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Find all real numbers $x$ satisfying:

a. $\lfloor x + 1/2 \rfloor = \lfloor x \rfloor$.

b. $\lfloor x + 1/2 \rfloor = \lceil x \rceil$.

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Begin with a few examples, such as $x = 1$, $x=1.2$, $x=1.5$, $x=1.7$, etc. Try to spot the pattern. –  Shaun Ault Aug 10 '12 at 18:59
    
First, consider a positive $x.$ Write $x$ as $k + f$ where $k$ is an integer, and $f$ is a real number between $0$ and $1$. –  user2468 Aug 11 '12 at 17:48
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3 Answers

This is homework, so you should really post what you have tried so far.

Nevertheless, I will give you the following hint: the floor function returns the integer part of any real number $x$, or in other word, it rounds every number down to the nearest integer.

So $\lfloor 5.9\rfloor = 5$ and $\lfloor 5.1\rfloor = 5$. If you added $1/2 = 0.5$ to $5.9$, would you still get $5$? What about if you added it to $5.1$?

Edit to provide further hints:

Let $5 \le x < 5.5$. Or, in other words, let $x \in [5,5.5)$. Does any value of $x$ in this set work for part (a)? Does every value of $x$ work?

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I start with let x = n + s, 0 <= s < n, n is integer then floor (x +1/2) = floor (x) that is equal to floor (n + s + 1/2) = floor (n +s) then that is same as n + floor (s + 1/2) = n + floor (s) then i cancelled n. That's the point I dunoo what to do. How can I cancel floor (s) ?? –  mth126 Aug 10 '12 at 19:11
    
You don't cancel floor. It seems like you're overcomplicating it. $\lfloor 5.1 + 0.5 \rfloor = 5 = \lfloor 5.1 \rfloor$, but $\lfloor 5.9 + 0.5 \rfloor = 6 \neq \lfloor 5.9 \rfloor$. Can you tell what the pattern is now? –  Arkamis Aug 10 '12 at 19:16
    
Wait. From my solution. floor (s) = 0. so my working solution is only floor (s +1/2) = 0. so now, how can i solve for s? –  mth126 Aug 10 '12 at 19:22
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You're not solving for $s$ in this problem. You asked to "find all real numbers satisfying...." So your solution will be a set of numbers, not a single value $x$. Do not try to "cancel out" floor operations or "solve for $x$." You are instead required to identify regions for which a pattern holds. –  Arkamis Aug 10 '12 at 19:25
    
I edited my question together with my solution. –  mth126 Aug 10 '12 at 19:33
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Given a real number $x$, write $x=n+s$, where $n\in\mathbb Z$ and $0\leq s<1$. That is, we have $\lfloor x\rfloor=n$. Consider the following two situations:

  1. $s\in[0,1/2)$;
  2. $s\in[1/2,1)$.

What can you say about $\lfloor x+1/2\rfloor$ in these two separate situations?

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Thats where I stopped? Am i right that floor (x +1/2) is not same as floor (x) + floor (1/2), x is real number? –  mth126 Aug 10 '12 at 19:23
    
In (a), I stopped in floor (s + 1/2) = 0. I dont know whats next. –  mth126 Aug 10 '12 at 19:25
    
@mth126 You are right: in general, the floor of $x+1/2$ is not the same as the floor of $x$ plus the floor of $1/2$ (which is simply 0). The problem with your decomposition $x=n+s$ is that you're not restricting $s$ enough. In my decomposition, I make sure that the floor of $x$ is exactly $n$. –  M Turgeon Aug 10 '12 at 19:28
    
@mth126 Hence, I separated $x$ into its integer part $n$ (which is exactly the floor of $x$) and its fractional part $s$. When I add $1/2$ to $x$, either the integer part of $x+1/2$ stays the same (when does this happen?) or it changes, in which case it increases the floor by exactly 1. –  M Turgeon Aug 10 '12 at 19:31
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Thanks so much! Thanks for the patience. –  mth126 Aug 10 '12 at 20:11
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You're just fine up to "$\lfloor s+\frac{1}{2}\rfloor=0\quad$ I don't know what's next". Now you should look at the $s$ values that will keep $s+\frac{1}{2}$ from going above $1$, since then $\lfloor s+\frac{1}{2} \rfloor$ would no longer be $0$. You'll find that there is a limited range of possibilities for $s$, which when you add the integer part $n$ back in, will give you the form of the $x$ answers you need. Once you see this, part (b) should be easy.

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No, the solution is not fine: the OP assumed what she was trying to prove! –  M Turgeon Aug 10 '12 at 20:08
    
So my solution have sense? Then I can conclude from that line that s = [0,1/2) –  mth126 Aug 10 '12 at 20:08
    
@MTurgeon. If you mean that the OP assumed that for integer $n$ and $s\in[0, 1)$ then $\lfloor n+s\rfloor=n$ then I would agree with you. However, my I read of the problem was that this result was already known to the OP and that the difficulty was that after using that result the answer was not yet clear. –  Rick Decker Aug 10 '12 at 20:33
    
@mth126. Yes. So your solutions will comprise all of the values you'd get by adding integers to those $s$ values. –  Rick Decker Aug 10 '12 at 20:39
    
@RickDecker I meant the line where the OP writes "floor($x$)=floor($x+1/2$)". –  M Turgeon Aug 10 '12 at 20:57
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