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~ MATHEMATICS JOURNAL
St. John's University
Jamaica, New York
VOLUME I SPRING 1965
Editor-in-Chief •••••••••••••••• John c. Maceli 1 65
Cover •••••••••••••••••••••••••• Joan Cummings '65
Moderator •••••••••••••••••••••• Professor Anthony Sarno
Contributors ••••••••••••••••••• James r. Dart, Jr.
Robert Stanton
frank J. Infante
Joy Cuttita
N, J. Biancardo
R. Thomas Maniscalo
ACKNOWLEDGMENT
I would like to thank Dean Kollintzas, the members of
the MATH CLUB and all those who have been instrumental
in making this publication possible. I would also like
to thank all those who have submitted articles to the
MATHEMATICS JOURNAL.
Sincerely,
John C. IYlaceli
Editor-in-Chief
MEASURE THEORY
James f. Dart, Jr.
(To conserve space, only important theorems shall be proven.)
- An extended real valued set function u defined on a class E
is finitely additive if, for every finite, disjoint class E , •• ,E of
l n
sets in E whose union is also in E:, we have
A measure is an extended real valued, non-negative, and countably
additive set function u, defined on a ring R, and such that u(O) = a.
We observe that, in view of the identity,
a measure is always finitely additive. A rather trivial example of a
measure may be obtained as follows. Let f be an extended real valued,
non-negative function of the points of a set X. Let the ring~ consist
of all finite subjects of X; define u by
u( { xl'.,. ,xn} ) :: 2 :=l f(x 1) and u (0) :: 0
<.~, ........ If us is a measure of a ring Ra set E in R is said to have a
finite measure if u ( E) < oe •
The measure u is called complete if EfR and rcE, u(E):: O;
this implies r~ R.
Example: ~ Let u = measure defined on R
u(E)<l)'.)for at least one E in R
Prove: u(O):: 0
Proof: u(E)~ o:,implies there exists E~"R' such that
E has a finite measure
therefore, u is a measure implies u(O) = 0 by definition.
Theorem 1. If { E1 ••• En} is a finite disjoint class of sets in P, and
contained in a given set E0 in P, then
Proof: We write Ei = (a1, b1), i:0,1, ••• ,n, and without any loss of
generality, we assume that: a1 "- ••• < a
- - n
It follows from the assumed properties of { E1, ••• ,En} that
ao ~ a1 ~bl!.•• ~ an~ bn~ bo'
and therefore
~~=l CJ(El.. )
_ ~n (b._a.)
::. ;£_ i=l l l.
+ ~ ~ (a. 1 - bl..) =
~ J.:l l.
= bn-a1 < b -a = u(E 0 ).
- 0 0 Q. E. D.
Theorem 2. If a closed interval F0 , where F
0
= [a0
, b
0
] is contained in
the union of a finite number of bounded, open intervals, Ui, ••• ,un,
Theorem 3. a sequence of sets in P such that,
then
Theorem 4. The set function u is countable additive on P.
Theorem 5. There exists a unique finite measure~ on the ring 'R, such
that whenever£ E' P,; (E) = u(E)
Properties of Measures
.,._
Theorem 6. If us is a measure on a ring R, then u is monotone and subtractive.
Proof: If E ~ R, r~R, and E~F", thel} F\.E~R and u(f") = u(E) u(F"-E).
2
The fact that u is monotone follows now from the fact that it is non-negative;
the fact that it is subtractive follows from the fact that u(E), if it is
finite, may be subtracted from both sides of the last written equation.
Q.E. o.
Theorem 7. If u is a measure on a ring 'R, if E~ R, and if f Ei1 is a finite
or infinite sequence of sets in R such that E~ui Ei then
u(E) ~ ~i u(Ei)
Theorem a. If u is a measure on a ring R, if E(R, and if\ Ei) is a finite
or infinite disjoint sequence of sets in R, such that
U i EiC. E, then Li u(Ei) ~ u(E)
Theorem 9. If u is a measure on a ring Rand if~ En1 is an increasing
sequence of sets in~ for which lim n En~~, then u(lim n En) = lim n u(En)
Theorem 10. If u is a measure on a ring R, and if i EnJ is a decreasing
sequence of sets in R' of which at least one has finite measure and for
which lim n E0 ~R, then u(lim n En) = lim n u(En)•
Proof:
then u(En) ~ u(Em)<ciofor n~m and therefore
u(lim m En)<«>. It follows from theorems 6 and 9 above, and the fact that
{Em-En} is an increasing sequence, that
u(Em) - u{limn En) = u(Em - limn En)
= u(lim n (E m - En) ) = lim n u(E m - En)
=limn {u {Em) - u {En) )
= u {Em)• limn u{En)
Since u(Em) ( oz, , the proof of the theorem is complete. - Theorem 11. Class"f is continuous from below at a set E ~n E if, for all
increasing sequences \ En} of sets in E for which lim n En :E, we have
limn u{En) = uCE) •
3
Similarly u is continuous from above at E if for all decreasing f En}
of sets in E for which \ u(Em) I .<:, o, for at least one value of m and for which
limn En= E, we have limn u(En) = u(E).
