# On a proof involving floor

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. – Thomas Andrews Jul 25 '12 at 5:02

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 $\lfloor 2k + 2f \rfloor = 2k + 1.$