Prove that $lfloor sqrt{(p-1)p} rfloor = p - 1$ and likewise $lceil sqrt{(p-1)p} rceil = p$.












1












$begingroup$


Here $p$ is prime but is not necessary for the problem just that $p ge 0$. I suspect that a statement like $p-1 le sqrt{(p-1)p} le p$ would be the case but I am not certain how to establish this condition.






elementary-number-theory maxima-minima floor-function ceiling-function







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$endgroup$












  • $begingroup$
    We need that $pge 1$, since otherwise the square root is not defined. That is also a sufficient condition.
    $endgroup$
    – I like Serena
    Dec 12 '18 at 17:48
















1












$begingroup$


Here $p$ is prime but is not necessary for the problem just that $p ge 0$. I suspect that a statement like $p-1 le sqrt{(p-1)p} le p$ would be the case but I am not certain how to establish this condition.






elementary-number-theory maxima-minima floor-function ceiling-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    We need that $pge 1$, since otherwise the square root is not defined. That is also a sufficient condition.
    $endgroup$
    – I like Serena
    Dec 12 '18 at 17:48














1












1








1





$begingroup$


Here $p$ is prime but is not necessary for the problem just that $p ge 0$. I suspect that a statement like $p-1 le sqrt{(p-1)p} le p$ would be the case but I am not certain how to establish this condition.






elementary-number-theory maxima-minima floor-function ceiling-function







share|cite|improve this question











$endgroup$




Here $p$ is prime but is not necessary for the problem just that $p ge 0$. I suspect that a statement like $p-1 le sqrt{(p-1)p} le p$ would be the case but I am not certain how to establish this condition.






elementary-number-theory maxima-minima floor-function ceiling-function




elementary-number-theory maxima-minima 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 Dec 12 '18 at 17:53









Bernard

120k740115




120k740115










asked Dec 12 '18 at 17:24









Lorenz H MenkeLorenz H Menke

10011




10011












  • $begingroup$
    We need that $pge 1$, since otherwise the square root is not defined. That is also a sufficient condition.
    $endgroup$
    – I like Serena
    Dec 12 '18 at 17:48


















  • $begingroup$
    We need that $pge 1$, since otherwise the square root is not defined. That is also a sufficient condition.
    $endgroup$
    – I like Serena
    Dec 12 '18 at 17:48
















$begingroup$
We need that $pge 1$, since otherwise the square root is not defined. That is also a sufficient condition.
$endgroup$
– I like Serena
Dec 12 '18 at 17:48




$begingroup$
We need that $pge 1$, since otherwise the square root is not defined. That is also a sufficient condition.
$endgroup$
– I like Serena
Dec 12 '18 at 17:48










2 Answers
2






active

oldest

votes


















2












$begingroup$

Hint: $(p-1)^2 le (p-1)p le p^2$ for $p ge 0$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    For $p>1$,$$(p-1)^2=p^2-2p+1<p^2-p<p^2$$
    and



    $$p-1<sqrt{(p-1)p}<p.$$



    As the extreme members are integer,



    $$p-1leleftlfloorsqrt{(p-1)p}rightrfloorleleftlceilsqrt{(p-1)p}rightrceille p.$$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

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      active

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      active

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      2












      $begingroup$

      Hint: $(p-1)^2 le (p-1)p le p^2$ for $p ge 0$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Hint: $(p-1)^2 le (p-1)p le p^2$ for $p ge 0$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Hint: $(p-1)^2 le (p-1)p le p^2$ for $p ge 0$






          share|cite|improve this answer









          $endgroup$



          Hint: $(p-1)^2 le (p-1)p le p^2$ for $p ge 0$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 17:29









          karakfakarakfa

          1,995811




          1,995811























              2












              $begingroup$

              For $p>1$,$$(p-1)^2=p^2-2p+1<p^2-p<p^2$$
              and



              $$p-1<sqrt{(p-1)p}<p.$$



              As the extreme members are integer,



              $$p-1leleftlfloorsqrt{(p-1)p}rightrfloorleleftlceilsqrt{(p-1)p}rightrceille p.$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                For $p>1$,$$(p-1)^2=p^2-2p+1<p^2-p<p^2$$
                and



                $$p-1<sqrt{(p-1)p}<p.$$



                As the extreme members are integer,



                $$p-1leleftlfloorsqrt{(p-1)p}rightrfloorleleftlceilsqrt{(p-1)p}rightrceille p.$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  For $p>1$,$$(p-1)^2=p^2-2p+1<p^2-p<p^2$$
                  and



                  $$p-1<sqrt{(p-1)p}<p.$$



                  As the extreme members are integer,



                  $$p-1leleftlfloorsqrt{(p-1)p}rightrfloorleleftlceilsqrt{(p-1)p}rightrceille p.$$






                  share|cite|improve this answer









                  $endgroup$



                  For $p>1$,$$(p-1)^2=p^2-2p+1<p^2-p<p^2$$
                  and



                  $$p-1<sqrt{(p-1)p}<p.$$



                  As the extreme members are integer,



                  $$p-1leleftlfloorsqrt{(p-1)p}rightrfloorleleftlceilsqrt{(p-1)p}rightrceille p.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 17:31









                  Yves DaoustYves Daoust

                  127k673226




                  127k673226






























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