Prove that \(2\lfloor y \rfloor > y\) when \(y \geq 1\)

Problem:

Prove that \(2\lfloor y \rfloor > y\) when \(y \geq 1\).

Proof:

Let \(y = \lfloor y \rfloor + r\), then \(0 \leq r < 1\).

Besides, since \(y \geq 1\), we have \(\lfloor y \rfloor \geq 1\), and thus \(\frac{1}{\lfloor y \rfloor} \leq 1\).

From \(r < 1\),

$$ \frac{r}{\lfloor y \rfloor} < \frac{1}{\lfloor y \rfloor} \leq 1 \\ \frac{r}{\lfloor y \rfloor} + 1 = \frac{r + \lfloor y \rfloor}{\lfloor y \rfloor} < 1+1 \\ \frac{y}{\lfloor y \rfloor} < 2 \\ y < 2 \lfloor y \rfloor \\ 2 \lfloor y \rfloor > y \quad \blacksquare $$