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
    22 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
    22 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 26 mins ago







Adam54

















asked 31 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
    22 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
    22 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
22 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
22 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
    14 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









      $endgroup$













        Your Answer





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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
          14 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
          14 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 17 mins ago









        Ilmari KaronenIlmari Karonen

        19.7k25183




        19.7k25183












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


















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
















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




        $begingroup$
        Thank you very much for the odd case, made very simple!
        $endgroup$
        – Adam54
        14 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 19 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 16 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









                        $endgroup$


















                          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









                          $endgroup$
















                            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









                            $endgroup$



                            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












                            share|cite



                            share|cite










                            answered 4 mins ago









                            Yves DaoustYves Daoust

                            125k671223




                            125k671223






























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                                Vallis Paradisi

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                                Coordinata: .mw-parser-output .geo-default,.mw-parser-output .geo-dms,.mw-parser-output .geo-dec{display:inline}.mw-parser-output .geo-nondefault,.mw-parser-output .geo-multi-punct{display:none}.mw-parser-output .longitude,.mw-parser-output .latitude{white-space:nowrap} 33°03′S 71°37′W  /  33.05°S 71.617°W  / -33.05; -71.617 Vallis Paradisi Res apud Vicidata repertae: Civitas : Chilia Numerus incolarum : 296 655 Zona horaria : UTC-4 Situs interretialis Nomen officiale : Valparaíso Gubernium Consilium : municipal council of Valparaíso Geographia Superficies : 401 600 000 Metrum quadratum Coniunctiones urbium Urbes gemellae : Ovetum , Badalona , Santa Fe , Ora longa , Sciamhaevum , Quantunia , Callao , Bat Yam , Melaka , Novorossiysk , Busan , Corduba , Odessa , Rosario , Malacca , Barcino , Soteropolis , Franciscopolis (California) , Guanajuato, Silao Tabula aut despectus Despectus in portum sinumque et urbem Vallis Paradisi Vallis Para...
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                                Tabula Rosettana

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                                Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a...
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