Ceil and floor function












2












$begingroup$


Let $a$ and $b$ be positive integers.



If $b$ is even, then we have $$leftlfloor frac{a-b}{2} rightrfloor + leftlceil frac{a+b}{2} rightrceil = a$$



I thing the equality also hold when $b$ is odd. What could be a proof for it?



Thank you






floor-function ceiling-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    Write $b$ as $2k + 1$, where $k$ is a nonnegative integer. Then look at two cases: $a = 2s$ ($s$ a positive integer), and $a = 2s + 1$ ($s$ a nonnegative integer). What do you get for the left-hand side in each case? [You can show your work by clicking "edit" beneath your question.]
    $endgroup$
    – John Hughes
    28 mins ago
















2












$begingroup$


Let $a$ and $b$ be positive integers.



If $b$ is even, then we have $$leftlfloor frac{a-b}{2} rightrfloor + leftlceil frac{a+b}{2} rightrceil = a$$



I thing the equality also hold when $b$ is odd. What could be a proof for it?



Thank you






floor-function ceiling-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    Write $b$ as $2k + 1$, where $k$ is a nonnegative integer. Then look at two cases: $a = 2s$ ($s$ a positive integer), and $a = 2s + 1$ ($s$ a nonnegative integer). What do you get for the left-hand side in each case? [You can show your work by clicking "edit" beneath your question.]
    $endgroup$
    – John Hughes
    28 mins ago














2












2








2





$begingroup$


Let $a$ and $b$ be positive integers.



If $b$ is even, then we have $$leftlfloor frac{a-b}{2} rightrfloor + leftlceil frac{a+b}{2} rightrceil = a$$



I thing the equality also hold when $b$ is odd. What could be a proof for it?



Thank you






floor-function ceiling-function







share|cite|improve this question











$endgroup$




Let $a$ and $b$ be positive integers.



If $b$ is even, then we have $$leftlfloor frac{a-b}{2} rightrfloor + leftlceil frac{a+b}{2} rightrceil = a$$



I thing the equality also hold when $b$ is odd. What could be a proof for it?



Thank you






floor-function ceiling-function




floor-function ceiling-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 32 mins ago







Adam54

















asked 37 mins ago









Adam54Adam54

615




615












  • $begingroup$
    Write $b$ as $2k + 1$, where $k$ is a nonnegative integer. Then look at two cases: $a = 2s$ ($s$ a positive integer), and $a = 2s + 1$ ($s$ a nonnegative integer). What do you get for the left-hand side in each case? [You can show your work by clicking "edit" beneath your question.]
    $endgroup$
    – John Hughes
    28 mins ago


















  • $begingroup$
    Write $b$ as $2k + 1$, where $k$ is a nonnegative integer. Then look at two cases: $a = 2s$ ($s$ a positive integer), and $a = 2s + 1$ ($s$ a nonnegative integer). What do you get for the left-hand side in each case? [You can show your work by clicking "edit" beneath your question.]
    $endgroup$
    – John Hughes
    28 mins ago
















$begingroup$
Write $b$ as $2k + 1$, where $k$ is a nonnegative integer. Then look at two cases: $a = 2s$ ($s$ a positive integer), and $a = 2s + 1$ ($s$ a nonnegative integer). What do you get for the left-hand side in each case? [You can show your work by clicking "edit" beneath your question.]
$endgroup$
– John Hughes
28 mins ago




$begingroup$
Write $b$ as $2k + 1$, where $k$ is a nonnegative integer. Then look at two cases: $a = 2s$ ($s$ a positive integer), and $a = 2s + 1$ ($s$ a nonnegative integer). What do you get for the left-hand side in each case? [You can show your work by clicking "edit" beneath your question.]
$endgroup$
– John Hughes
28 mins ago










4 Answers
4






active

oldest

votes


















0












$begingroup$

If $a$ and $b$ are both odd (or both even), then $a-b$ and $a+b$ are both even, and thus $$leftlfloor frac{a-b}{2} rightrfloor = frac{a-b}{2} quadtext{and}quad leftlceil frac{a+b}{2} rightrceil = frac{a+b}{2}.$$



Otherwise, if exactly one of $a$ and $b$ is odd, then $a-b$ and $a+b$ are both odd, and thus $$leftlfloor frac{a-b}{2} rightrfloor = frac{a-b}{2} - frac12 quadtext{and}quad leftlceil frac{a+b}{2} rightrceil = frac{a+b}{2} + frac12.$$



In either case, it's easy to check that your equation holds.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much for the odd case, made very simple!
    $endgroup$
    – Adam54
    20 mins ago



















2












$begingroup$

The equality only depends on the parity of $a+b$, as it is the same as that of $a-b$. Then



$$leftlfloorfrac02rightrfloor+leftlceilfrac02rightrceil=0$$
and
$$leftlfloorfrac12rightrfloor+leftlceilfrac12rightrceil=1$$
are enough as a proof.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    If $a$ and $b$ are both odd then we have $a=2m+1$ and $b=2n+1$ where $m$ and $n$ are positive integers.



