Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$[x]-2[x/2]\leq 1$$ Equivalently, $[x]-2[x/2]$ assumes only the values 0 and 1.

It seems easy, but I don't know how to prove it...

share|improve this question
    
If $x$ is an integer, then your inequality is the same as saying $x \bmod 2 \leq 1$ –  J. M. Nov 21 '11 at 14:35

3 Answers 3

up vote 5 down vote accepted

Let $f(x)=[x]-2[x/2]$.

Then $$f(x+2)= [x+2]-2[(x+2)/2]=[x]+2-2[x/2+1]=[x]+2-2[x/2]-2=f(x) \,.$$

Thus $f$ is periodic with period $2$. It suffices to prove that $f(x) \leq1$ for all $x \in [0,2)$. But this is obvious since then $[x] \leq 1, [x/2] \geq 0$.

share|improve this answer
    
+1 Very neat. :-) –  Srivatsan Nov 21 '11 at 16:41

Say $x=n+r$, with $0\leq r\lt 1$, and $n\in\mathbb{Z}$.

If $n=2k$ is even, then $\frac{x}{2} = k+(r/2)$, $0\leq (r/2)\lt 1$, so $[x]=2k$, $[x/2]=k$, and $[x]-2[x/2] = 2k-2k = 0$.

If $n=2k+1$ is odd, then $x/2 = k+(0.5+(r/2))$. Since $0\leq r\lt 1$, then $0\leq \frac{r}{2}\lt 0.5$, so $[x/2] = k$. Hence $[x]-2[x/2] = 2k+1 - 2k = 1$.

share|improve this answer
    
It should be $n=2k+1$ is odd? –  Kou Nov 21 '11 at 14:31
    
@Kou: Indeed; fixed. Thanks! –  Arturo Magidin Nov 21 '11 at 14:33

Let $y = \frac{x}{2}$ i.e. $x = 2y$. We then have $y = [y] + r$, where $0 \leq r <1$ and $[y] \in \mathbb{Z}$.

If $0 \leq r < \frac12$, then $0 \leq 2r<1$ and $2y = 2[y] + 2r$, where $2[y] \in \mathbb{Z}$.

Hence, $[2y] = 2[y] \implies [x] = 2 \left[\frac{x}2\right]$.

If $\frac12 \leq r < 1$, then $0 \leq 2r-1 <1$ and $2y = 2[y] + 1 + 2r-1$, where $2[y] + 1 \in \mathbb{Z}$.

Hence, $[2y] = 2[y] + 1 \implies [x] = 2 \left[\frac{x}2\right] + 1$.

Hence, $[x]- 2 \left[ \frac{x}2\right] \in \{0,1\}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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