The idea of a measure is also introduced into the theory of probability.
When we are considering the outcome of an experiment, if there are only a finite
number of logical possibilities, there will only be a finite number of
truth sets and hence the process of assigning probabilities is a finite one. ·
We proceed in three sets:
l. We first determine pa possibility sat that is the set of possibilities.
2. To each subset x of p we assign a number called a measure m(x) such
that sum of the numbers is one. The measure of the empty set is O.
The properties of a probability measure ares
A) m(x) = o if and only if x = empty set
8) ofm(x) ~ 1 for any set x
C) for two sets x and y, m(xvy) = m(x) + m(y), if x and y are disjoint.
Example 1
An example would be with an ordinary dice cube. If we are concerned
with the probability that the number which turns up is less than 4. Here the
possibility set is {1,2,3,4,5,6,}. The symmetry of the die suggests that
each face should.have the same probability of turning up. To make this so, we
assign a number 1/6 to each of the outcomes. The truth set of the statement,
"The number which turns up is less than 4" is 1,2,3. Hence the probability
of this statement is 3/6 = 1/2, the sum of the number of the elements in the
truth set.
4
THE SLIDING BLOCK PUZZLE
Robert Stanton
In the f,larch issue of ~,tific American, Martin Gardner discusses a
type of sliding block puzzle. The problem is to slide the blocks in figu~e
A until the position in figure Bis reached.
8 7 6 l 2 3 '
5 4 3 4 5 6
2 l 7 8
A B
Moves are made by sliding a square into the adjacent empty space. It is easy
to see that a solution exists, either by trial and error or the following
general method. Switch pairs of blocks until the desired pattern is reached.
If the number of switches is even (as in this case) a solution is always
possible.
While it is a simple matter to find a solution, it is desirable to know
the least number of moves in which a solution can be found. Gardner shows
that it is impossible to do the problem in less than 26 moves, based on an
analysis of the shortest path of each square and the first nine moves. Further
analysis undoubtably will raise this limit. Gardner states that the best
recorded solution is in fact 36 moves, a solution in Henry Ernest Dudeney 1s
Puzzles and Curious Problems. His solution, in order of the squares moved,
is 12543 12376 12376 12375 48123 65765 84785 6. Gardner notes that
in any solution the number of moves is necessarily even, as the empty space
returns to its original position.
The following is from my letter to Mr. Gardner, shedding new light on
the problem.
"The solution to the sliding-block puzzle, number 9 in the March 1965
issue, can be obtained in less than 36 moves. The general method is to move
5
in one direction (I chose counterclockwise) around the center, inserting
squares in their logical order in the rotation. Note that in the original
position, 1,2 and 8,7 ere in proper order. i~ves should aim to put the 4
in front of 1 and 3,6 in back of 2. The 5 is inserted and removed from
the center for convenience. I found one basic solution of 34 moves (with
variations in the order of moves) and one solution of 32 moves. My 34
move solution is the following: 12587 63463 41263 41265 87412 35874 1236.
It is interesting that after 26 moves, each square is one move away from
its proper position. By using the transformation n" = 9-n, and reversing
the moves, we obtain a solution: 36785 21467 85214 37856 37856 35632 1478.
This is the transformation for those who prefer the clockwise direction.
Any solution can be reversed in this manner. My 32 move solution is as
follows: 12587 43123 87461 23861 23857 41236 58. This can be transformed
into a clockwise solution as before, although the first move is
counterclockwise.
"The fastest solution, however, takes only nine moves. This consists
of sweeping all the squares off the board in one move, and placing them one
by one in their proper places."
For the interested reader, Mr. Gardner will publish the results of
his readers• findings in the May or June issue of Scientific American.
