Why is $leftlfloorfrac{lfloor apirfloor}{a}rightrfloor=3$ for $a>0$?












-1












$begingroup$


Why is this true?
$$leftlfloorfrac{lfloor apirfloor}{a}rightrfloor=3 text{, for } a>0$$



I need this to solve the Ukraine Math Olymipiad 1999. "$lfloorcdotrfloor$" indicates the floor function.






floor-function pi







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$endgroup$












  • $begingroup$
    Is $a$ supposed to be an integer?
    $endgroup$
    – Blue
    Jan 12 at 14:49
















-1












$begingroup$


Why is this true?
$$leftlfloorfrac{lfloor apirfloor}{a}rightrfloor=3 text{, for } a>0$$



I need this to solve the Ukraine Math Olymipiad 1999. "$lfloorcdotrfloor$" indicates the floor function.






floor-function pi







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is $a$ supposed to be an integer?
    $endgroup$
    – Blue
    Jan 12 at 14:49














-1












-1








-1





$begingroup$


Why is this true?
$$leftlfloorfrac{lfloor apirfloor}{a}rightrfloor=3 text{, for } a>0$$



I need this to solve the Ukraine Math Olymipiad 1999. "$lfloorcdotrfloor$" indicates the floor function.






floor-function pi







share|cite|improve this question











$endgroup$




Why is this true?
$$leftlfloorfrac{lfloor apirfloor}{a}rightrfloor=3 text{, for } a>0$$



I need this to solve the Ukraine Math Olymipiad 1999. "$lfloorcdotrfloor$" indicates the floor function.






floor-function pi




floor-function pi






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 at 14:45









Blue

48k870153




48k870153










asked Jan 12 at 14:15









mathheromathhero

235




235












  • $begingroup$
    Is $a$ supposed to be an integer?
    $endgroup$
    – Blue
    Jan 12 at 14:49


















  • $begingroup$
    Is $a$ supposed to be an integer?
    $endgroup$
    – Blue
    Jan 12 at 14:49
















$begingroup$
Is $a$ supposed to be an integer?
$endgroup$
– Blue
Jan 12 at 14:49




$begingroup$
Is $a$ supposed to be an integer?
$endgroup$
– Blue
Jan 12 at 14:49










3 Answers
3






active

oldest

votes


















2












$begingroup$

The equation is equivalent to



$$0le{pi}-frac{{api}}a<1.$$



The right inequality is always verified. The left one is certainly verified when



$$0le{pi}-frac1a,$$ or $$agefrac1{{pi}},$$ which is a little more than $7$.



Assuming that $a$ is restricted to be a natural, it remains to try $a=1,2,cdots7$. And as ${pi}<dfrac17$, all these values will work.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @JohnDoe: ooops, yes of course.
    $endgroup$
    – Yves Daoust
    Jan 12 at 15:05










  • $begingroup$
    Also, what do you mean when you say it suffices to try $a=cdots$? Are you saying these are the only values of $a$ for which it works?
    $endgroup$
    – John Doe
    Jan 12 at 15:05










  • $begingroup$
    @JohnDoe: no, these are the values for which the bounding argument doesn't work.
    $endgroup$
    – Yves Daoust
    Jan 12 at 15:06










  • $begingroup$
    but $[6timespi]=[18.8...]=18$, $[18/6]=[3]=3$ does work. In fact it should always work for a natural number. I don't think those numbers are special
    $endgroup$
    – John Doe
    Jan 12 at 15:08












  • $begingroup$
    @JohnDoe: I didn't say that these don't work, I said that they needed to be confirmed with another argument.
    $endgroup$
    – Yves Daoust
    Jan 12 at 15:09



















1












$begingroup$

It isn't true! a= 1/2 is a counter example. If a= 1/2 then $api$ is 1.5707... and the floor or that is 1. Dividing that by 1/2 gives 2 which, of course, has floor 2, not 3.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Perhaps the question intends that $a$ is an integer.
    $endgroup$
    – Blue
    Jan 12 at 14:47



















0












$begingroup$

Let $a=b/pi$ for some $binBbb R$.



