Advanced book on Mathematics Olympiad

436 Algebra

is irrational for any nonnegative rational numbersa 1 ,a 2 ,...,annot all equal to zero.
Denote the other elements ofSbyxn,xn+ 1 ,...,x 2 n− 1 and assume that the property does
not hold forn. Then for eachk= 0 , 1 ,...,n−1 we can find rational numbersrksuch that
(n− 1
∑

i= 1

bikxi

)

+ckxn+k=rk

withbik,cksome nonnegative integers, not all equal to zero. Because linear combinations
of thexi’s,i= 1 , 2 ,...,n−1, with nonnegative coefficients are irrational, it follows
thatckcannot be equal to zero. Dividing by the appropriate numbers if necessary, we
may assume that for allk,ck=1. We can writexn+k=rk−

∑n− 1
i= 1 bikxi. Note that the
irrationality ofxn+kimplies in addition that for a fixedk, not all thebik’s are zero.
Also, for thennumbersxn,xn+ 1 ,...,x 2 n− 1 , we can find nonnegative rationals
d 1 ,d 2 ,...,dn, not all equal to zero, such that

n∑− 1

k= 0

dkxn+k=r,

for some rational numberr. Replacing eachxn+kby the formula found above, we obtain

∑n−^1

k= 0

dk

(

−

∑n−^1

i= 1

bikxi+rk

)

=r.

It follows that

∑n−^1

i= 1

(n− 1

∑

k= 0

dkbik

)

xi

is rational. Note that there exists a nonzerodk, and for that particularkalso a nonzero
bik. We found a linear combination ofx 1 ,x 2 ,...,xn− 1 with coefficients that are positive,
rational, and not all equal to zero, which is a rational number. This is a contradiction.
The conclusion follows.

Second solution: LetVbe the span of 1,x 1 ,x 2 ,...,x 2 n− 1 overQ. ThenVis a finite-
dimensionalQ-vector space insideR. Choose aQ-linear functionf :V →Qsuch
thatf( 1 )=0 andf(xi) =0. Such anfexists since the space of linear functions with
f( 1 )=0 has dimension dimV−1 and the space of functions that vanish on 1 andxi
has dimension dimV−2, and becauseQis infinite, you cannot cover anm-dimensional
vector space with finitely many(m− 1 )-dimensional subspaces. By the pigeonhole
principle there arenof thexifor whichf(xi)has the same sign. Sincef(r)=0 for all
rationalr, no linear combination of thesenwith positive coefficients can be rational.
(second solution by R. Stong)