# Full text of "The Transcendence of |pi and e"

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THE

AMERICAN
MATHEMATICAL MONTHLY.

Entered at the Post-office at Springfield, Missouri, as second-class matter.

Vol. XI. DECEMBER, 1904. No. 12.

THE TRANSCENDENCE OF * AND e.

By DB. OSWALD VEBLEN, The University oi Chicago.

§1. The proof that * is a transcendental number is ordinarily arranged as
follows. If it should satisfy any algebraic equation, so would tt.j/— 1. But it
is well known that

e^-i^-l (A).

Hence if *. j/— 1 is one of the m roots «, , z 2 , , z m , of an algebraic equation,

we must have

(e*i+l)(e*«+ 1) (e*»+l)=0 (£)

since one of its factors is zero. On expanding (B) we obtain

c+e*i+e x '+ +e«»=0 (0)

where x lt x t , , x n are the n roots of an algebraic equation and where c is a

whole number not zero. The rest of the argument consists in showing that
equation (0) is impossible.

The proof* that ( 0) is impossible is so difficult for most students that it

*The principal references in English on the subject of the transcendence of ir and e seem to be the
translation by W. W. Beman of the chapter on Transcendental Numbers in Weber's Algebra published in
the Bulletin of the American Mathematical Society, Vol. 3 (1897), p. 174, and the translation by Beman and
Smith of Klein's Famous Problems of Elementary Geometry (Ginn & Co., Boston). A good elementary
treatment in the German language Is that by Weber and Wellstein, Encyclopadie der Elementarmathema-
tile, Vol. I, pp. 418-482. (B. G. Teubner, Leipzig).

220

seems worth while to publish the simplified arrangement of the argument that is
given below. The simplification consists in leaving out one factor ordinarily
multiplied into the function <£(#) and in the device of adding together the terms
of equation (3) first by diagonals and then by columns.
§2. Our task is to show that

c+e x i+e x > + +e x » (1)

cannot be zero if c is an integer not zero and x lt x 2 , , x n are the roots

of an equation

f(x)=a +a i x + a i x* + +a„x»=0 (2)

with integral coefficients, a ?±0, a n7 ±0.

The scheme of proof is to find a number N such that when we multiply it
into (1) the resulting expression becomes equal to a whole number plus a quan-
tity numerically less than unity, a sum which surely cannot be zero. To find
this multiplier N, we study the series for e xk where x k is any one of the roots of
/(*)=0.

+ 1! h 2! + 3! +

Multiplying this successively by arbitrary factors, we obtain the equations called
(3):

e-.l l.b 1= b t .1 ! +&x**(l+ -^ + Jj+ )

«-.2!.6,=6 i .2!(l + ^) + M» , (l+ T + + }

e**.Zl.b s =b a .d\(l+^+ X ^)thz k \l+% + X £ 5 + )

-■.i-^MKl+^ + ^-F ^)+W(l+^ 1 + ^ ! | w + )

Now bj , , b, can be regarded as coefficients of an arbitrary polynomial

<Kx)=b 4-b 1 x+b 2 x"- + +b e z<.

Differentiating, we have

*'(*)=&, +b. 2 .2.x+ + b a .s.x-\

and in general

221

^("•)( a; )=5 TO .m!+5 m+1 .^±lI ! a:+ +6 ,.^_£-L^

a—m

If we add together the equations (3), we evidently obtain as the sum of the terms

£ 8-1

in the main diagonal, from 5 1 1! to &,.s!. , the polynomial <t>'(x k ); as the

sum of the terms in the next lower diagonal <j>"(z k ), etc. We therefore have

e-»(l!& 1 + 2!6, + .-4«!6,)=*'(**)+*"(**)+--+* w (**)+ ^ &m** m -«*m (*)

m=l

in which B km =l-\ ^r+ —. — —^-. — j-cvr +

m m +1 (w4-l)(>»+2)

Suppose now that <K#), which is perfectly arbitrary, be chosen as below so that
^(**)=o, *"(**)=o, , ^*- 1 )(**)=0,

for every #*, p<s. By returning to the arrangement of (3) and leaving out the
terms due to </>'(%*), > ^ ip ~ 1 K x k)> we could then rewrite (4) in the form

«"»(! !&, +2 !6 4 + +s !&,)= *S l m x k m K km

m—l

+ Vi> !

+6 p+1 .( J) + l)!(l+^. | )

+ 5 p _ 2 .(p+2)!(l+^ l + ^ 7 )+

1 r 2 ! '

+M!(l + ^+f- 1+ +^ T ). (5).