Then $[api]=[b]$



$$[[b]/a]=left[picdotfrac{[b]}{b}right]$$
For this to equal 3, we need $$frac{[b]}{b}gefrac3piimplies [b]ge3b/piapprox 0.95 b$$



This fails for $$bin{0}cupleft(frac{(n-1)pi}{3},nright)$$ for $nin{1,2,cdots,22}$, and thus for $$ain{0}cupleft(frac{n-1}3,frac npiright)$$ For all other values of $a$, this holds.






share|cite|improve this answer











$endgroup$













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    The equation is equivalent to



    $$0le{pi}-frac{{api}}a<1.$$



    The right inequality is always verified. The left one is certainly verified when



    $$0le{pi}-frac1a,$$ or $$agefrac1{{pi}},$$ which is a little more than $7$.



    Assuming that $a$ is restricted to be a natural, it remains to try $a=1,2,cdots7$. And as ${pi}<dfrac17$, all these values will work.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @JohnDoe: ooops, yes of course.
      $endgroup$
      – Yves Daoust
      Jan 12 at 15:05










    • $begingroup$
      Also, what do you mean when you say it suffices to try $a=cdots$? Are you saying these are the only values of $a$ for which it works?
      $endgroup$
      – John Doe
      Jan 12 at 15:05










    • $begingroup$
      @JohnDoe: no, these are the values for which the bounding argument doesn't work.
      $endgroup$
      – Yves Daoust
      Jan 12 at 15:06










    • $begingroup$
      but $[6timespi]=[18.8...]=18$, $[18/6]=[3]=3$ does work. In fact it should always work for a natural number. I don't think those numbers are special
      $endgroup$
      – John Doe
      Jan 12 at 15:08












    • $begingroup$
      @JohnDoe: I didn't say that these don't work, I said that they needed to be confirmed with another argument.
      $endgroup$
      – Yves Daoust
      Jan 12 at 15:09
















    2












    $begingroup$

    The equation is equivalent to



    $$0le{pi}-frac{{api}}a<1.$$



    The right inequality is always verified. The left one is certainly verified when



    $$0le{pi}-frac1a,$$ or $$agefrac1{{pi}},$$ which is a little more than $7$.



    Assuming that $a$ is restricted to be a natural, it remains to try $a=1,2,cdots7$. And as ${pi}<dfrac17$, all these values will work.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @JohnDoe: ooops, yes of course.
      $endgroup$
      – Yves Daoust
      Jan 12 at 15:05










    • $begingroup$
      Also, what do you mean when you say it suffices to try $a=cdots$? Are you saying these are the only values of $a$ for which it works?
      $endgroup$
      – John Doe
      Jan 12 at 15:05










    • $begingroup$
      @JohnDoe: no, these are the values for which the bounding argument doesn't work.
      $endgroup$
      – Yves Daoust
      Jan 12 at 15:06










    • $begingroup$
      but $[6timespi]=[18.8...]=18$, $[18/6]=[3]=3$ does work. In fact it should always work for a natural number. I don't think those numbers are special
      $endgroup$
      – John Doe
      Jan 12 at 15:08












    • $begingroup$
      @JohnDoe: I didn't say that these don't work, I said that they needed to be confirmed with another argument.
      $endgroup$
      – Yves Daoust
      Jan 12 at 15:09














    2












    2








    2





    $begingroup$

    The equation is equivalent to



    $$0le{pi}-frac{{api}}a<1.$$



    The right inequality is always verified. The left one is certainly verified when



    $$0le{pi}-frac1a,$$ or $$agefrac1{{pi}},$$ which is a little more than $7$.



    Assuming that $a$ is restricted to be a natural, it remains to try $a=1,2,cdots7$. And as ${pi}<dfrac17$, all these values will work.






    share|cite|improve this answer











    $endgroup$



    The equation is equivalent to



    $$0le{pi}-frac{{api}}a<1.$$



    The right inequality is always verified. The left one is certainly verified when



    $$0le{pi}-frac1a,$$ or $$agefrac1{{pi}},$$ which is a little more than $7$.



