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$$\lfloor x \rfloor \leq n \iff x < n+1, \\\\\\\ \lfloor x\rfloor < n \iff x <n .$$ These are the two inequalities given by my book. But why are they so?

Suppose $x = 2.3\quad \& \quad n = 3$, then of course $\lfloor x\rfloor < n \iff x <n$. But $\lfloor x \rfloor < n+1 \implies 2.3< 3+1$ which is true. So, why only $\lfloor x\rfloor < n \iff x <n$ but not also $\lfloor x\rfloor < n \iff x <n +1$?? Can anyone please explain me why the inequalities are what they are? What is the reason?? Isn't my approach correct? Plz help.

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    $\begingroup$ The content of these statements are in the $\iff$ (it means that the inequalities are always true or false together). You can't use just a single example to justify a statement; instead, you must think about it for all possible cases. $\endgroup$ – Michael Burr Aug 16 '15 at 18:11
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For the first,

  • $\lfloor x\rfloor \le n\Rightarrow \lfloor x\rfloor \le n\lt n+1\Rightarrow \lfloor x\rfloor \lt n+1$

  • $\lfloor x\rfloor \lt n+1\Rightarrow \lfloor x\rfloor \le n$. Note that $\lfloor x\rfloor $ is an integer less than $n+1$.

For the second,

  • $x\ge n\Rightarrow \lfloor x\rfloor \ge n$, so $\lfloor x\rfloor \lt n\Rightarrow x\lt n$.

  • $x\lt n\Rightarrow \lfloor x\rfloor \le x\lt n\Rightarrow \lfloor x\rfloor \lt n$.

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    $\begingroup$ The only complete demonstration $\endgroup$ – user261263 Aug 16 '15 at 18:32
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If $x\geqslant n$ then $[x]\geqslant n$. Contradiction

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  • $\begingroup$ Okay, why did you call it contradiction? $\endgroup$ – user142971 Aug 16 '15 at 18:08
  • $\begingroup$ We prove that if $[x]<n$ then $x<n$. Let's $x\geqslant n$ then $[x]\geqslant n$ but $[x]<n$ by condition. Converse is analogue $\endgroup$ – ZFR Aug 16 '15 at 18:14
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The first one is true, mainly because $1>x-\lfloor{x}\rfloor\geq0$ $$\lfloor x\rfloor \leq n\implies \lfloor x\rfloor +1\leq n+1 \implies \lfloor x\rfloor + \left(x - \lfloor x\rfloor\right)<n+1\implies x < n+1$$

The second one is only true when $n$ is an integer, for example $x = 0.7, n = 0.5$... $$\lfloor x\rfloor = \lfloor 0.7\rfloor = 0 < 0.5 = n$$ $$x = 0.7 \not< 0.5 = n$$ If $n$ is an integer, this is true because the floor function is always within $1$ of $x$, and the fact that $n$ is an integer means that $x$ cannot be larger than $n$ when $\lfloor x\rfloor<n$

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Let's consider the second statement. If $x<n$, then since $\lfloor x\rfloor\leq x$ (as we are rounding down) we can combine inequalities to get $\lfloor x\rfloor<n$. On the other hand (using the contrapositive) if $x\geq n$, then rounding $x$ down results in $n$, at least. Therefore, $\lfloor x\rfloor\geq n$.

For the first statement, if $x<n+1$, then by the second statement, $\lfloor x\rfloor<n+1$, but since the first integer less than $n+1$ is $n$, $\lfloor x\rfloor\leq n$. On the other hand, if $\lfloor x\rfloor\leq n<n+1$, then by the second statement, we know that $x<n+1$.

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I'm only going to give an explicit counterexample for a part of the question, namely "why $\dots$ not also $\lfloor x\rfloor<n\iff x<n+1$?" Consider $n=2$ and either $x=2$ or $x=2.5$; in either case, the right side of the proposed equivalence, $x<n+1$, is true but the left side, $\lfloor x\rfloor<n$, is false. So these two statements are not equivalent.

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  • $\begingroup$ Wait! Some confusion still lurks; if I insert 2.5, floor function gives 2 which implies 2 < 2 which is not possible & thus you said it is not equivalent to $x < n+1$. But it is neither equivalent to $x < n$ also, isn't it? $\endgroup$ – user142971 Aug 16 '15 at 19:20
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    $\begingroup$ When $x=2.5$ and $n=2$, then, as we've both said, $\lfloor x\rfloor<n$ is false. Also, $x<n$ is false for these values of $x$ and $n$. So the two statements are equivalent because they're both false. $\endgroup$ – Andreas Blass Aug 16 '15 at 19:24
  • $\begingroup$ @ Andreas Blass: Thanks, sir for the info; so $\iff$ means if one side is true, the other side must be true & if one-side is false, the other side must be false, right? So, this is what @ Michael Burr wanted to tell, I see:) $\endgroup$ – user142971 Aug 17 '15 at 2:50
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$\lfloor x \rfloor \leq x $ & $\lfloor x \rfloor \leq n $

both do not imply that any particular relation between $x$ and $n$. Since there is no strict inequality. For eg. take $x=3.6 ; n=3 $ in one case and $x=2.6 ; n=3$ in other case.

while, $ x-1 \leq \lfloor x \rfloor \leq x \; \& \; \lfloor x \rfloor \leq n $ do imply that $x-1 \leq n \implies x \leq n+1$

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