A choice of ^(a:) that satisfies the conditions just required is

of which every x A is ajj-tuple root, by (2). Herep is still perfectly arbitrary,
but s=np+p— 1, the degree of 0(<c). Expanding ^(a;), we find on account of
the factor zp- 1

222

& o =0, l> t =0, , V-2=0,

I„ , I.

J, "> J, -"-P 7,

u p— 1 ( m IN I) "j) /•„ 1\ I) ! °8

(p-1)!' "» (j>-l)!' ' f_ (p-l) !'

where ip, , J„ are all integers.

Now the coefficient of e xh in (5) evidently becomes

**=** + iA-r pl +(p^i^+ 1 > , + + arnfi-

If the arbitrary p is taken as a prime number greater than a , this expression is
the sum of a *, which cannot contain p as a factor, plus a number of other inte-
gers each of which does contain the factor p. W p is therefore not zero and not
divisible by p.

Further, since (j>+&) !-h[(.P— 1) 1 &!] is an integer divisible byp, it fol-
lows that all of the coefficients of the last block of terms in (5) contain p as a
factor. On adding the columns of (5) we have :

N p e^=p\_P +P 1 as t +P t *»« + +P.- P («*) 8 - p ]+ 2 b m ^fi tm , (6)

m=l

where P , P i , , P,_fc are integers.

Before completing our argument we need only to show that by choosing
as p a prime number sufficiently large, the last term of (6) can be made as small
as we please. If a is a number greater than unity and greater than any of the
n roots x k of /(a;),

I *- I = I ! *sq4+ o.+l?(»+2) + l<{1 + TT + ^T + ' •

/. I Rkm I <e a •

Now since the coefficients b m in (6) are the coefficients of <j>(x) and since
each coefficient of <£(#) is numerically less than or equal to the corresponding
coefficient of

(P-l)l

we have the inequality, <? denoting a constant,

g n p -l /• ov

|JS b m x h m B km | <e* • (p _ 1) | ( I «o I + I «i I «+ + I «» I a Y < (p_i) i -

The last expression, designated Sp, is the pth term of the series for Qe® and
therefore approaches zero as p is increased indefinitely.

223

We now choose the arbitrary prime number p~>\ so that it shall be larger
that a , larger than G, and also so that S i ,<l/». The number N P is the required
multiplier 2f.

For if we multiply lf p into (1) in follows directly from equation (6) that

JT P ( O+t* +e» + + «"■) =N,O+pCP +P 1 Si +P 2 S s +

+P.-.p&_ p ) +r, +r 2 + + r„ (7)

where r k = 3 l m (x k )™R km <l/n, S t =sfi 1 +a? 4 + -fa?„.

But from Newton's formulas*

S I +o 1 =0, « 4 +a l «i+2a 8 =0,

it follows that #, , # 2 , , 8 a - p are whole numbers. Hence the second term

of the right-hand member of (7) is an integer divisible by p. On the contrary,
2f p and G are not divisible by p. The sum of these terms therefore is a whole

number greater than +1 or less than — 1 ; and since the sum r x +r 2 + +r„

is less than unity the right-hand member of (7) cannot be zero. Hence the left-
hand member of (7) is not zero and hence (1) cannot be zero.

§3. The proof that e is a transcendental number can be effected by almost
precisely the same argument as that given above. It is required to show that
the algebraic equation with integral coefficients

c+c^+c^-t + c„e»=0 (1')

is impossible. Evidently no generality is lost by assuming c^O and c n5 *sO. Let

f(x)=(x-l)(x-2) <*_n)=a +a 1 *+a i *» + -f «ge». (2')

The argument now is exactly like that of §2 from equation (2) to the sentence
introducing equation (7). At this point we observe that since all the roots of
/(«) are integers, (6) may be written

N p e xk =pW k + r k ,

where W k is a whole number and r k is less than 1/n. We therefore have

2¥ p (c + c 1 e+._.4c„e'')-(!J p +KF 1 + F J + + W n )+r i + r2 + +*«. (7')

In the right-hand member, the first term is not divisible by p, the second term
is divisible by p and the third term is numerically less than unity. From this
it follows as before that the left-hand member of (7') cannot be zero and hence
that (1') is impossible. Therefore e cannot satisfy an algebraic equation.

*Cf. Burnside and Panton, Theory of Equation*, Chapter VIII, or any book on higher algebra.

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