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

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


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). 


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 

+ 1! h 2! + 3! + 

Multiplying this successively by arbitrary factors, we obtain the equations called 

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 


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


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 (*) 


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 


+ 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 


& 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) 


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 


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. 


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.