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VOL. 78, NO. 5, DECEMBER2005
with centerF and radius-,SF. TF. The desiredpoint P is the intersectionof this
circle and the slant line (FIGURE9b).
REFERENCES
1. James Stewart,Calculus,4th ed., Brooks/Cole,Pacific Grove,CA, 1999.
2. JamesR. Munkres,Topology,A First Course,Prentice-Hall,EnglewoodCliffs, NJ, 1975.
3. JamesR. Smart,Modem Geometries,5th ed., Brooks/Cole,Pacific Grove,CA, 1998.
ProofWithoutWords:
AlternatingSums of Odd Numbers
1)(-1)n-k
n even
n odd
-ARTHUR
T. BENJAMIN
HARVEY MUDD COLLEGE
CA 91711
CLAREMONT,
A ShortProofof Chebychev'sUpper Bound
Kimberly Robertson
William Staton
University of Mississippi
University, MS 38677
[email protected]
Examining7 (n), the numberof primesless thanor equalto n, is surelyone of the
most fascinating projects in the long history of mathematics. In 1852, Chebychev [3]
proved that there are constants A and B so that, for all naturalnumbers n > 1,
In(n)
Later,in 1896, with argumentsof analysis, the Prime Number Theorem was proved,
showing that for n sufficiently large, A and B may be taken arbitrarilyclose to 1. Es-
Mathematical Association of America
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386
MATHEMATICS
MAGAZINE
tablishingthe PrimeNumberTheoremis difficultandthe proofis often omittedfrom
texts in elementarynumbertheory.Chebychev'sarguments,by contrast,were elementaryand wonderfullyclever,using propertiesof the middlebinomialcoefficients
(2n).Ourpurposehere is to providewhatwe believe is a brief and elegantapproach
to Chebychev'supperbound,a proof accessiblein its brevityto even the beginning
numbertheorystudent.We were initiallymotivatedby Bollobas'lovely Englishpresentation[2] of PaulErd6s'proofof Bertrand'sPostulate.
We begin with threesimplelemmas,in whichn alwaysdenotesa positiveinteger.
LEMMA. For all n,
nr(2 n)-((n)
< (2n)
Proof There are exactly 7r(2n)- r(n) primesbetweenn and 2n. Each appears
preciselyonce in the numeratorof the factorialexpressionfor (2n)andnone appearsat
all in the denominator.So each is a factorof (2n),andof courseeach is biggerthann.
LEMMA. For all n > 8, 7r(2n) < n - 2.
Proof By induction: 7r(16) = 6 < 8 - 2. If r(2k) < k - 2, then xr(2k + 2) < (k -
2) + 1 since 2k + 2 is certainlynotprime.So 'r(2k+ 2) < (k + 1) - 2 andthe lemma
U
follows.
LEMMA. For all n, we have (2n2)< 1(4n).
Proof Begin a proof by induction, by noting that, for n = 1, (2) = 2 = (41).
by (2n)yields
Dividing(2n+12)
n+
< (4n), then
< 4. Hence,if (2n)
(n+1
W
andthe lemmafollows.
We will prove our theoremby a variantof induction,moving from k to both 2k
and2k - 1. Thatthis schemeis adequateis easily seen by notingthatthe truthof the
statementfor all k up to 2r thenimpliesthe truthof the statementfor all k up to 2r+1
THEOREM.For all n > 1, n(n)
< 8n.
Proof Forn < 8 thisis clearsinceboththebasesandthe exponentscomparein the
desired way. Furthermore,since 16"r(16) = 166 - 88 < 89, the statement holds for 9 <
n < 16. Now suppose n > 8 and nr(n) < 8n. Then, applying our various lemmas, we
have (2n - 1)r(2n-1) < (2n)r(2n) = 2J(2n)nr(2n) =- 2(22)nr(2n)-r(n)nJ(n) < 2n-2(2n)8n
< (2n-2)(4n/2)(8n) = 82n-1 < 82n.Hence the inequality holds for 2n - 1 and 2n and
U
the theorem follows.
COROLLARY. (CHEBYCHEV, 1852) There is a constant B so that for every posi-
tive integer n > 1,
In(n)
Proof n(n1) < 8n. Taking natural logarithms yields w(n) In(n) < n ln(8) and the
U
inequality is established with B = ln(8).
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2005
We see the potentialvalue of this proof as twofold.First, it appearscleanerand
shorterthan what is found in most texts. And our constant,ln(8), is modest comparedto Sierpinski's4 [7], Apostol's6 [1], or the 32 ln(2) offeredin earliereditionsof
Niven andZuckerman[6]. LeVeque[5], HardyandWright[4], and the latestedition
of Niven andZuckerman[6] give no particularconstant,merelyprovingthatone exists. Chebychev[3] achieveda muchsmallerconstantthanours,butwithconsiderably
moreeffort.Wehopethatourshortproofwill be foundto havepedagogicalvalue.
REFERENCES
1. Tom M. Apostol, Introductionto AnalyticNumberTheory,Springer-Verlag,New York, 1976.
2. BelaBollobas,Erd6sfirstpaper:A proofof Bertrand's
1998.
Postulate,
pre-print,
3. P.L.Chebychev,Memoiresur les nombrespremiers,J. Math.Pures et Appl. 17, (1852), 366-390.
4. G.H.HardyandE.M.Wright,AnIntroduction
totheTheoryofNumbers,
5thEdition,Clarendon
Press,Oxford,
1979.
5. William J. LeVeque,Fundamentalsof NumberTheory,Dover Publications,New York, 1977.
AnIntroduction
6. I. NivenandH.S.Zuckerman,
to theTheoryof Numbers,3rd,4th,5thEditions,Wiley,New
York,1972,1980,1991.
7. Waclaw Sierpinski,ElementaryTheoryof Numbers,PWN, Warsaw,1964.
Recountingthe Odds
of an EvenDerangement
ARTHUR T. BENJAMIN
HarveyMuddCollege
Claremont,CA91711
[email protected]
CURTIS T. BENNETT
LoyolaMarymountUniversity
LosAngeles,CA90045
[email protected]
FLORENCE NEWBERGER
California State University
LongBeach,CA90840-1001
[email protected]
Oddas it maysound,whenn examsarerandomlyreturnedto n students,the probability thatno studentreceiveshis or herown examis almostexactly 1/e (approximately
0.368), for all n > 4. We call a permutationwith no fixedpoints,a derangement,and
we let D(n) denotethe numberof derangementsof n elements.For n > 1, it can be
shownthat D(n) = -_,(-1)kn!/k!, andhence the odds thata randompermutation
of n elementshas no fixedpointsis D(n)/n!, whichis within 1/(n + 1)! of 1/e [1].
Permutationscome in two varieties:even and odd. A permutationis even if it can
be achievedby makingan even numberof swaps;otherwiseit is odd.Thus,one might
evenbe interestedto knowthatif we let E (n) andO(n) respectivelydenotethe number
of even andoddderangementsof n elements,then(oddlyenough),
E (n) =
D(n) + (n - 1)(-1)n-1
2