Pregled bibliografske jedinice broj: 277683
On explicit bijection between nonnegative integers and rationals and Knuth's problem
On explicit bijection between nonnegative integers and rationals and Knuth's problem // International Congress of Mathematicians 2002
Peking, Kina, 2002. (predavanje, međunarodna recenzija, sažetak, znanstveni)
CROSBI ID: 277683 Za ispravke kontaktirajte CROSBI podršku putem web obrasca
Naslov
On explicit bijection between nonnegative integers and rationals and Knuth's problem
Autori
Svrtan, Dragutin
Vrsta, podvrsta i kategorija rada
Sažeci sa skupova, sažetak, znanstveni
Skup
International Congress of Mathematicians 2002
Mjesto i datum
Peking, Kina, 20.08.2002. - 28.08.2002
Vrsta sudjelovanja
Predavanje
Vrsta recenzije
Međunarodna recenzija
Ključne riječi
bijection; rationals; Knuth's problem; sawtooth function
Sažetak
We show that a sequence, defined by $q_0=0$, $q_{;2m};={;q_m\over 1+q_m};$, $m>0$, and $q_{;2m+1};=1+q_m$ generates (without repetitions) all nonnegative rational numbers. A shorter recurrence is $1/q_{;n};=\lfloor q_{;n-1};\rfloor +1-\{;q_{;n-1};\};$ ). Then $\lfloor q_{;n-1};\rfloor = \varepsilon_2(n)$ ($=k$ if $n$ is divisible by $2^k$ but not by $2^{;k+1};$) leads to a solution of a recent Knuth's problem. A bijective discontinuous function $h:\R^+_0\rightarrow\R^+$, $h(x)=1/(2\lfloor x\rfloor +1-x)$ is discovered, which realizes the sequence $(q_n)_{;n\in\N_0};$ as one injective trajectory (through $0$) since $(\underbrace{;h\circ h\circ\cdots\circ h};_m)(0)=h^m(0)=q_m$ ($m\geq 0$). It would be interesting to explore other trajectories of $h$ in various number fields. In terms of the {;\it sawtooth}; function ((x)) one can write $1/h(x)=x-2((x))+\delta(x)$. The inverse of $h$ is given by $h^{;-1};(x)=1/x-2((1/x))-\delta (1/x)$. These results and some other conjectures are done jointly with I.Urbiha
Izvorni jezik
Engleski
Znanstvena područja
Matematika
POVEZANOST RADA
Projekti:
0037117
Ustanove:
Prirodoslovno-matematički fakultet, Matematički odjel, Zagreb