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$ \left\lfloor { - x} \right\rfloor = - \left\lfloor x \right\rfloor $ if $x \in \mathbb{Z}$ and

$ \left\lfloor { - x}\right\rfloor = - \left\lfloor x \right\rfloor -1 $ otherwise

This is an exercise from Tom Apostol's book "Calculus Volume I" section 1.11 number 4. He defined $\left\lfloor x \right\rfloor$ as the greatest integer $\leqslant x$.

I have tried it but I don't get it. Could you help me?.

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  • $\begingroup$ Direct from the definition. $\endgroup$ – OR. Jul 18 '13 at 22:06
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Use that you define $[x]$ such that $[x]\leq x<[x]+1$. Now replace this into all the statements you want to prove.

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  • $\begingroup$ Thanks. I was able to prove that with your definiton. However I think, because Apostol's definition is "the greatest integer $\leqslant x$", I must prove $\left[ x \right] \leqslant x < \left[ x \right] + 1$ as a consequence of his definition, and that $\left[ x \right]$ is unique. $\endgroup$ – ILikeMath Jul 19 '13 at 18:25
  • $\begingroup$ I think saying that $[x]$ is an integer such that $[x]\leq x <[x]+1$ is saying that it is the greatest integer $\leq x$. This is because the next integer is $[x]+1$ and you are saying that this is already strictly greater than $x$. $\endgroup$ – OR. Jul 19 '13 at 23:34
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Let, $x = y+z$ where y is an integer number and $1>z>0$. So $y = \left\lfloor x\right\rfloor$. multiplying both sides of $x = y+z$ with -1, we get,

$-x=-y-z$

So $-x<-y$ and the difference between $-y$ and $-x$ is $-z$ where, $-1<z<0$ So, $\left\lfloor -z\right\rfloor=-1$. $ \left\lfloor { - x} \right\rfloor = -y-1= - \left\lfloor x \right\rfloor-1 $ as $\left\lfloor {x} \right\rfloor = y$

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Lets make 2 cases:
CASE 1 :**$ \left\lfloor { - x} \right\rfloor = - \left\lfloor x \right\rfloor $ if $x \in \mathbb{Z}$
Whenever we have an Integer inside this function it remains unchanged so bringing the minus sign out is completely legal.
**CASE 2:
$ \left\lfloor { - x}\right\rfloor = - \left\lfloor x \right\rfloor -1 $ otherwise.
We know [-x] = -x - {-x} where{.} denotes the fractional part of x
Take minus common in RHS to get [-x] = -(x+{-x})
Now we know that {-x} = 1-{x} when x is not an integer (By the very nature of{.})
Now substitute this to the above equation to get the desired result.

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  • $\begingroup$ I think the statement {-x} = 1-{x} should be proved. It does not seem different than writing without proof $\left\lfloor { - x}\right\rfloor = - \left\lfloor x \right\rfloor -1$ by the very nature of the floor function. $\endgroup$ – Taladris Jul 19 '13 at 5:04

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