How can we find the solution set of $\lceil x\rceil^2+\lceil x+1\rceil^2=25$ where $\lceil x\rceil$ is the ceiling function?
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yes it is the same thing... thanks for editing .. – Sachin Mar 30 '13 at 15:27
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I'm not sure what you call integral part, but I'm sure it does not make an accurate definition here. A possible definition is: $\lceil x\rceil$ is the smallest integer $n$ such that $n\geq x$. Note this agrees with the floor function on integers, but nowhere else. – Julien Mar 30 '13 at 15:34
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You can check that the only solution to $$ n^2 + m^2 = 5^2 $$ for $n$ and $m$ integers are $n = \pm 3, m = \pm 4$ or $n = \pm 4, m = \pm 3$ or $n= 0, m =5$ or $n = 5, m =0$. So since the ceiling function does give you integers and you need $ m = n+1$ (or the other way around), you have that $\lceil x\rceil = 3$ or $\lceil x\rceil = -4$.
Thomas
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1down-vote without showing the mistake really is really annoying. – lab bhattacharjee Mar 30 '13 at 15:35
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1I was not necessarily planning to upvote, but I will to fix the annoyingly mute downvote. – Julien Mar 30 '13 at 15:37
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Observe that $\lceil x+1\rceil =\lceil x\rceil +1$. So setting $n=\lceil x\rceil$, it amounts to solving for $n$ in $$n^2+(n+1)^2=25\;\iff\;n^2+n-12=0\;\iff\;(n+4)(n-3)=0\;\iff\; n\in\{-4,3\}.$$
The case $n=\lceil x\rceil=-4$ gives you $-5<x\leq-4$ while $n=\lceil x\rceil=3$ yields $2<x\leq 3$. So the solution set is $$ (-5,-4]\cup(2,3].$$
Julien
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