Advanced book on Mathematics Olympiad

452 Algebra

Using this and the associativity and commutativity of∗, we obtain

2 nr(y)= 0 ∗y∗ 2 y∗···∗ 2 ny

=( 0 ∗ny )∗(y∗(n+ 1 )y)∗( 2 y∗(n+ 2 )y)∗···∗(ny∗ 2 ny )

=r(ny)∗(y∗(y+ny ))∗( 2 y∗( 2 y+ny ))∗···∗(ny∗(ny+ny)).

The first formula we have proved implies that this is equal to

( 0 +r(ny))∗(y+r(ny))∗···∗(ny+r(ny)).

The distributive-like property of∗allows us to transform this into

( 0 ∗y∗ 2 y∗···∗ny )+r(ny)=nr(y)+r(ny).

Hence 2nr(y) = nr(y)+r(ny) ,orr(ny)=nr(y). Replacingybyxn, we obtain

r(xn)=n^1 r(x), and hencer(mnx)=mnr(x), as desired.

Next, note thatr◦r=r; henceris the identity function on its image. Also,

r(z)= 0 ∗z=(−z+z)∗( 0 +z)=(−z)∗ 0 +z=r(−z)+z,

orr(z)−r(−z)=z. Hence forz =0, one ofr(z)andr(−z)is nonzero. Letybe this
number. Sincer(y)=y, we havey=r(y)−r(−y)=y−r(−y),sor(−y)=0.
Also, ifx=mny, thenr(x)=mnr(y)=mny=x, andr(−x)=mnr(−y)=0. Ify>0,
thenr(y)=max(y, 0 )and consequentlyr(x)=x=max(x, 0 ), for allx>0, while
r(x)= 0 =max(x, 0 )for allx<0. Similarly, ify<0, thenr(y)=min(y, 0 ), and then
r(x)=min(x, 0 )for allx∈Q. The general case follows from(a−b+b)∗( 0 +b)=
(a−b)∗ 0 +b.
(American Mathematical Monthly, proposed by H. Derksen, solution by J. Dawson)
278.Forx∈Gandx′its left inverse, letx′′∈Gbe the left inverse ofx′, meaning that
x′′x′=e. Then

xx′=e(xx′)=(x′′x′)(xx′)=x′′(x′x)x′=x′′(ex′)=x′′x′=e.

Sox′is also a right inverse forx. Moreover,

xe=x(x′x)=(xx′)x=ex=x,

which proves thateis both a left and right identity. It follows thatGis a group.

279.Lete∈Gbe the identity element. Setb=ein the relation from the statement.

Then

a=a∗e=(a⊥a)⊥(a⊥e)=(a⊥a)⊥a,