A few readers program computers to solve these problems, and it is
possible that the minimum numer of moves may be definitely established by
that time.
6
AN INTRODUCTION TO OPERATIONS RESEARCH
Frank J. Infante
When approaching a new subject it is often advantageous to start
with a survey which focuses on broad outlines rather than details. This
survey seeks to answer general questions about the subject. Some of
these ares What is it7 What does it do7 How does it do this7 Its
object is to give the student a basic familiarity with the subject in
general ·--its aims, the kinds of problems with which it is concerned, and
the principal concepts and techniques which it uses. The result of the
survey is a frame of reference for the student who may wish to examine
the subject more closely. This survey will, it is hoped, provide such
a result.
It is well to begin by saying that Operations .Research is hot a
'new subject' in the broad sense of being a radical innovation. It is,
instead, another phase during this century in the development of the
stream of ideas associated with what is known as 'scientific management.'
Moreover, many of the techniques now identified with Operations Research
have existed independently for some time. In a narrower sense, however,
Operations Research is a new discipline. Its methods, co:i~spts, and
techniques, although derived from the physical sciences end mathematics,
had, before World War II, very rarely been applied in an organized and
systematic manner to the operating problems of human organizations,
whether military, political, or industrial. Thus Operations Research
may be defined as the application of the Scientific method to the solution
of the overall operating problems of an organization.
If we assume this definition to be true then it should be clear that
the legitjmate subject matter for Operations Research includes almost
7
all aspects of any complex organization of men and/or equipment. more
specifically, Operations Research is concerned with the effectiveness of
the complex organization in achieving its goals, whatever they might be.
Now the manager of the organization is also concerned with, as
well as responsible for its effectiveness. You may well ask, "Then what
is the difference between the tasks of the manager and those of the Operations
Researcher?" Actually there is no difference. However, Operations
Research acts as a supplement to the work of the manager. Operations Research
makes no decisions about running the organization. All it does is
make recommendations for change. It is up to the manager to decide whether
or not to use them. In effect, the Operations Research team says to the
manager, "We don't believe you are running your organization as effectively
as it could be run in terms of your objectives. Let us study your
operation and we believe we can improve it." Here again Operations Research
is closely analogous to the field of Management Consulting; the difference
between the two being the fa~t that Operations Research groups consist
largely of men tr.ained as scientists while management consultants generally
have business backgrounds.
In any one Operations Research team we may expect to find men with
many different scientific backgrounds--physicists, chemists, engineers,
and above all~ mathematicians. The work of the mathematician is especially
important for it is with mathematical techniques that most of the problems
which the team encounters are solved.
The main job of the mathematician is to represent the physical
operation or •real world' situation with a mathematical model. This model
usually consists of a series of equations or relationships and must resemble
the 'real' situation in such a way that when solutions are found in the
8
model, analo~ous solutions tdll be found for the actual operation. The
formulation of a mathematical model is a difficult task. The mathematician
as an Operations Researcher must understand fully the operation to be duplicatsd
and include in his model only those factors which are related to
the problem while ignoring the trivial ones. He then builds his model
with the •tools of his trade.' Besides using algebra, trigonometry, analytic
geometry, and calculus as basics, the Operations Researcher also usess
differential ~guations, especially for military applications; probability
theory and statistics, if events in the model are probabilistic in nature
or if statistical data is being used in the model; and computer programming,
for a great many of the problems met by an Operations Researcher will have
facets suitable for solution by a computer. At this point it is safe to
say that the mathematical model is~ most useful and important technique
at the disposal of the Operations Research team. However, one indispensable
factor must always be present in a model. for, essential to the model is a
conGistent statement of the fundamental goals of the operation. Just as
the model cannot make 2 and 2 add up to 5, so it is impossible to relate
fundamentally inconsistent objectives and produce consistent and meaningful
results.
As has been previously stated, legitimate areas of investigation for
Operations Research teams include virtually all aspects of an organization.
Thus in the area of industrial operatiohs the team may face problems ranging
from accounting procedures to analysis of warehouse facilities, from
credit extension to inventory control, from scheduling of airline flights
to proper coordination of advertising campaigns. The military, where
Operations Research first found practical application, :::onstantly pl'ovides
important problems since its effectiveness depends upon continuous analysis
and re-analysis of its operations.