    Assuming that $a$ is restricted to be a natural, it remains to try $a=1,2,cdots7$. And as ${pi}<dfrac17$, all these values will work.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 12 at 15:19

























    answered Jan 12 at 14:59









    Yves DaoustYves Daoust

    125k671223




    125k671223












    • $begingroup$
      @JohnDoe: ooops, yes of course.
      $endgroup$
      – Yves Daoust
      Jan 12 at 15:05










    • $begingroup$
      Also, what do you mean when you say it suffices to try $a=cdots$? Are you saying these are the only values of $a$ for which it works?
      $endgroup$
      – John Doe
      Jan 12 at 15:05










    • $begingroup$
      @JohnDoe: no, these are the values for which the bounding argument doesn't work.
      $endgroup$
      – Yves Daoust
      Jan 12 at 15:06










    • $begingroup$
      but $[6timespi]=[18.8...]=18$, $[18/6]=[3]=3$ does work. In fact it should always work for a natural number. I don't think those numbers are special
      $endgroup$
      – John Doe
      Jan 12 at 15:08












    • $begingroup$
      @JohnDoe: I didn't say that these don't work, I said that they needed to be confirmed with another argument.
      $endgroup$
      – Yves Daoust
      Jan 12 at 15:09


















    • $begingroup$
      @JohnDoe: ooops, yes of course.
      $endgroup$
      – Yves Daoust
      Jan 12 at 15:05










    • $begingroup$
      Also, what do you mean when you say it suffices to try $a=cdots$? Are you saying these are the only values of $a$ for which it works?
      $endgroup$
      – John Doe
      Jan 12 at 15:05










    • $begingroup$
      @JohnDoe: no, these are the values for which the bounding argument doesn't work.
      $endgroup$
      – Yves Daoust
      Jan 12 at 15:06










    • $begingroup$
      but $[6timespi]=[18.8...]=18$, $[18/6]=[3]=3$ does work. In fact it should always work for a natural number. I don't think those numbers are special
      $endgroup$
      – John Doe
      Jan 12 at 15:08












    • $begingroup$
      @JohnDoe: I didn't say that these don't work, I said that they needed to be confirmed with another argument.
      $endgroup$
      – Yves Daoust
      Jan 12 at 15:09
















    $begingroup$
    @JohnDoe: ooops, yes of course.
    $endgroup$
    – Yves Daoust
    Jan 12 at 15:05




    $begingroup$
    @JohnDoe: ooops, yes of course.
    $endgroup$
    – Yves Daoust
    Jan 12 at 15:05












    $begingroup$
    Also, what do you mean when you say it suffices to try $a=cdots$? Are you saying these are the only values of $a$ for which it works?
    $endgroup$
    – John Doe
    Jan 12 at 15:05




    $begingroup$
    Also, what do you mean when you say it suffices to try $a=cdots$? Are you saying these are the only values of $a$ for which it works?
    $endgroup$
    – John Doe
    Jan 12 at 15:05












    $begingroup$
    @JohnDoe: no, these are the values for which the bounding argument doesn't work.
    $endgroup$
    – Yves Daoust
    Jan 12 at 15:06




    $begingroup$
    @JohnDoe: no, these are the values for which the bounding argument doesn't work.
    $endgroup$
    – Yves Daoust
    Jan 12 at 15:06












    $begingroup$
    but $[6timespi]=[18.8...]=18$, $[18/6]=[3]=3$ does work. In fact it should always work for a natural number. I don't think those numbers are special
    $endgroup$
    – John Doe
    Jan 12 at 15:08






    $begingroup$
    but $[6timespi]=[18.8...]=18$, $[18/6]=[3]=3$ does work. In fact it should always work for a natural number. I don't think those numbers are special
    $endgroup$
    – John Doe
    Jan 12 at 15:08














    $begingroup$
    @JohnDoe: I didn't say that these don't work, I said that they needed to be confirmed with another argument.
    $endgroup$
    – Yves Daoust
    Jan 12 at 15:09




    $begingroup$
    @JohnDoe: I didn't say that these don't work, I said that they needed to be confirmed with another argument.
    $endgroup$
    – Yves Daoust
    Jan 12 at 15:09