    Then, we have begin{align}leftlfloorfrac{a-b}2rightrfloor+leftlceilfrac{a+b}2rightrceil &= leftlfloorfrac{(2m+1)-(2n+1)}2rightrfloor+ leftlceilfrac{(2m+1)+(2n+1)}2rightrceil\
    &=leftlfloorfrac{2m-2n}2rightrfloor+ leftlceilfrac{2m+2n+2}2rightrceil\
    &=lfloor m-nrfloor + lceil m+n+1rceil\
    &= m-n + m+n+1tag{$*$}\
    &= 2m+1\
    &=aend{align}



    We can get to $(*)$ because $m$ and $n$ are integers and so their floor (or ceiling) is just the number inside (two integers added or subtracted will always give an integer answer).



    If $a$ is even and $b$ is odd, then we have $a=2m$ and $b=2n+1$, again for $m$ and $n$ integers.



    So we have begin{align}leftlfloorfrac{a-b}2rightrfloor+leftlceilfrac{a+b}2rightrceil &= leftlfloorfrac{(2m)-(2n+1)}2rightrfloor+ leftlceilfrac{(2m)+(2n+1)}2rightrceil\
    &=leftlfloorfrac{2m-2n-1}2rightrfloor+ leftlceilfrac{2m+2n+1}2rightrceil\
    &=leftlfloor m-n-frac 12rightrfloor + leftlceil m+n+frac12rightrceil\
    &= m-n -1+ m+n+1tag{$dagger$}\
    &= 2m\
    &=aend{align}



    Here we get to $(dagger)$ because we round the first half down and the second half up to their respective nearest integers






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      As $s:=a+b$ and $a-b$ have the same parity, we can write



      $$leftlfloorfrac{s-2b}2rightrfloor+leftlceilfrac s2rightrceil=leftlfloorfrac s2rightrfloor+leftlceilfrac s2rightrceil-b=s-b=a.$$






      share|cite|improve this answer









      $endgroup$













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        4 Answers
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        4 Answers
        4






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        active

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        active

        oldest

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        0












        $begingroup$

        If $a$ and $b$ are both odd (or both even), then $a-b$ and $a+b$ are both even, and thus $$leftlfloor frac{a-b}{2} rightrfloor = frac{a-b}{2} quadtext{and}quad leftlceil frac{a+b}{2} rightrceil = frac{a+b}{2}.$$



        Otherwise, if exactly one of $a$ and $b$ is odd, then $a-b$ and $a+b$ are both odd, and thus $$leftlfloor frac{a-b}{2} rightrfloor = frac{a-b}{2} - frac12 quadtext{and}quad leftlceil frac{a+b}{2} rightrceil = frac{a+b}{2} + frac12.$$



        In either case, it's easy to check that your equation holds.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Thank you very much for the odd case, made very simple!
          $endgroup$
          – Adam54
          20 mins ago
















        0












        $begingroup$

        If $a$ and $b$ are both odd (or both even), then $a-b$ and $a+b$ are both even, and thus $$leftlfloor frac{a-b}{2} rightrfloor = frac{a-b}{2} quadtext{and}quad leftlceil frac{a+b}{2} rightrceil = frac{a+b}{2}.$$



        Otherwise, if exactly one of $a$ and $b$ is odd, then $a-b$ and $a+b$ are both odd, and thus $$leftlfloor frac{a-b}{2} rightrfloor = frac{a-b}{2} - frac12 quadtext{and}quad leftlceil frac{a+b}{2} rightrceil = frac{a+b}{2} + frac12.$$



        In either case, it's easy to check that your equation holds.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Thank you very much for the odd case, made very simple!
          $endgroup$
          – Adam54
          20 mins ago














