Why does $\lfloor x \rfloor \leq n \iff x < n+1$ but $\lfloor x\rfloor < n \iff x< n$?

Question

$$\lfloor x \rfloor \leq n \iff x < n+1, \\\\\\\ \lfloor x\rfloor < n \iff x <n .$$ 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.

Answer 1

If $x\geqslant n$ then $[x]\geqslant n$. Contradiction

Answer 2

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

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

Answer 3

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 $\lfloor x\rfloor<n$

Answer 4

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 $x<n+1$.

Answer 5

$\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 $x-1 \leq n \implies x \leq n+1$

Answer 6

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.