9
The entry of Operations Research as a scientific discipline into the
domain of organization management is highly significant. It represents
the trend of the modern business world toward increased reliance on sound
scientific principles to guide its organizations rather than solely on the
intuition or 1 feel of the situation' of a few top executives with little
scientific background. Of course Operations Research does not promise
to be a cure-all for the problems of managers. There is no guarantee
that the Operations Research team will provide the absolutely correct
answer to any problem. However, as the record of significant success of
Operations Research continues to grow, it will become clear to many
managers that Operations Research is the key to efficient decision making
in organization management,
10
LOOKING AT LATTICES
Joy Cuttita
A lattice is one of the most elementary of all algebraic systems.
It is defined ass A set of elements partially ordered with respect to
some binary relation and where for every pair of elements x, y, ( S there
is a 11meet, 11 x f'I y, and a II join" xUy. Let us further analyze this definition.
What do we mean by a partial ordering? Any set is partially
ordered with respect to some relation R if the relation is reflexive,
transitive, and antisymmetric. i.e. V x,y,z, '- S
l) xRx
2) xRy and yRz ;) xRz
3) xRy and yRx > x=y
Thus the ordering, due to a lack of symmetry is not an equivalence relation.
It is not necessary that either xRy . or yRx hold for any two
distinct elements of S, in a partial ordering. If either xRy or yRx
exists for every x., y ( S we have a simple ordering or a chain. This
is a special type of partial ordering
~ In II neither
b bRc nor cRb
(. b C
exists. In I
I d II elements are
d
e.g. fig I is a chain whereas fig II is not. related
~ The relation R will be denoted by? although it need not mean
set inclusion. It can have several definitions, "divides," "is a
all
subset of," "is less than or equal to," etc. Lattices can be illustrated
by means of Hasse diagrams. Lat's define xRy as y divides x and consider
the set of divisors of 12.
11
1
In this dia;ram, the points correspond to
the elements of our set. x ~ y will maen
either x:y or we can go from x toy by mov-ing
only downward along a series of lines,
Let's consider the concepts of meet and join of our original definition.
If fer elements x,y,z E S x~ y and x~ z then the x is an upper bound
for y and z. There can be more than one upper bound for each pair of ale-ments.
However each pair must have a unique least upper bound or "join."
The least upper bound his defined for elements x and y if:
1)
2)
h is an upper bound for x and y. i.e. h ~x and
If d is any upper bound of x and y then d? h.
h ~ y.
In our example, although both 12 and 6 are upper bounds for the pair 3
and 2, 12~ 6 therefore 6 is the l.u.b. for this pair. 6:3V 2
Likewise, each pair x, y, E: S must have a unique greatest lower bound
or "meet." VJe define g as the g.l.b. for elements x and y if:
l) g is a lower bound for x and y. i.e. x'=g and y"tg.
2) If pis any lower bound of x and y then g~p.
In our example, 1 is the only lower bound for elements 3 and 2, thus it
is the unique g.l.b. for this pair, 1:3 (\ 2. It is easy to see that the
g.l.b. for any pair must be unique. Suppose there were two, p and g,
g=xfly and P=X"Y· Then, from 2) above, g must be below p, p~g
also p must be below g, g~p. Now due to antisymmetry p:g. A similar
argument can be used to show the uniqueness of the l.u.b. for any pair
of elements in S,
It is of further interest to note that two different lattices may
have the same Hasse diagram,
12
This diagram represents the subsets of a set
of card three, where the relation is inclusion
and also the set of divisors of 30 where ~
means divides.
These two lattices are isomorphic and their elements can be set up in a
1-1 correspondence. A lattice may have more than one representation,
providing the ordering of the elements is not distorted. e.g.
a
(
Q
These diagrams all represent the same lattice.
e
In every lattice the idempotent, absorption, commutative, and as-sociative
laws are satisfied and completely characterize the structure.
The distributive law does not necessarily hold. If every subset of e
lattice has a g.l.b. and a l.u.b. the lattice is called "complete.tt
In particular, L itself has a l.u.b. I called the supremum and a g.l.b.
0 called the infinum. Thus in a lattice, I is an upper bound which is a
lower bound for the set of all upper bounds; 0 is a lower bOtJnd which
1
is also an upper bound for the set of all lower bounds.
By a complement of an element x of a lattice L which has Sup. I
and Inf. 0 we mean an element y{ L xv'y:I and xf\ y:O. If all
the elements of L have complements, Lis called a complemented lattice.
A complGmented, distributive lattice forms a special algebraic system
known as a Boolean Algebra.