        0












        0








        0





        $begingroup$

        If $a$ and $b$ are both odd (or both even), then $a-b$ and $a+b$ are both even, and thus $$leftlfloor frac{a-b}{2} rightrfloor = frac{a-b}{2} quadtext{and}quad leftlceil frac{a+b}{2} rightrceil = frac{a+b}{2}.$$



        Otherwise, if exactly one of $a$ and $b$ is odd, then $a-b$ and $a+b$ are both odd, and thus $$leftlfloor frac{a-b}{2} rightrfloor = frac{a-b}{2} - frac12 quadtext{and}quad leftlceil frac{a+b}{2} rightrceil = frac{a+b}{2} + frac12.$$



        In either case, it's easy to check that your equation holds.






        share|cite|improve this answer









        $endgroup$



        If $a$ and $b$ are both odd (or both even), then $a-b$ and $a+b$ are both even, and thus $$leftlfloor frac{a-b}{2} rightrfloor = frac{a-b}{2} quadtext{and}quad leftlceil frac{a+b}{2} rightrceil = frac{a+b}{2}.$$



        Otherwise, if exactly one of $a$ and $b$ is odd, then $a-b$ and $a+b$ are both odd, and thus $$leftlfloor frac{a-b}{2} rightrfloor = frac{a-b}{2} - frac12 quadtext{and}quad leftlceil frac{a+b}{2} rightrceil = frac{a+b}{2} + frac12.$$



        In either case, it's easy to check that your equation holds.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 23 mins ago









        Ilmari KaronenIlmari Karonen

        19.7k25183




        19.7k25183












        • $begingroup$
          Thank you very much for the odd case, made very simple!
          $endgroup$
          – Adam54
          20 mins ago


















        • $begingroup$
          Thank you very much for the odd case, made very simple!
          $endgroup$
          – Adam54
          20 mins ago
















        $begingroup$
        Thank you very much for the odd case, made very simple!
        $endgroup$
        – Adam54
        20 mins ago




        $begingroup$
        Thank you very much for the odd case, made very simple!
        $endgroup$
        – Adam54
        20 mins ago











        2












        $begingroup$

        The equality only depends on the parity of $a+b$, as it is the same as that of $a-b$. Then



        $$leftlfloorfrac02rightrfloor+leftlceilfrac02rightrceil=0$$
        and
        $$leftlfloorfrac12rightrfloor+leftlceilfrac12rightrceil=1$$
        are enough as a proof.






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          The equality only depends on the parity of $a+b$, as it is the same as that of $a-b$. Then



          $$leftlfloorfrac02rightrfloor+leftlceilfrac02rightrceil=0$$
          and
          $$leftlfloorfrac12rightrfloor+leftlceilfrac12rightrceil=1$$
          are enough as a proof.






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            The equality only depends on the parity of $a+b$, as it is the same as that of $a-b$. Then



            $$leftlfloorfrac02rightrfloor+leftlceilfrac02rightrceil=0$$
            and
            $$leftlfloorfrac12rightrfloor+leftlceilfrac12rightrceil=1$$
            are enough as a proof.






            share|cite|improve this answer









            $endgroup$



            The equality only depends on the parity of $a+b$, as it is the same as that of $a-b$. Then



            $$leftlfloorfrac02rightrfloor+leftlceilfrac02rightrceil=0$$
            and
            $$leftlfloorfrac12rightrfloor+leftlceilfrac12rightrceil=1$$
            are enough as a proof.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 25 mins ago









            Yves DaoustYves Daoust

            125k671223




            125k671223























                2












                $begingroup$

                If $a$ and $b$ are both odd then we have $a=2m+1$ and $b=2n+1$ where $m$ and $n$ are positive integers.



                Then, we have begin{align}leftlfloorfrac{a-b}2rightrfloor+leftlceilfrac{a+b}2rightrceil &= leftlfloorfrac{(2m+1)-(2n+1)}2rightrfloor+ leftlceilfrac{(2m+1)+(2n+1)}2rightrceil\
                &=leftlfloorfrac{2m-2n}2rightrfloor+ leftlceilfrac{2m+2n+2}2rightrceil\
                &=lfloor m-nrfloor + lceil m+n+1rceil\
                &= m-n + m+n+1tag{$*$}\
                &= 2m+1\
                &=aend{align}



                We can get to $(*)$ because $m$ and $n$ are integers and so their floor (or ceiling) is just the number inside (two integers added or subtracted will always give an integer answer).



                If $a$ is even and $b$ is odd, then we have $a=2m$ and $b=2n+1$, again for $m$ and $n$ integers.



