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

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STOP Early Journal Content on JSTOR, Free to Anyone in the World This article is one of nearly 500,000 scholarly works digitized and made freely available to everyone in the world by JSTOR. Known as the Early Journal Content, this set of works include research articles, news, letters, and other writings published in more than 200 of the oldest leading academic journals. The works date from the mid-seventeenth to the early twentieth centuries. We encourage people to read and share the Early Journal Content openly and to tell others that this resource exists. People may post this content online or redistribute in any way for non-commercial purposes. Read more about Early Journal Content at http://about.jstor.org/participate-jstor/individuals/early- journal-content . JSTOR is a digital library of academic journals, books, and primary source objects. JSTOR helps people discover, use, and build upon a wide range of content through a powerful research and teaching platform, and preserves this content for future generations. JSTOR is part of ITHAKA, a not-for-profit organization that also includes Ithaka S+R and Portico. For more information about JSTOR, please contact support@jstor.org. 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.