Ceiling of any Real Number Proof

Question

Prove or disprove:

For any real number $x$, if $⌈x⌉ − x ≥ 1/2 $ then $ ⌈2x⌉ = 2⌈x⌉ − 1$.

How would one solve this question, because i cant seem to solve it just as is, am i missing something?

Answer 1

Note that for any $x$, we have $\lceil x\rceil$ is the unique integer such that $x+1 > \lceil x \rceil \geq x$.

Note that if $\lceil x \rceil \geq x+\frac 12$ then $2\lceil x \rceil \geq 2x+1$. On the other hand, of course it is true that $2\lceil x \rceil < 2x+2$, so $2x+1 \leq 2\lceil x \rceil \leq 2x+2$, hence it follows that $2\lceil x \rceil = \lceil 2x+1\rceil$.

Now , $2x+1 \leq \lceil 2x+1 \rceil < 2x+2 \implies 2x \leq \lceil 2x+1 \rceil -1 < 2x + 1$, and therefore it follows that $\lceil 2x+1\rceil - 1 = \lceil 2x\rceil$. Therefore, by the equality from the previous paragraph, $2\lceil x \rceil - 1 = \lceil 2x\rceil$.

Answer 2

$⌈x⌉ − x ≥ 1/2 $

Let $\lceil x \rceil = n $, so $n-1 < x \le n$.

We are given $n-x \ge \frac12$, so that $n-\frac12 \ge x$.

If $m = \lceil 2x \rceil$, then $m-1 < 2x \le m$.

We want to show that $m = 2n-1$.

We have $n-1 < x \le n-\frac12$, so $2n-2 < 2x \le 2n-1$.

Since $m-1 < 2x \le m$, $m-1 < 2n-1$ and $2n-2 < m$ so that $m < 2n$ and $m > 2n-2$.

Since $m$ and $n$ are integers, these imply $m \le 2n-1$ and $m \ge 2n-1$ so that $m = 2n-1$ as desired.

Note that the basic inequality for integers is that $a > b$ implies $a-1 \ge b$.

Answer 3

Let $\lceil x\rceil=:n$. Then $\lceil x\rceil-x\geq{1\over2}$ implies $n-1<x\leq n-{1\over2}$, hence $2n-2<2x\leq 2n-1$, as claimed.