Let $x \bmod 1\in \Bbb R$ be fractional part of $x \in \Bbb R$.

For what values of $x \in \Bbb R$ is $x^2 \bmod 1 = (x \bmod 1)^2$? Naturally if $|x| < 1$ we have the equality. However, there are some numbers like $x \in 0.5r \pm \Bbb N$ where $r \in \{0, 1\}$ which satisfy this equality. What is the entire solution set?

What about for higher powers?

  • $\begingroup$ Well, in $-1< x < 1$ then $0\leqx^2 $\endgroup$
    – AD.
    Aug 18 '13 at 14:43

Written $$x=[x]+\{x\}$$ we have $$\{x^2\}=\{[x]^2+2[x]\{x\}+\{x\}^2\}=\{2[x]\{x\}+\{x\}^2\}$$

So the necessary and sufficient condition for $\{x^2\}=\{x\}^2$ is that $2[x]\{x\}$ is an integer.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.