MHB Problem of the Week # 210 - April 5, 2016

[Ackbach]

Here is this week's POTW:

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Let $S$ be a set of real numbers which is closed under multiplication (that is, if $a$ and $b$ are in $S$, then so is $ab$). Let $T$ and $U$ be disjoint subsets of $S$ whose union is $S$. Given that the product of any three (not necessarily distinct) elements of $T$ is in $T$ and that the product of any three elements of $U$ is in $U$, show that at least one of the two subsets $T,U$ is closed under multiplication.

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[Ackbach]

Re: Problem Of The Week # 210 - April 5, 2016

This was Problem A-1 in the 1995 William Lowell Putnam Mathematical Competition.

Congratulations to Fallen Angel and Deveno for their correct solutions! Here is Fallen Angel's solution:

Spoiler

Let's prove it by contradiction.

Suppose neither $U$ nor $T$ is closed under multiplication; i.e., there exists $a,b\in T$ such that $n=ab\in U$ and $x,y\in U$ such that $m=xy\in T$.

Now $nxy\in U$ because they all are in $U$, and similarly $abm\in T$. But $abm=nxy,$ so $U$ and $T$ are not disjoint, contrary to our assumption.