Advanced book on Mathematics Olympiad

Algebra 451

For an example of such a binary operation consider any setSendowed with the

operationa∗b=afor anya, b∈S.

276.Using the first law we can write

y∗(x∗y)=(x∗(x∗y))∗(x∗y).

Now using the second law, we see that this is equal tox. Hencey∗(x∗y)=x.

Composing withyon the right and using the first law, we obtain

y∗x=y∗(y∗(x∗y))=x∗y.

This proves commutativity.

For the second part, the setSof all integers endowed with the operationx∗y=−x−y

provides a counterexample. Indeed,

x∗(x∗y)=−x−(x∗y)=−x−(−x−y)=y

and

(y∗x)∗x=−(y∗x)−x=−(−y−x)−x=y.

Also,( 1 ∗ 2 )∗ 3 = 0 and 1∗( 2 ∗ 3 )=4, showing that the operation is not associative.
(33rd W.L. Putnam Mathematical Competition, 1972)
277.Definer(x)= 0 ∗x,x∈Q. First, note that

x∗(x+y)=( 0 +x)∗(y+x)= 0 ∗y+x=r(y)+x.

In particular, fory=0 we obtainx∗x=r( 0 )+x= 0 ∗ 0 +x=x.

We will now prove a multiplicative property ofr(x), namely thatr(mnx)=mnr(x)

for any positive integersmandn. To this end, let us show by induction that for allyand

all positive integersn,0∗y∗···∗ny=nr(y). Forn=0 we have 0= 0 ·r(y), and

forn=1 this follows from the definition ofr(y). Assume that the property is true for

k≤nand let us show that it is true forn+1. We have

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

=( 0 ∗y∗···∗ny )∗(ny∗(n+ 1 )y)

=(n( 0 ∗y))∗(( 0 +ny )∗(y+ny ))

=( 0 ∗y+(n− 1 )( 0 ∗ y))∗( 0 ∗y+ny )

=(n− 1 )r(y)∗ny+ 0 ∗y.

Using the induction hypothesis,(n− 1 )r(y)∗ny= 0 ∗y∗···∗(n− 1 )y∗ny=nr(y)

(this works even whenn=1). Hence 0∗y∗···∗(n+ 1 )y=nr(y)+r(y)=(n+ 1 )r(y),

which proves the claim.