                So we have begin{align}leftlfloorfrac{a-b}2rightrfloor+leftlceilfrac{a+b}2rightrceil &= leftlfloorfrac{(2m)-(2n+1)}2rightrfloor+ leftlceilfrac{(2m)+(2n+1)}2rightrceil\
                &=leftlfloorfrac{2m-2n-1}2rightrfloor+ leftlceilfrac{2m+2n+1}2rightrceil\
                &=leftlfloor m-n-frac 12rightrfloor + leftlceil m+n+frac12rightrceil\
                &= m-n -1+ m+n+1tag{$dagger$}\
                &= 2m\
                &=aend{align}



                Here we get to $(dagger)$ because we round the first half down and the second half up to their respective nearest integers






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  If $a$ and $b$ are both odd then we have $a=2m+1$ and $b=2n+1$ where $m$ and $n$ are positive integers.



                  Then, we have begin{align}leftlfloorfrac{a-b}2rightrfloor+leftlceilfrac{a+b}2rightrceil &= leftlfloorfrac{(2m+1)-(2n+1)}2rightrfloor+ leftlceilfrac{(2m+1)+(2n+1)}2rightrceil\
                  &=leftlfloorfrac{2m-2n}2rightrfloor+ leftlceilfrac{2m+2n+2}2rightrceil\
                  &=lfloor m-nrfloor + lceil m+n+1rceil\
                  &= m-n + m+n+1tag{$*$}\
                  &= 2m+1\
                  &=aend{align}



                  We can get to $(*)$ because $m$ and $n$ are integers and so their floor (or ceiling) is just the number inside (two integers added or subtracted will always give an integer answer).



                  If $a$ is even and $b$ is odd, then we have $a=2m$ and $b=2n+1$, again for $m$ and $n$ integers.



                  So we have begin{align}leftlfloorfrac{a-b}2rightrfloor+leftlceilfrac{a+b}2rightrceil &= leftlfloorfrac{(2m)-(2n+1)}2rightrfloor+ leftlceilfrac{(2m)+(2n+1)}2rightrceil\
                  &=leftlfloorfrac{2m-2n-1}2rightrfloor+ leftlceilfrac{2m+2n+1}2rightrceil\
                  &=leftlfloor m-n-frac 12rightrfloor + leftlceil m+n+frac12rightrceil\
                  &= m-n -1+ m+n+1tag{$dagger$}\
                  &= 2m\
                  &=aend{align}



                  Here we get to $(dagger)$ because we round the first half down and the second half up to their respective nearest integers






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    If $a$ and $b$ are both odd then we have $a=2m+1$ and $b=2n+1$ where $m$ and $n$ are positive integers.



                    Then, we have begin{align}leftlfloorfrac{a-b}2rightrfloor+leftlceilfrac{a+b}2rightrceil &= leftlfloorfrac{(2m+1)-(2n+1)}2rightrfloor+ leftlceilfrac{(2m+1)+(2n+1)}2rightrceil\
                    &=leftlfloorfrac{2m-2n}2rightrfloor+ leftlceilfrac{2m+2n+2}2rightrceil\
                    &=lfloor m-nrfloor + lceil m+n+1rceil\
                    &= m-n + m+n+1tag{$*$}\
                    &= 2m+1\
                    &=aend{align}



                    We can get to $(*)$ because $m$ and $n$ are integers and so their floor (or ceiling) is just the number inside (two integers added or subtracted will always give an integer answer).



                    If $a$ is even and $b$ is odd, then we have $a=2m$ and $b=2n+1$, again for $m$ and $n$ integers.



                    So we have begin{align}leftlfloorfrac{a-b}2rightrfloor+leftlceilfrac{a+b}2rightrceil &= leftlfloorfrac{(2m)-(2n+1)}2rightrfloor+ leftlceilfrac{(2m)+(2n+1)}2rightrceil\
                    &=leftlfloorfrac{2m-2n-1}2rightrfloor+ leftlceilfrac{2m+2n+1}2rightrceil\
                    &=leftlfloor m-n-frac 12rightrfloor + leftlceil m+n+frac12rightrceil\
                    &= m-n -1+ m+n+1tag{$dagger$}\
                    &= 2m\
                    &=aend{align}



                    Here we get to $(dagger)$ because we round the first half down and the second half up to their respective nearest integers






                    share|cite|improve this answer









                    $endgroup$



                    If $a$ and $b$ are both odd then we have $a=2m+1$ and $b=2n+1$ where $m$ and $n$ are positive integers.



