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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. –  Thomas Andrews Jul 25 '12 at 5:02

1 Answer 1

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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This is nice, but I would have written the chunk between "Now, we have to..." and "You have to be careful" in the opposite order, without the question marks, and at the very bottom, under the last bit. –  user22805 Jul 25 '12 at 8:40
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 help –  user979616 Jul 25 '12 at 22:06

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