism
of the structure in Theorem 2 forwhich ~*(1, e)
= K.Set ~(a) = ~*(0, a).
Note that0 ~ ~(a) ~*(1, e)
= K. It is obviousthat a ~
b iff~(a) ~ ~(b).
ConsiderBut
~ Thus,
Since i
197
The
necessity
of the axioms is routine toverify. QED
6. Proof of Theorem 2
Suppose ~A, ~, 0~
satisfies Axioms 1-3 of Definition 1 and C is defi- ned onA/~
as in the statement of the theorem. We show the four axioms of Definition 2.1,
2. Trivial3.
Suppose C(a, b, c)
andC(a,
c,d). Suppose
a >-b ~
c. Since bothe ~ d ~
aand d ~
c ~ a lead to the contradiction a >- a, we know a ~ c ~d,
and so a ~b ~ d,
whenceC(a, b, d).
Theargument
issimilar in the other six cases.
4.
Suppose C(a, b, c). If a » b »
c, thenby
Axiom 3 of Definition 1 either(i) a d ~ b d or (ii) b o d » d » a o d
and(iii) b o d » c o d or
(iv) c d ~ d ~ b d. Clearly (i)
and(iii) imply C(a d,
bd, c d)
and(ii)
and(iv)
areimpossible
sincethey imply d ~
d. If(i)
and(iv),
theneither
c d ~ a d ~ d,
whichimplies
c » a(Lemma 1), contrary
toassumption,
orc d ~ d ~ a d ~ b d,
whenceC(a
od, b
od, c d).
The
argument
is similar for(ii)
and(iii)
and for the two other casesb ~
c » a and c >- a » b. The fact that~A, 0)
is abelian establishes the other conclusion.Assume,
inaddition,
Axiom 4 and anidentity.
For a EA, a ~ e
since ife » a, Axiom 4
implies
there is apositive integer n
suchthat e »
ne - e,which is
impossible.
IfC(e,
a,b),
then theonly possibility
is a »b ~
e,whence
by
Axiom4,
there is apositive integer n
suchthat a »
na ~ eand nb ~
b »
e, whenceC(e,
a,na)
andC(e, nb, b).
Conversely,
suppose~A, 0)
is an abeliansemigroup
withidentity e
and
cyclic
order C. Since - is =, weignore
it. We prove Axioms 1 and 3 of Definition 1(2
is one of thehypotheses).
1.
If a ~
b andb ~
c, thenC(e,
a,b)
andC(e, b, c). By
Axiom 3 ofDefinition
2, C(e,
a,c),
so a ~ c.By
Axiom1, either a ~
b orb ~ a for a ~
b.3.
If a ~ b, C(e,
a,b)
and soby
Axiom4, C(e
o c, a oc, b
oc)
=C(c,
a o c, b o
c).
Thisimplies
Axiom 3 of Definition 1 if we can showC(a, b, c)
implies a ~ b ~ c, b ~ c ~ a, or c ~ a ~ b. Suppose
not, then one of the other threepossibilities
must hold.If a ~
c ~b,
thenC(e,
a,c),
C(e,
a,b),
andC(e,
c,b). By
Axiom2, C(c,
e,a)
andC(c, b, e),
whenceby
Axiom3, C(c, b, a).
Soby
Axiom2, C(a,
c,b),
which isimpossible
by
Axiom 1. Theremaining
two cases are similar.If C is
Archimedean,
Axiom 4 of Definition 1 followsimmediately.
QED
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L. FUCHS
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O. HÖLDER
[4] Die Axiome der Quantität und die Lehre vom Mass. Ber. D. Säch., Gesellsch.
D. Wiss., Math-Phy. Klasse, 1901, 53, 1-64.
H. LENZ
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F. ROBERTS & R. D. LUCE
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(Oblatum 13-IV-70) Professor R. Duncan Luce Program in the Social Sciences The Institute for Advanced Study Princeton, New Jersey 08540 U.S.A.