There are numerous non-mathematical examples of partially ordered
sets in the world. Every hierarchy in nature is essentially a partial
13
ordering. Lattices can be constructed to represent every-day situations.
Consider the lattice which represents the possible orders for breakfast in
2
a particular restaurant. Orders are limited to coffee, orange juice, eggs
and pancakes, with an additional stipulation. Everyone must order and each
order must include coffee.
(c,o,E.P)
(C)
In the lattice:
C - coffee
0 - orange juice
E - eggs
p - pancakes
(c) is the infinum and (coEp) the
supremum for the lattice.
The theory of lattices is a very concise one. All of its concepts
can be defined in terms of one undefined relation?. However, it is
also dependent upon two dual, binary operations f\ and U which in many
properties resemble the+ and X of ordinary arithmetic. It is this
analogy which makes lattice theory a very interesting branch of algebra.
3
l
Rober R. Stoll, Sets. looic and Axiomatic Theories(San Francisco, 1961) p.53.
2Roy Dubisch, l&itices to Looi~(New York, 1964) p. 4.
3carrett Birkhoff, ~attics Theory(New York, 1948).
14
THE mm PARADOX
N.J. Biancardo and
R. Thomas Maniscalco
"The time which we have at our disposal every day is elastic"
Marcel Proust
Remembrance of Things Past
As true as the elasticity of time pertains to the realm of
classical prose, it likewise commands consideration in the science
of mathematics. And, as in the case of the twin paradox of the Theory
of Relativity, we surely must regard times' unique and theoretical
elastic qualities exhibited in those instances which both overtly
display themselves in the paradox, as well as implied occasions which
are afforded to those interested in them.
In the years following Einstein's publication of the Theory of
Relativity, mathematicians the world over have been discussing the
possibility of a hidden paradox; a hidden paradox that actually does
not exist, at least for those who are interested enough to delve deeply
into the matter.
We must therefore consider each of the two sides of the situation
separately. Vet, before we go any further, we find it may be necessary
to define the problem posed by the twin paradox.
If a man were to be sent into the void of interstellar space,
in a ve'hicle that was capable of approaching the speed of light, would
he or his stay-at-home twin brother be the older upon his return?
The Theory of Relativity states that the younger of the two would be
the space traveler.
Jay Orear, noted physicist, claims that:
Since the advent of space exploration, it has become co~mon knowledge
that space travelers will not age as fast as their bn:it!"1ers on earth.
In fact, if a space traveler could travel at the speeu of light, he
would not age at all.l
15
However, the Theory of Relativity also states that all motion is relative
to its surroundings. Charles Steinmetz expounds on this view to the extent
that:
All phenomena of space, time and motion are relative: that is, there
is no absolute motion, etc., but only motion and time relative to some
other motion, time, etc.2
Hence, could it not be possible to assume the earth to be in motion,
and the receding rocket to be at rest? And, as such, sould not the twin
in the rocket be the older of the two? This is the paradox. It is our
contention that there is no paradox. That no matter the point of view, the
twin in the rocket will always be the younger of the two. Various situations
exist, and it therefore becomes necessary for us to make use of illustrative
examples in properly evolving a solution to this problem.
Let us first take the position that the earth is at rest, and the
rocket is in a state of motion relative to the earth. We shall equip each
of our twins with a radio transmitter and a receiver. The transmitter will
be set up so that it can only propagate radio waves of 20vps. We shall call
the earthmen A, and the space traveler B. Let us now send Bon his journey.
He will have a velocity of 99.0 per cent of the velocity of light.
During the period of B's recession from A, A will turn on his transmitter,
and B his receiver. The effect of B's transmitter will be comparable
t~ the Doppler Shift of light, except that it will deal with radio waves.
In explaining the Doppler Shift, George Gamow states that:
When the source of light is approaching the observerr light waves
are shortened due to the motion of the source, and t~~ opposite is
true for the recession of a source of light. When the light (All
electromagnetic) waves are shortened, more vibration~ per second
are received, again, when the light waves are lengthened, less
vibrations per second are received.3
Therefore, B will receive less than the 20vps. transmitted by A. We
will give this a convenient value of lSvps. On the return trip, B will
16
receive more than the 20vps (Doppler Shift) again, we will give it a convenient
value of 25vps. So, if according to 8, the trip took 400 sec.
(200 each way), the total number of waves received by 8 will be:
15•200 + 2s·200 = BOOOv/400 sec.
