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My professor wrote this proof but i didn't understand some parts:

Suppose $x$ is a real number such that $x$ is not an integer. let $⌊x⌋=n$. By definition of floor and since $x$ is not an integer, $n<x<n+1$.
Then by multiplying $-1$ from each side of the previous equations, we get $-n>-x>-n-1$.
Note that $-n-1$ is an integer. Thus, by definition of floor, $⌊-x⌋=-n-1$.
Therefore, $⌊x⌋+⌊-x⌋=n+(-n-1)=n-n-1=-1$.

1- Why is $-1$ multiplied from each side?

2- How did we know that $-n-1$ is an integer?

3- Why is $⌊-x⌋$ equal to $-n-1$ yet we see that $-x$ is not equal to $-n-1$?

4- Why is $⌊x⌋+⌊-x⌋=n+(-n-1)=n-n-1=-1$?

keep in mind that I am still learning the basics of floor and ceiling functions. so, I would appreciate any help.

Omar
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    I suggest: take a particular value for $x$. try $x=3.1$ or $x=-9.7$, or whatever you like (so long as it isn't an integer). Go through the given argument line by line using the value you chose. That might clarify the steps for you. Also a good idea to try an integer value for $x$ to see where the proof breaks down (after all, the claim is false for integers). – lulu Aug 30 '23 at 13:20
  • This actually helped me a lot. thanks. – Omar Aug 30 '23 at 17:10

1 Answers1

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  1. We multiply by $-1$ on each side to get information about $-x$.
  2. $\lfloor x \rfloor = n$, so $n$ is an integer. Thus $-n-1$ is an integer as well.
  3. I'm not sure what this is asking; the floor of $-x$ is $-n-1$. That doesn't mean that $-x$ has to equal $-n-1$. For example, $\lfloor 3.5 \rfloor = 3$ but $3.5 \ne 3$.
  4. We know that $\lfloor x \rfloor = n$, and $\lfloor -x \rfloor = -n-1$ by the previous parts. Thus $$\lfloor x \rfloor + \lfloor -x \rfloor = n + (-n-1).$$
kipf
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  • so, for your first point: what do you mean by get more information about -x? – Omar Aug 30 '23 at 13:25
  • the third point: I was asking the reason we got ⌊−x⌋=−n−1 from −n>−x>−n−1. – Omar Aug 30 '23 at 13:35
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    @Omar The first point is that we need information about the floor of $-x$ in order to get the sum we want. And we do this (proof strategy) by going via $-x$ itself. The second point is that the floor of a real number is the greatest integer less than or equal to that number. When a real number lies between two integers which differ by one, as in this case, then the floor is the lower of those two integers. – Mark Bennet Aug 30 '23 at 14:05
  • ok. thanks for the clarification. – Omar Aug 30 '23 at 17:09