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prove the following statement:

$\forall x \in \mathbb{R}, \lfloor x-1 \rfloor = \lfloor x \rfloor - 1$

suppose $x \in \mathbb{Z}$, then $\lfloor x-1 \rfloor = x-1 $ and $ \lfloor x \rfloor -1 = x-1 $ since the floor of any integer is itself.

suppose $x \in \mathbb{R} $, then $\lfloor x-1 \rfloor$ will give an integer that is also given when taking $\lfloor x \rfloor -1$.

eg. $\lfloor 1.5-1 \rfloor = \lfloor .5\rfloor = 0 = \lfloor 1.5 \rfloor - 1$

i think ive almost got this proof correct but something about it just doesnt seem quite right. Can someone please help me verify?

-thanks

kr1s
  • 311

3 Answers3

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Good start.

You have shown that for $x \in \mathbb{Z}$, $\lfloor x-1 \rfloor = x-1 $ and $ \lfloor x \rfloor -1 = x-1 $.

For $x \in \mathbb{R} $, let $x=y+\delta$, where $y \in \mathbb{Z}$ and $0 \le \delta <1$.

Then $\lfloor x-1 \rfloor = \lfloor y+ \delta -1 \rfloor = \lfloor y-1 + \delta \rfloor = y-1$.

and $\lfloor x \rfloor-1 = \lfloor y+ \delta \rfloor -1= y-1$.

tomi
  • 9,594
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You have simply restated what you want to prove.

By definition, $n=\lfloor y\rfloor$ if and only if $n$ is an integer and $n\le y<n+1$.

Let $m=\lfloor x\rfloor$. Then $m$ is an integer and $m\le x<m+1$, so $m-1$ is an integer and $$ m-1\le x-1<m $$ Therefore $m-1=\lfloor x-1\rfloor$.

More generally, the same argument shows that if $k$ is an integer, then for every real $x$ we have $$ \lfloor x+k\rfloor = \lfloor x\rfloor + k $$

egreg
  • 238,574
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Just label things.

Let $[x] = n$. that means $n$ is the unique integer $n \le x < n+1$. (Now, technically, we have proven that there is such an integer nor that it is unique but... I think we can take that as given. It was certainly assumed when it was declared that $[x]$ was a well-defined function.)

Let $[x - 1] = k$. That means $k \le x - 1 < k+1$ where $k$ is an integer.

Then $k + 1 \le x < k + 2$. But that would mean $[x] = k+ 1 =n$ so $k = [x-1] = n-1 = [x]-1$.

fleablood
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