Prove that the set of real numbers can be partitioned in (disjoint) sets of two elements each.
Problem
Source: Romanian JBMO TST 2006, Day 3, Problem 4
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JBL
16.05.2006 22:04
This is cute. The problem with the most obvious ways of doing it is that if you have "too much continuity" when you assign points to each other, you end up assigning a point to itself. I'd like to see if other people have drastically different solutions than I.
The easiest way I found to get around this problem is just to use an idea inspired by the floor function. For each integer $n$ and each real $\varepsilon, 0 \leq \varepsilon < \frac12$ make a set of $n + \varepsilon$ and $n + \varepsilon + \frac12$.
Valentin Vornicu
16.05.2006 22:40
I have one (which took about .36 seconds to cook up )
For all non-integers $x$ we consider the set $\{x,-x\}$. For integers we consider the sets $\{2k,2k+1\}$, for all integers $k$
perfect_radio
17.05.2006 02:27
Actually JBL's idea is very obvious. I found a solution in $3.6 \, - \, 36$ seconds or so ( ) :
I first thought of pairing $x$ with $x+1$, but that didn't work. Now it is "obvious" that we should write $\mathbb R = \bigcup_{k \in \mathbb Z} \left[ 2k,2k+2 \right)$. If an element $a$ is in $A_k = \left[ 2k,2k+2 \right)$, then we put it in a set with $a \pm 1 \in A_k$.
Of course, this is the same as JBL's solution. But I wanted to post here, because I don't see too often TST problems (even for juniors) that can be solved only in mind.
JBL
17.05.2006 02:54
Valentin Vornicu wrote: I have one (which took about .36 seconds to cook up )
For all non-integers $x$ we consider the set $\{x,-x\}$. For integers we consider the sets $\{2k,2k+1\}$, for all integers $k$
Okay, nice. I was expecting there to be something like this. Actually, the problem this made me think of is to construct a bijection between the intervals $[0, 1)$ and $(0, 1)$ (or equivalently between the points of a line and of a circle), and your solution is much more like the simplest solution of that problem than mine.
Cristi Sav
26.05.2006 15:46
My idea is to invert the reals line into a circle and to form the sets of diametral oposite points. Is it right ?
perfect_radio
26.05.2006 22:32
Does it really work? If I try to make a bijection circle - real line, I'm leaving a point uncovered.
hsgser
23.05.2014 12:59
i think {x;-x+1} for all x.