<s>[INST] Show that $\, 0 \leq \left \lfloor{\frac{2a}{b}}\right \rfloor - 2 \left \lfloor{\frac{a}{b}}\right \rfloor \leq 1 $ How can I prove that, for $a,b \in \mathbb{Z}$ we have $$ 0 \leq \left \lfloor{\frac{2a}{b}}\right \rfloor - 2  \left \lfloor{\frac{a}{b}}\right \rfloor \leq 1 \, ? $$ Here, $\left \lfloor\,\right \rfloor$ is the floor function.  I tried the following: say that $\frac{2a}{b} = x$, and $ \left \lfloor{\frac{2a}{b}}\right \rfloor = m$, with $0 \leq x - m \leq 1$. I tried the same for $ 2 \left \lfloor{\frac{2a}{b}}\right \rfloor $, and then combining the two inequalities. It did not seem to help, though. 
 [/INST] $0 \leq \left \lfloor{\frac{2a}{b}}\right \rfloor  \leq  \frac{2*a}{b} < 2*(\left \lfloor{\frac{a}{b}}\right \rfloor + 1)$ = $2*\left \lfloor{\frac{a}{b}}\right \rfloor$ + 2
$\left \lfloor{\frac{2a}{b}}\right \rfloor$ is an integer, so being < than an other integer means being $\leq$ than this integer -1 
=> $\left \lfloor{\frac{2a}{b}}\right \rfloor$ $\leq $ $2*\left \lfloor{\frac{a}{b}}\right \rfloor$ + 1
Edit:
As for the other part of the inequality:
$$\frac{2a}{b}= 2\left \lfloor{\frac{a}{b}}\right \rfloor + 2(\frac{a}{b}) $$ with $(x)$ being the fractional part of $x$.
All that's left to consider is whether $2(\frac{a}{b})$ is smaller or greater than $1$ to conclude.
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