                    Then, we have begin{align}leftlfloorfrac{a-b}2rightrfloor+leftlceilfrac{a+b}2rightrceil &= leftlfloorfrac{(2m+1)-(2n+1)}2rightrfloor+ leftlceilfrac{(2m+1)+(2n+1)}2rightrceil\
                    &=leftlfloorfrac{2m-2n}2rightrfloor+ leftlceilfrac{2m+2n+2}2rightrceil\
                    &=lfloor m-nrfloor + lceil m+n+1rceil\
                    &= m-n + m+n+1tag{$*$}\
                    &= 2m+1\
                    &=aend{align}



                    We can get to $(*)$ because $m$ and $n$ are integers and so their floor (or ceiling) is just the number inside (two integers added or subtracted will always give an integer answer).



                    If $a$ is even and $b$ is odd, then we have $a=2m$ and $b=2n+1$, again for $m$ and $n$ integers.



                    So we have begin{align}leftlfloorfrac{a-b}2rightrfloor+leftlceilfrac{a+b}2rightrceil &= leftlfloorfrac{(2m)-(2n+1)}2rightrfloor+ leftlceilfrac{(2m)+(2n+1)}2rightrceil\
                    &=leftlfloorfrac{2m-2n-1}2rightrfloor+ leftlceilfrac{2m+2n+1}2rightrceil\
                    &=leftlfloor m-n-frac 12rightrfloor + leftlceil m+n+frac12rightrceil\
                    &= m-n -1+ m+n+1tag{$dagger$}\
                    &= 2m\
                    &=aend{align}



                    Here we get to $(dagger)$ because we round the first half down and the second half up to their respective nearest integers







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 22 mins ago









                    lioness99alioness99a

                    3,6842727




                    3,6842727























                        0












                        $begingroup$

                        As $s:=a+b$ and $a-b$ have the same parity, we can write



                        $$leftlfloorfrac{s-2b}2rightrfloor+leftlceilfrac s2rightrceil=leftlfloorfrac s2rightrfloor+leftlceilfrac s2rightrceil-b=s-b=a.$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          As $s:=a+b$ and $a-b$ have the same parity, we can write



                          $$leftlfloorfrac{s-2b}2rightrfloor+leftlceilfrac s2rightrceil=leftlfloorfrac s2rightrfloor+leftlceilfrac s2rightrceil-b=s-b=a.$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            As $s:=a+b$ and $a-b$ have the same parity, we can write



                            $$leftlfloorfrac{s-2b}2rightrfloor+leftlceilfrac s2rightrceil=leftlfloorfrac s2rightrfloor+leftlceilfrac s2rightrceil-b=s-b=a.$$






                            share|cite|improve this answer









                            $endgroup$



                            As $s:=a+b$ and $a-b$ have the same parity, we can write



                            $$leftlfloorfrac{s-2b}2rightrfloor+leftlceilfrac s2rightrceil=leftlfloorfrac s2rightrfloor+leftlceilfrac s2rightrceil-b=s-b=a.$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 10 mins ago









                            Yves DaoustYves Daoust

                            125k671223




                            125k671223






























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                                1 I would like to know if and how is it possible to know if a AD user can login interactively (on a server) in a domain. I need to know if I can find it out using a LDAP search. Thanks. windows active-directory login user-permissions user-profile share | improve this question edited 1 hour ago yagmoth555 ♦ 11.6k 3 17 42 asked 1 hour ago Luigi Luigi 6 1 New contributor
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                                TypeError: fit_transform() missing 1 required positional argument: 'X'

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                                0 I am trying to do Feature Scaling in a dataset, but I get an error and have no idea how to proceed: > Traceback (most recent call last): > > File "<ipython-input-10-71bea414b4d0>", line 22, in <module> > x_train = sc_X.fit_transform(x_train) > > TypeError: fit_transform() missing 1 required positional argument: 'X' and here is my code: import pandas as pd # Importing the dataset dataset = pd.read_csv('Data.csv') X = dataset.iloc[:, :-1].values y = dataset.iloc[:, 3].values # Taking care of missing data from sklearn.preprocessing import Imputer imputer = Imputer(missing_values="NaN", strategy="mean", axis=0) imputer = Imputer.fit(imputer,X[:,1:3]) X[:, 1:3] = Imputer.transform(imputer,X[
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