Proof that $\left\lfloor -x\right\rfloor =-\left\lceil x\right\rceil$

Question

I wanted to ask if this kind of reasoning for proving the result in the title could be considered correct:

We know that: $\left\lceil x\right\rceil =n$ if and only if $n-1<x\leq n$

Then $-\left\lceil x\right\rceil =-n $ if and only if $-n-1<x\leq-n$

Then multiplying by $-1$ the formula $-n-1<x\leq-n$ we get $n+1>-x\geq n$ , inverting the sign.

But $n+1>-x\geq n$ is equivalent to $n\leq-x<n+1$.

We know that $\left\lfloor x\right\rfloor = n$ if and only if $ n\leq x<n+1$.

So from $n\leq-x<n+1$ we can infer that $\left\lfloor -x\right\rfloor =$ $n=-\left\lceil x\right\rceil$

Answer 1

Your reasoning is quite involved, I think. Try to use the definitions of floor and ceiling directly instead.

By definition, $\lfloor y \rfloor = k$ if $k$ is the greatest integer such that $k \leq y$, and $\lceil y \rceil = k$ if $k$ is the least integer such that $y \leq k$. Moreover, we know that $x_1 \leq x_2$ if and only if $-x_2 < -x_1$.

So if $\lfloor -x \rfloor = k$, then $k \leq -x$ and therefore $x \leq -k$. Moreover, $-k$ must be the least such integer, since $k$ is the greatest integer such that $k \leq -x$.

Answer 2

From the definitions, $$\lfloor x\rfloor=-n\iff -n-1<x\le-n$$ and $$\lfloor-x\rfloor=n\iff n\le-x<n+1.$$

These are equivalent statements.

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Alternatively, WLOG the integer part is $0$ (because the floor/ceiling commute with addition of an integer), and the fractional part is $0$ or $\in (0,1)$, say $0.5$.

We have $$\lfloor-0\rfloor=-0=-\lceil-0\rceil$$

and

$$\lfloor-0.5\rfloor=-1=-\lceil-0.5\rceil.$$

Answer 3

Go straight at it, but make careful and clear steps: $$ \begin{align} \lfloor -x\rfloor =n &\iff n\leq -x<n+1\\ &\iff -n\geq x>-n-1 \end{align} $$ and $$ \begin{align} -\lceil x\rceil=n&\iff \lceil x\rceil=-n\\ &\iff -n-1 < x \leq -n \end{align} $$