    1












    $begingroup$

    It isn't true! a= 1/2 is a counter example. If a= 1/2 then $api$ is 1.5707... and the floor or that is 1. Dividing that by 1/2 gives 2 which, of course, has floor 2, not 3.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Perhaps the question intends that $a$ is an integer.
      $endgroup$
      – Blue
      Jan 12 at 14:47
















    1












    $begingroup$

    It isn't true! a= 1/2 is a counter example. If a= 1/2 then $api$ is 1.5707... and the floor or that is 1. Dividing that by 1/2 gives 2 which, of course, has floor 2, not 3.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Perhaps the question intends that $a$ is an integer.
      $endgroup$
      – Blue
      Jan 12 at 14:47














    1












    1








    1





    $begingroup$

    It isn't true! a= 1/2 is a counter example. If a= 1/2 then $api$ is 1.5707... and the floor or that is 1. Dividing that by 1/2 gives 2 which, of course, has floor 2, not 3.






    share|cite|improve this answer









    $endgroup$



    It isn't true! a= 1/2 is a counter example. If a= 1/2 then $api$ is 1.5707... and the floor or that is 1. Dividing that by 1/2 gives 2 which, of course, has floor 2, not 3.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 12 at 14:27









    user247327user247327

    10.6k1515




    10.6k1515












    • $begingroup$
      Perhaps the question intends that $a$ is an integer.
      $endgroup$
      – Blue
      Jan 12 at 14:47


















    • $begingroup$
      Perhaps the question intends that $a$ is an integer.
      $endgroup$
      – Blue
      Jan 12 at 14:47
















    $begingroup$
    Perhaps the question intends that $a$ is an integer.
    $endgroup$
    – Blue
    Jan 12 at 14:47




    $begingroup$
    Perhaps the question intends that $a$ is an integer.
    $endgroup$
    – Blue
    Jan 12 at 14:47











    0












    $begingroup$

    Let $a=b/pi$ for some $binBbb R$.



    Then $[api]=[b]$



    $$[[b]/a]=left[picdotfrac{[b]}{b}right]$$
    For this to equal 3, we need $$frac{[b]}{b}gefrac3piimplies [b]ge3b/piapprox 0.95 b$$



    This fails for $$bin{0}cupleft(frac{(n-1)pi}{3},nright)$$ for $nin{1,2,cdots,22}$, and thus for $$ain{0}cupleft(frac{n-1}3,frac npiright)$$ For all other values of $a$, this holds.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Let $a=b/pi$ for some $binBbb R$.



      Then $[api]=[b]$



      $$[[b]/a]=left[picdotfrac{[b]}{b}right]$$
      For this to equal 3, we need $$frac{[b]}{b}gefrac3piimplies [b]ge3b/piapprox 0.95 b$$



      This fails for $$bin{0}cupleft(frac{(n-1)pi}{3},nright)$$ for $nin{1,2,cdots,22}$, and thus for $$ain{0}cupleft(frac{n-1}3,frac npiright)$$ For all other values of $a$, this holds.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Let $a=b/pi$ for some $binBbb R$.



        Then $[api]=[b]$



        $$[[b]/a]=left[picdotfrac{[b]}{b}right]$$
        For this to equal 3, we need $$frac{[b]}{b}gefrac3piimplies [b]ge3b/piapprox 0.95 b$$



        This fails for $$bin{0}cupleft(frac{(n-1)pi}{3},nright)$$ for $nin{1,2,cdots,22}$, and thus for $$ain{0}cupleft(frac{n-1}3,frac npiright)$$ For all other values of $a$, this holds.






        share|cite|improve this answer











        $endgroup$



        Let $a=b/pi$ for some $binBbb R$.



        Then $[api]=[b]$



        $$[[b]/a]=left[picdotfrac{[b]}{b}right]$$
        For this to equal 3, we need $$frac{[b]}{b}gefrac3piimplies [b]ge3b/piapprox 0.95 b$$



        This fails for $$bin{0}cupleft(frac{(n-1)pi}{3},nright)$$ for $nin{1,2,cdots,22}$, and thus for $$ain{0}cupleft(frac{n-1}3,frac npiright)$$ For all other values of $a$, this holds.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 12 at 15:07

























        answered Jan 12 at 14:51









        John DoeJohn Doe

        11.1k11238




        11.1k11238






























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