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I have a question in my text that asks me to prove that if $x- \lfloor x\rfloor \geq \frac{1}{2}$, then the $\lfloor 2x\rfloor=2\lfloor x\rfloor+1$.

I understand the proof up to the point where they obtain that:

2*floor of x+1 is less then or equal to 2x is less then or equal to 2*the floor of x+2. But then it simply states, by the definition of floor, the desired conclusion follows. Can anyone explain why this is in some detail? I know what floor means, but perhaps i'm just not seeing something small

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    Note that the phrase "2 times floor of x + 1" is ambiguous, meaning either $2\lfloor x+1\rfloor$ or $2\lfloor x\rfloor +1$. The problem means the latter.2012-07-25

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You might not like this proof, but anyways:

For a positive integer $x,$ the condition $x - \lfloor x \rfloor \ge \frac{1}{2},$ means that we can write $x$ as $x = k + f,$ where $k$ is a natural number and $f$ is a fraction such that $\frac{1}{2} \le f < 1. \tag{1}$ (Double check: $x - \lfloor x \rfloor = (k+f) - k = f \ge \frac{1}{2}$.)

Now, we have to show that $ \lfloor 2x \rfloor \stackrel{?}{=} 2 \lfloor x \rfloor + 1\\ \lfloor 2(k+f) \rfloor \stackrel{?}{=} 2 \lfloor k+f \rfloor + 1\\ \text{i.e. } \lfloor 2k + 2f \rfloor \stackrel{?}{=} 2 k + 1. $ You have to be careful when you reason about the LHS. From inequality $(1)$ above, we have $1 \le 2f < 2 \\ 2k + 1 \le 2k + 2f < 2k + 2.$ So $\lfloor 2k + 2f \rfloor = 2k + 1.$

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    Thank you for the help. But my question is really just why i'm able to jump from the step I outlined to the conclusion. If you could maybe help me understand floor a little better, that would help2012-07-25
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If $x- \lfloor x\rfloor \geq \frac 12$, then $ x = \lfloor x \rfloor + \epsilon$ where $\frac 12 \le \epsilon < 1 \implies 1 \le 2 \epsilon < 2 \implies \lfloor 2\epsilon \rfloor = 1$

So \begin{align} \lfloor 2x \rfloor &= \lfloor 2\lfloor x \rfloor + 2 \epsilon \rfloor \\ &= 2\lfloor x \rfloor + \lfloor 2 \epsilon \rfloor & \text{(If $n$ is an integer, then $\lfloor n + \xi \rfloor = n + \lfloor \xi \rfloor$)}\\ &=2\lfloor x\rfloor+1 \\ \end{align}

Added 12/25/2017

If $x- \lfloor x\rfloor \geq \frac{1}{2}$

Then $2x- 2\lfloor x\rfloor \geq 1$

So $2\lfloor x \rfloor + 1 \le 2x$


Since $ x - \lfloor x \rfloor < 1$

Then $2x \le 2\lfloor x \rfloor + 2$


So $2\lfloor x \rfloor + 1 \le 2x < 2\lfloor x \rfloor + 2$

Since $2\lfloor x \rfloor + 1$ is an integer, then $2\lfloor x \rfloor + 1 \le \lfloor 2x \rfloor$

Since $2\lfloor x \rfloor + 2$ is an integer, then $\lfloor 2x \rfloor < 2\lfloor x \rfloor + 2$

So $(2\lfloor x \rfloor + 1) \le \lfloor 2x \rfloor < (2\lfloor x \rfloor + 1) + 1$

By the definition of the floor function, then $2\lfloor x \rfloor + 1 = \lfloor 2x \rfloor$