Find all real numbers $x$ satisfying:
a. $\lfloor x + 1/2 \rfloor = \lfloor x \rfloor$.
b. $\lfloor x + 1/2 \rfloor = \lceil x \rceil$.
Finding $x$ such that $\lfloor x + 1/2 \rfloor = \lfloor x \rfloor$
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0Begin with a few examples, such as $x = 1$, $x=1.2$, $x=1.5$, $x=1.7$, etc. Try to spot the pattern. – 2012-08-10
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0First, consider a positive $x.$ Write $x$ as $k + f$ where $k$ is an integer, and $f$ is a real number between $0$ and $1$. – 2012-08-11
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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0I 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) ?? – 2012-08-10
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0You 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? – 2012-08-10
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0Wait. 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? – 2012-08-10
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1You'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. – 2012-08-10
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0I edited my question together with my solution. – 2012-08-10
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0All you have done is reduced the general problem to a specific problem, where $x \in [0, 1/2)$. You still need to address the question asked in my most recent edit to my answer in order to solve your problem. You have done a lot of work for no true gain; please look at my answer and edit again, and try to describe the exact scenario and questions I pose -- they will lead you immediately to your answer. And then you will see how your current approach reduces the general problem to a specific case. – 2012-08-10
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0yes letter (a) works when let x∈[5,5.5) and when even we change the integral part. – 2012-08-10
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0Right! You have correctly determined that $x$ solves part (a) whenever $x \in [k, k+1/2)\ \forall k = 0,1,2,3,\ldots$. Now use the same reasoning for when $k$ is negative. Also, you can use the same reasoning for part (b). Remember, you're not trying to solve for specific values of $x$; you're trying to find the SET of values that hold true for any $x$ in that set. – 2012-08-10
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0So I dont need my solutions? ANyway, thank you so much! – 2012-08-10
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:
- $s\in[0,1/2)$;
- $s\in[1/2,1)$.
What can you say about $\lfloor x+1/2\rfloor$ in these two separate situations?
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0Thats where I stopped? Am i right that floor (x +1/2) is not same as floor (x) + floor (1/2), x is real number? – 2012-08-10
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0In (a), I stopped in floor (s + 1/2) = 0. I dont know whats next. – 2012-08-10
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0@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 $n$. – 2012-08-10
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0@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. – 2012-08-10
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0I edited my question. That's my solution. Which part is not clear in my solution? – 2012-08-10
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0@mth126 The only thing you proved in your solution is that, if the floor of $x$ is equalt to the floor of $x+1/2$, then the floor of $x$ is equalt to the floor of $x+1/2$. That is, you assumed what you want to prove. – 2012-08-10
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0Looking at your first comment again, case 1 works for (a) and case 2 works for (b) – 2012-08-10
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0@mth126 Yes, that is correct! – 2012-08-10
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1Thanks so much! Thanks for the patience. – 2012-08-10
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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0No, the solution is not fine: the OP assumed what she was trying to prove! – 2012-08-10
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0So my solution have sense? Then I can conclude from that line that s = [0,1/2) – 2012-08-10
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0@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. – 2012-08-10
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0@mth126. Yes. So your solutions will comprise all of the values you'd get by adding integers to those $s$ values. – 2012-08-10
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0@RickDecker I meant the line where the OP writes "floor($x$)=floor($x+1/2$)". – 2012-08-10
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0@MTurgeon. I see. A more felicitous wording (which I provided in my reading) would have been "$\lfloor x+1/2\rfloor = \lfloor x\rfloor$ if and only if..." and had continued in that vein until the conclusion "therefore, $x$ must have the form ...". – 2012-08-10
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0@RickDecker Oh, I see! Fair enough, that would have been a valid argument then! – 2012-08-10