Let us now compare these findings with those obtained by letting 8
do the transmitting. When 8 sends out his 20vps.(during the recession), A
will receive less than 20vps. for the same reason that~ did on his trip.
However, at the moment 8 turns around, A will not instantly receive 25vps,
since it takes time for those last few impulses from B's transmitter to
reach A's receiver. So, A will receive the lSvps. for a longer period of
time than the 25vps. This is due to the fact that while those last few
impulses from B's transmitter are approaching the earth, so is B •. It follows
then, that the number of vps. received by A will be less than B's.
However, we must still account for the fact that B did send out 8000v,
even though A did not receive them in the 400 sec. time period. The only
way to resolve this is to accept the fact that the trip must hav~ lasted
longer for A than it did for B, as well as the fact that B could not have
traveled faster than the electromagnetic waves he was transmitting.
" ••• light is the absolute velocity, which cannot be exceeded by anything
else in nature.
4
Rather, B could not have traveled faster than the radio
waves he was ~ending, because light and radio waves ere both electromagnetic
wave forms and behave according to the same laws of nature. Therefore,
A must have received the 8000v for the above reason. And since the journey
lasted for only 400 seconds for B, A must have aged more than B. This is
the first part of the paradox, and it parallels what Einstein said when he
claimed that: 11a moving man will age less than his stay-at-home brother.
5
17
We must now consider what will happen if we allow B to remain at rest.
At this point, the main difference between A and Bis that B must have a
tremendous amount of energy to escape the gravitational force of the earth,
while A does not. In other words, 8 will remain accelerated relative to
an inertial frame of reference, while A is not. This inertial frame of
reference can be compared to the distant matter in the universe. So,
again, 8 accelerated relative to distant matter while A did not.
Let us now turn to Issac Newton's classic Law of Motion with regard
to the relationship b.etween a water bucket, the ground, and a quantity of
water contained in the bucket. Newton states that when we tie the bucket
to a pole, securely attached to the ground, and wind up the string and
let the bucket .90, the bucket will revolve relative to the ground. But
the water in the bucket will not. Instead, it will revolve relative to
the bucket until the friction of the bucket and the water on each other
will reach a point where the water will also start to revolve.
We can compare the bucket to the rocketship, the water to the earth,
and the distant matter to the ground. If we say that the bucket (8) is
at rest, then we must concede that not only will the earth (water) accelerate,
but so will the distant matter (ground). This distant matter
exerts a gravitational force (in the case of Newton's experiment it would
be the ground). All acceleration exerts a gravitational force on the
objects over which it presides. Therefore, the assumption that 8 is at
rest would mean not only that distant matter is accelerating, but so is
twin A. Now, this distant matter (ground) and also A (water) leaves the
area of 8; 2) When A is in the process of turning around; 3) when A is
on the return trip; or on acceleration, deceleration and acceleration respectively.
At the start of the trip, both clocks will be affected in
18
precisely the same way. The same is true when A is at home. However,
when A is called upon to turn around in the middle of the trip, A and B
will each be as far apart from each other in relationship to the distant
matter contained in the vast expanse of the universe.
Therefore, the gravitional force produced by the distant matter in
the universe will affect the clocks of both twin A and B differently.
During the period of the turn-around, the gravitional force of distant matter
will still be acting, and at this particular point, A's clock will tick
faster than B's. This is due to the fact that the force of gravity of
distant matter will be exerting a force on the clock of twin B because of
the fact that only 8 is moving relative to the inertial frame, while A is
not. Therefore, .A's clock will not be affected but B's will. Consequently
A's clock will tick faster than B's since "As a consequence of its motion,
6
a clock will tick more slowly than when at rest. 11 Since A's clock ticks
faster than B's, B will register a shorter length of time for the duration
of the turn-around. At the end of the trip, the time duration of the
clocks for the trip will differ. And since A's clock ticked faster than
B's, A will register a longer period of time for the duration of the trip.
Therefore, A will age more than 8 even though A was considered to be the
moving party.
1 Jay Orear, Fundamental Phvrsics, 1st. ed., p. 233.
2 Charles Steinmetz, Relativity and Space, 1st ed., p. l.
3 George Gamow, The Creat:i.on of the Unive~ p. 31
4 Herman Nurnberg, Einstein(Thsory of Relativity) for Everybody, 1st ed., p.38.
5 D. ~. Sciama, The Unity of the Universe, 2nd ed., p. 151.
'
6 Albert Einstein, Relativity, 2nd ed., p. 44.
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