Competition problem — Number theory

Library / /341 of 520

Number theory Difficulty 6.3 National olympiad Prove it

Property 1 The function $f(x)=[x]$ is a non-decreasing function, i.e., for any $x_{1} \leqslant x_{2}$ , we have $\left[x_{1}\right] \leqslant\left[x_{2}\right]$ ; while the function $f(x)=\{x\}$ is a periodic function, with each integer $n$ being its period, i.e., $\{n+x\}=\{x\}$ .

Solution

None

Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.

Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". Here is the formatted output as requested:

None

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.